Pre Algebra Notes 2013

Free learning resources from Saddleback College.

Pre-algebra Lecture Notes August 2013 Online videos, worksheets, class notes, quizzes, and much more!

Whether you are entering the world of algebra for the first time or just need to brush up on your fractions, we have the tools to help you succeed!

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For additional information contact Larry Perez in the Department of Mathematics.

Phone: (949) 582-4826

Email: Lperez@saddleback.edu

www.Algebra2go.com

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Objective 1

Place value

Determine the place value of a digit in a number.

The position of a digit in a number determines its place value.

Example 1: Find a pattern for place value in a whole number.

3 9 0 , 5 8 4 , 7 2 6.

Ones place.

Tens place.

Hundreds place.

One-millions place.

Ten-millions place.

Hundred-millions place.

One-thousands place.

Ten-thousands place.

Hundred-thousands place.

Answer the following homework questions using the whole number given in Example 1.

1) What pattern occurs with place values?

2) How are the commas being used?

3) What is the place value of the 9?

4) The 4 is in which thousands place?

5) The digit to the left of the decimal is which place value?

6) How many digits are always placed between two commas?

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Objetivo 1

Valor por posición

Determinar el valor de un dígito por su posición en un número dado.

La posición de un dígito en un número determina su valor.

Ejemplo 1: Encuentra el patrón por su posición en un número entero cualquiera.

3 9 0 , 5 8 4 , 7 2 6.

Unidades

Unidades de

millones Decenas de millones Centenas de millones

Decenas Centenas Unidades de miles Decenas de miles Centenas de miles

Tarea: Contesta las preguntas siguientes usando el número entero del Ejemplo 1.

1) ¿Qué patrón es evidente debido al valor por su posición?

2) ¿Cómo están siendo usadas las comas?

3) ¿Cuál es el valor de 9?

4) ¿En cuál lugar ‘de miles’ está el 4?

5) ¿Cuál es el valor del dígito a la izquierda de las decenas?

6) ¿Cuántos dígitos encontramos en medio de dos comas?

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Objective 2

Learn how to say and write numbers using digits.

Note: Each group of three digits in a number is called a period.

Example 2: Define each period in the number below.

34,436,781,525,372

Example 3: Write each of the following word statements as a number using digits. a) Four thousand, three hundred twelve. b) Eighty seven thousand, one hundred forty-one. c) Seven hundred fifty three thousand, five hundred ninety-eight.

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Objetivo 2

Aprender a leer y escribir números usando dígitos.

Nota: A cada grupo de tres dígitos en un número se le llama ‘período’.

Ejemplo 2: Define cada periodo en el número siguiente.

34,436,781,525,372 Ones Thousands Millions Billions Trillions

Ejemplo 3: Escribe cada uno de los enunciados siguientes como números usando dígitos. a) Cuatro mil, trescientos doce.

4 comma 3 1 2 b) Ochenta y siete mil, ciento cuarenta y uno.

8 7 comma 1 4 1 c) Setecientos cincuenta y tres mil, quinientos noventa y ocho.

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7 5 3 comma 5 9 2

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Example 4: Write each of the following numbers as a word statement. a) 3 , 5 0 1 b) 4 3 , 8 6 2 c) 7 0 4 , 0 1 6 Objective 3

Learn how to write numbers in expanded form.

Understanding place value allows us to represent a large number as a simple addition problem. Seeing numbers in this form is a good way to develop good addition techniques!

Example 5: Write each of the following numbers in expanded form. a) 856 b) 1,397 Page 3 of 4

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Ejemplo 4: Escribe cada uno de los números siguientes como enunciados. a) 3 , 5 0 1 Three thousand, five hundred one. b) 4 3 , 8 6 2 Forty three thousand, eight hundred sixty-two.

c) 7 0 4 , 0 1 6

Seven hundred four thousand, sixteen.

Objetivo 3

Aprender a escribir números en su forma ‘expandida’.

Es posible representar números grandes como sumas simples, si entendemos el concepto ‘valor’. Ver los números como sumas simples es un método efectivo para practicar ‘adición’.

Ejemplo 5: Escribe cada uno de los números siguientes en su forma ‘expandida’. a) 856 800 + 50 + 6

b) 1,397 Página 3 de 4

1,000 + 300 + 90 + 7

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C) 23,512 d) 105,408 Example 5: Calculate the number written in expanded from by performing each addition problem. a) 500 + 80 + 7 = b) 2,000 + 300 + 50 + 9 = c) 40,000 + 2,000 + 40 + 1 = d) 200,000 + 7,000 + 600 = Answer the following homework questions.

7) Write the following word statement as a number using digits; Two thousand, seven hundred thirty-eight.

8) Write the following number as a word statement; 406,348.

9) Write the number 315,006 in expanded form.

10) What number is equal to 70,000 + 5,000 + 80 + 3?

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C) 23,512

20,000 + 3,000 + 500 + 10 + 2

d) 105,408

100,000 + 5,000 + 400 + 8

Ejemplo 6: Calcula el número escrito en su forma ‘expandida’ resolviendo cada uno de los problemas de adición a continuación. a) 500 + 80 + 7 = 587 b) 2,000 + 300 + 50 + 9 = 2,359 c) 40,000 + 2,000 + 40 + 1 = d) 200,000 + 7,000 + 600 = 207,600 Tarea: Contesta las siguientes preguntas.

7) Escribe el enunciado siguiente como un número usando dígitos: Dos mil, setecientos treinta y ocho.

8) Escribe el número siguiente como un enunciado: 406,348.

9) Escribe el número 315,006 en su forma expandida.

10) ¿A qué número es igual 70,000 + 5,000 + 80 + 3?

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Objective 1

Addition and Perimeter Understand the Number Line -7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

On the number line, zero is sometimes called the ________ of the number line. The numbers to the right of the zero are ________ and the numbers to the left of zero are ________.

Objective 2

Understand addition on a Number Line Note: When adding positive numbers we move to the right on the number line.

Example 1: Evaluate the expression -6 + 5.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

Start at -6 and move to the right 5 units.

-6 + 5 = Example 2: Evaluate the expression -4 + 10.

-7 -6 -5 -4 -3 -2 -1 Page 1 of 7

0

1

2

3 4

5

6

7

Start at -4 and move to the right 10 units.

-4 + 10 =

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Example 3: Use the number line to determine what number must be added to -3 to get 7.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

Answer the following homework questions.

In Exercises 1 - 15, use a number line to evaluate each expression.

1) -5 + 10

6) 3 + 4

11) 5 + 4 + 5

2) -2 + 10

7) 7 + 2

12) 3 + 8 + 7

3) -8 + 10

8) 4 + 5

13) -8 + 6 + 8

4) -1 + 10

9) 5 + 1

14) -4 + 2 + 4

5) -9 + 10

10) 6 + 0

15) -15 + 10 +20

Objective 3 Write a mathematical expression using words.

Definition

The sum of two numbers a and b is written a + b. The word sum indicates addition.

Example 4: Using the word sum, write “4 + 9” as a word statement. Find the value of the sum.

Answer: The sum of four and nine. The value of the sum is 13.

Example 5: Using the word sum, write “-3 + 8” as a word statement. Find the value of the sum.

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Answer: The sum of negative three and eight. The value of the sum is 5.

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Objective 4 Perform Addition using the Vertical Format

When finding sums of large numbers, we can use the vertical format to help us organize our work.

Example 6: Calculate the sum of 311, 54, and 32.

We will use the vertical format to get the result. Be sure to line up the numbers in columns according to place value.

Note: In this problem we did not have any 311 “carry over”. When the sum of the digits in 54 a column is greater than 9, we must “carry over” to the next place value column + 32 moving to the left. This process is 397 demonstrated in the next example.

Example 6: Calculate 364 + 178 + 95.

2 1

36 4 17 8 + 85 627 Page 3 of 7

These are the numbers that have been carried over. Remember, this will occur when the sum of the digits in any column is greater than 9.

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Objective 5 Understand Perimeter

The perimeter of a shape is defined to be the sum of its side lengths. We often use the capital letter ___ to represent perimeter.

For rectangles: This side length is labeled as the _________ of the rectangle.

This side length is labeled as the _________ of the rectangle.

Example 6: Find the perimeter of the rectangle.

3 in 7 in

P = ____ +____ +____ +____ = ________ Page 4 of 7

Note: Don’t forget to include the units of measurement in your final answer!

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Example 7: Find the perimeter of the figure. ? ?

5 cm

9 cm

2 cm

14 cm Notice that we must first find the two missing side lengths of the figure before we find its perimeter.

Let’s begin by finding the missing horizontal side length. Since the sum of the missing length and the 9 cm length must equal 14 cm, we ask ourselves “what number do we add to 9 to get 14?” The answer is _____. ,

Next we find the missing vertical side length. Since the sum of the missing length and the 2 cm length must equal 5 cm, we ask ourselves “what number do we add to 2 to get 5?” The answer is _____. ,

,

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,

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Now we can label the missing side lengths.

5 cm 3 cm

5 cm

9 cm

2 cm

14 cm To find the perimeter we sum up all the side lengths.

P= Answer the following homework questions.

16) The word “sum” is used to represent ______________.

17) Write “the sum of 8 and 3” using math symbols.

18) Write “the sum of x and y” using math symbols.

19) Using words, write “-7 + 13” using the word sum and evaluate the expression.

In Exercises 20 – 25, write in the correct number to make the equation true.

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20) 2 + ___ = 7

22) 8 + ___ = 11

24) 13 + ___ = 21

21) -2 + ___ = 7

23) -8 + ___ = 11

25) 17 + ___ = 24

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26) Find the perimeter of the figure below.

16 f t

3 ft

7 ft

4 ft

27) Find the perimeter of the figure below.

3m 5m 14 m 11 m 3m 35 m

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Objective 1

Rounding Numbers

Understand Rounding Numbers using a Number Line Consider the number line below. Explain what it means to round to the nearest 10?

-70 -60 -50 -40 -30 -20 -10

0

10

20

30 40

50 60

70

Example 1: Use the number line above to round 34 to the nearest 10.

The number 34 is between ___ and ___ on the number line. Since 34 is closer to ___ than to ___ , 34 rounded to the nearest 10 is ___. ,

Note: In the cases where the number falls in the middle, we round up to the nearest 10!

Answer the following homework questions.

In Exercises 1 - 9, use the number line to round the given number to the nearest 10.

-70 -60 -50 -40 -30 -20 -10

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0

10

20

30 40

1) 59

4) -26

7) -4

2) 44

5) -5

8) 5

3) 65

6) -45

9) 74

50 60

70

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Next, let’s use a number line to determine how to round a number to the nearest hundred. In this case, we require a number line that is labeled with only hundreds.

39,300

39,400

39,500

39,600

39,700

39,800

39,900

40,000

Example 2: Use the number line above to round 39,455 to the nearest hundred.

The number 39,455 is between ______ and ______ on the number line. Since 39,455 is closer to ______ than to ______ , 39,455 rounded to the nearest hundred is ______. ,

Objective 2

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Learn how to Round Numbers

Sometimes we only want an approximation to a given quantity, or maybe we want to approximate a sum.

Suppose we want to approximate the sum 485 + 337 +196. If we rounded each number to the nearest hundred, we would have 500 + 300 + 200 = 1,000.

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The exact value of 485 + 337 +196 is equal to 1,018 and our approximation of 1,000 is relatively close to the actual value.

The procedure for rounding positive whole numbers is summarized below.

Rounding Positive Whole Numbers First locate the digit that is one place to the right of the place value you want to round to.

If that digit is less than 5, replace it and all digits to the right with zeros.

If the digit is greater than or equal to 5, replace it and all digits to the right with zeros and add 1 to the digit to its left.

Example 3: Round the following numbers to the nearest one-thousand.

The 2 is in the a) 32,439 The digit to the one-thousands right is less than 5.

place value.

32,439

Remains unchanged.

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32,000

Replace with zeros.

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b) 10,506

The 0 is in the one-thousands place value.

The digit to the right is greater than or equal to 5.

10,506

Add 1.

Replace with zeros.

11,000 Example 4: Round the following numbers to the nearest hundred.

The digit to the The 6 is in the right is greater a) 895,684 hundreds place than or equal to 5.

value.

895,684 Add 1.

Replace with zeros.

895,700 b) 5,960

The 9 is in the hundreds place value.

The digit to the right is greater than or equal to 5.

5,960 Replace with zeros.

Add 1.

6,000 +1

Note: Adding a 1 to the 9 requires that we carry over a 1 to the onethousands place value.

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Answer the following homework questions.

In Exercises 10 – 15, round each number to the nearest ten-thousand.

10) 381,520

12) 4,605,267

14) 8,742,097,404

11) 100,095

13) 50,909,613

15) 6,058,355,000

In Exercises 16 – 21, round each number to the nearest ten.

16) 76

18) 187

20) 5,999

17) 5

19) 395

21) 13,999

In Exercises 22 – 25, first find the sum. Next, estimate the sum by first rounding each number to the nearest hundred. Compare your results.

22)

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376 104 + 285

23)

945 363 + 807

24)

154 26 + 12

25)

6,952 7,805 + 5,481

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Objective 1

Subtraction

Understand Subtraction on a Number Line

Using a number line let’s demonstrate the subtraction process using the problem 7 – 5.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3

4

5

6

7

Using the number line above, start at 7 and move left 5 units. You end up at 2. Therefore 7 – 5 = 2.

Next, let’s demonstrate the subtraction process using the problem 5 – 7.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3

4

5

6

7

Using the number line above, start at 5 and move left 7 units. You end up at –2. Notice that 7 – 5 = 2 and 5 – 7 = –2. Think about how these two problems are related. This can help you with basic subtraction problems that have negative results!

If 35 – 10 = 25, what do you think 10 – 35 is equal to? It must equal –25.

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Example 1: Use the number line below to perform each subtraction problem.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1

a) 10 – 5 b) 5 – 10 c) 10 – 8 d) 8 – 10 e) 7 – 4 f) 4 – 7 Objective 2

0

1

2

g) 9 – 8 h) 8 – 9 i) 4 – 8 j) 1 – 10 k) 3 – 6 l) 2 – 5

3 4

5

6

7

8

9 10

m) 7 – 0 n) 0 – 7 o) 8 – 16 p) 6 – 12 q) 7 – 16 r) 8 – 18

Perform Subtraction Problems using the Vertical Format (no borrowing).

Suppose we want to subtract 24 from 56. In this case we would need to calculate 56 – 24.

Performing this calculation on a number line would allow us to visually demonstrate the process. In this case we start at 56 and move left a total of 24 units. The result is 32.

4

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32 36

10

10

46

56

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Performing subtraction problems on a number line can help us develop our “mental math” skills. But when the numbers are relatively large, the vertical format is most often used.

Example 2: Calculate 56 – 24 using the vertical format.

56 Note: Be sure to line up the numbers in columns – 24 according to place value.

32 Note: Performing subtraction using the vertical format cannot give us negative results.

Example 3: Calculate 33 – 48.

In this case we will first calculate 48 – 33 using the vertical format. From Example 1, we can conclude that our answer is the negative result of 48 – 33.

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48 – 33 15

Therefore 33 – 48 = –15.

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Objective 3

Perform Subtraction Problems using the Vertical Format with Borrowing.

Sometimes using the vertical format requires a technique called “borrowing”. This occurs when subtracting two numbers in a column gives a negative result. To prevent the negative result, we borrow from the adjacent column to the left. The process of “borrowing” is demonstrated in the following example.

Example 4: Calculate 302 – 175.

Here we will use the vertical format which requires us to use the “borrowing” technique.

302 – 1 75 Notice that 2 – 5 in the ones place value gives a negative result.

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302 – 1 75 Because we have a zero in the tens column, we must move to the hundreds column to borrow.

2

1

302 – 1 75 Here we have borrowed a 100 and carried it over to the tens column.

Therefore, we now have ten 10’s.

Our result is 302 – 175 = 127.

2 9

1 1

302 – 1 75 127 We now borrow a 10 and carry it to the ones column. We now have twelve 1’s.

302 is now written as 200+90+12. We can now subtract.

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Answer the following homework questions.

In Exercises 1 – 9, use a number line to perform each subtraction problem.

1) 9 – 7

4) 15 – 7

7) 7 – 15

2) 8 – 5

5) 13 – 8

8) 8 – 13

3) 6 – 4

6) 11 – 6

9) 6 – 11

In Exercises 10 – 15, perform each subtraction problem using the vertical format. Note: These problems do not require borrowing.

10) 48 – 13

12) 138 – 126

14) 3,508 – 1,207

11) 96 – 52

13) 627 – 405

15) 7,096 – 5,084

In Exercises 16 – 21, perform each subtraction problem using the vertical format. Note: These problems require borrowing.

16) 15 – 7

18) 600 – 429

20) 59 – 73

17) 13 – 8

19) 1,000 – 837

21) 48 – 61

In Exercises 22 – 27, write in the correct number to make the equation true.

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22) 9 – ___= 5

24) 48 – ___= 38

26) 21 – ___= 13

23) ___ – 9= –5

25) ___ – 48= –38

27) ___ – 21= –13

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Objective 4 Write a mathematical expression using words.

Definition

The difference of two numbers a and b is written a ‒ b.

The word difference indicates subtraction. If a is larger than b, the difference is positive. If a is smaller than b, the difference is negative.

Example 5: Using the word difference, write “8 ‒ 6” as a word statement, and find the value of the difference.

We first begin our sentence by defining the mathematical operation first and then define the numbers. Notice how the word “and” is used.

The word statement is written as: “The difference of eight and six.”

The value of the difference is 2.

Example 6: Using the word difference, write “–7 ‒ 5” as a word statement, and find the value of the difference.

The word statement is written as: “The difference of negative seven and five.”

The value of the difference is -12.

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Answer the following homework questions.

28) The word “difference” is used to represent ______________.

29) Write “the difference of 8 and 3” using math symbols.

30) Write “the difference of x and y” using math symbols.

31) Using the word difference, write “-7 – 13” as a word statement and find the value of the difference.

In Exercises 32 – 35, find each difference.

33) 504 32) 608 34) 9,014 – 387 – 405 – 152

35)

8,000 – 3,618

36) Find the perimeter of the figure below.

6m 10 m 28 m 22 m 6m 70 m

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Objective 1

Multiplication

Understand the Meaning and Notation of Multiplication

There are different ways to indicate 3 multiplication. Here are a few.

3 × 4 3 ⋅ 4 3(4) (3)(4) × 4 But what does 3 ⋅ 4 or “3 times 4” actually mean?

3 ⋅ 4= 3 + 3 + 3 + 3 = 12 So what does 24⋅ 10 or “24 times 10”actually mean?

24 ⋅ 10=24+24+24+24+24+24+24+24+24+24 24 ⋅ 10=240 Notice there are ten 24’s being added together!

Now, how can we easily calculate 24 ⋅ 11?

Since we know that 24⋅ 10=240, and that 24 ⋅ 10 represents ten 24’s being added together, we just need to add one more 24 to 240!

24 ⋅ 10=240 24 ⋅ 11=240 + 24 = 264 Page 1 of 7

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24 Notice that when using the vertical format to calculate 24⋅ 11, you end × 1 1 24 up having to find the sum of 24 +2 4 0 and 240!

264 24 ⋅ 11=240 + 24 = 264 Note: The vertical format uses the “Distributive Property” and the “Expanded Form” of a number.

24 ⋅ 11=24(10+1)= 24(10+1)=240 + 24 The Expanded Form of 11 = 264 Example 1: Use the distributive property and the expanded form of a number to perform each multiplication problem. a) 18 ⋅ 11=18(10+1)= b) 9 ⋅ 11=9(10+1)= c) 7 ⋅ 12=7(10+2)= d) 5 ⋅ 13=5(10+3)= e) 13 ⋅ 12=13(10+2)= f) 8 ⋅ 7=8(5+2)= Page 2 of 7

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Now, how can we easily calculate 24 ⋅ 9?

Since we know that 24⋅ 10=240, and that 24 ⋅ 10 represents ten 24’s being added together, we just need to subtract 24 from 240!

24 ⋅ 11=240 + 24 = 264 24 ⋅ 10=240 24 ⋅ 9 =240 – 24 = 216 Objective 2

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Understand the Multiplication Table

There are certain multiplication problems that we can sometimes easily recall, like 8 ⋅ 5=40.

Knowing this, we can now use a pattern to calculate other multiplication problems with the number 8 as well as with other numbers!

8 ⋅ 5 =40 9 ⋅ 5 =45 12 ⋅ 5 =60 8 ⋅ 6 =48 9 ⋅ 6 =54 12 ⋅ 6 =72 8 ⋅ 7 =56 9 ⋅ 7 =63 12 ⋅ 7 =84 8 ⋅ 8 =64 9 ⋅ 8 =72 12 ⋅ 8 =96 8 ⋅ 9 =72 9 ⋅ 9 =81 12 ⋅ 9 =108 8 ⋅ 10=80 9 ⋅ 10=90 12 ⋅ 10=120 8 ⋅ 11=88 9 ⋅ 11=99 12 ⋅ 11=132

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Example 2: Construct a 12 by 12 multiplication table and write down any patterns that you notice.

Objective 3

Perform Multiplication using the Vertical Format Recall that the vertical format is based on using the distributive property and the expanded form of a number.

36 ⋅ 23=36(20+3)= 36(20+3)=720 + 108 The Expanded Form of 23

Example 3: Calculate 36 ⋅ 23 using the vertical format.

Finally add the results

First multiply 36 by 3.

1

36 3 × 108 Page 4 of 7

Next, multiply 36 by 20.

1

36 × 20 720

36 × 23 108 + 7 20 828

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Answer the following homework questions.

In Exercises 1 – 9, perform each multiplication problem.

1) 9 ⋅ 4

4) 7 ⋅ 5

7) 6 ⋅ 4

2) 9 ⋅ 5

5) 7 ⋅ 6

8) 6 ⋅ 5

3) 9 ⋅ 6

6) 7 ⋅ 7

9) 6 ⋅ 6

In Exercises 10 – 15, first rewrite each multiplication problem as an addition problem, then find the sum.

Example: 3 ⋅ 4= 3 + 3 + 3 + 3 = 12 10) 5 ⋅ 4

12) 9 ⋅ 3

14) 12 ⋅ 4

11) 6 ⋅ 4

13) 8 ⋅ 6

15) 150 ⋅ 4

In Exercises 16 – 21, perform each multiplication problem using the vertical format.

16) 17 ⋅ 9

18) 37 ⋅ 15

20) 60 ⋅ 20

17) 12 ⋅ 11

19) 45 ⋅ 12

21) 55 ⋅ 24

In Exercises 22 – 27, write in the correct number to make the equation true.

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22) 9 ⋅ ___= 54

24) 7 ⋅ ___= 42

26) 12 ⋅ ___= 108

23) ___ ⋅ 9= 36

25) ___ ⋅ 8= 24

27) ___ ⋅ 6= 72

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Objective 4 Write a mathematical expression using words.

Definition

The product of two numbers a and b is written a ⋅ b. The word product indicates multiplication.

Example 4: Using the word product, write “9 ⋅ 7” as a word statement, and find the value of the product.

We first begin our sentence by defining the mathematical operation first and then define the numbers. Notice how the word “and” is used.

The word statement is written as: “The product of nine and seven.”

The value of the product is 63.

Example 5: Using the word product, write “16 ⋅ 7” as a word statement, and find the value of the product.

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Answer the following homework questions.

28) The word “product” is used to represent ______________.

29) Write “the product of 12 and 7” using math symbols.

30) Write “the product of p and q” using math symbols.

31) Using the word product, write “19 ⋅ 14” as a word statement and find the value of the product.

In Exercises 32 – 35, find each product.

33) 52 32) 28 34) 320 × 27 × 30 × 15

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35)

600 × 400

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Objective 1

The Distributive Property

Understand the Distributive Property The Distributive Property The Distributive Property states that multiplication can be distributed across addition and subtraction.

X(a+b) = ax + bx a(x − y) = ax − ay 4(10+1) = 40 + 4 13(10 − 2)=130 − 26

The distributive property can be helpful when performing multiplication.

8 ⋅ 7=8(5+2)= 40+16= 7 ⋅ 13=7(10+3)=70+21= 12 ⋅ 9=12(10 − 1)= 120 − 12 = 13 ⋅ 8=13(10 − 2)=130 − 26= 9(314)=9(300+10+4)=2,700+90+36= Page 1 of 2

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Recall that when performing multiplication using the vertical format, you are using the distributive property!

Example 1: Find the product of 256 and 8 using the vertical format for multiplication.

Notice how it relates to the calculation below where the Distributive Property is being used.

8(256)=8(200+50+6)=1,600+400+48=2,048

256 8 × Answer the following homework questions.

In Exercises 32 – 35, find each product.

2) 212 1) 436 3) × 7 9 ×

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508 × 6

4)

978 × 4

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Properties of Multiplication and Addition Objective 1

Understand the Associative Properties The Associative Property For addition, this property states that: (a+b)+c = a+(b+c) (3+4)+6 = 3+(4+6) For multiplication, this property states that: (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) (7 ⋅ 5) ⋅ 2 = 7 ⋅ (5 ⋅ 2)

The associative property can be helpful when performing basic arithmetic calculations.

Notice how the calculations below are somewhat simplified by applying the associative property.

(35+17)+3 35+(17+3) 35+20 55 Page 1 of 4

(13 ⋅ 5) ⋅ 2 13 ⋅ (5 ⋅ 2) 13 ⋅ 10 130

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Objective 2

Understand the Commutative Properties The Commutative Property

For addition, this property states that: a+b = b+a 3+7 = 7+3 For multiplication, this property states that: a⋅ b=b⋅ a 5⋅ 8=8⋅ 5

The associative and commutative properties for addition, provides us the ability to add numbers in any order. Therefore, if all our numbers are being added, we can rearrange them in order we see fit! We do not have to work left to right in these cases!

Suppose we want to calculate 7+8+3+2. We can rearrange the problem to be 7+3+8+2 We can now simplify this problem to 10+10 20 Page 2 of 4

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Suppose we want to calculate 15+23+5+17.

We can rearrange the problem to be 15+5+23+17 We can now simplify this problem to 20+40 60 The associative and commutative properties for multiplication, provides us the ability to multiply numbers in any order. Therefore, if all our numbers are being multiplied, we can rearrange them in order we see fit! We do not have to work left to right in these cases either!

Suppose we want to calculate 12 ⋅ 2 ⋅ 5 ⋅ 5.

We can rearrange the problem to be 12 ⋅ 5 ⋅ 2 ⋅ 5 We can now simplify this problem to 60 ⋅ 10 600 Page 3 of 4

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Answer the following homework questions.

In Exercises 1 – 9, perform each addition problem. Apply the associative and commutative properties when performing the calculations. Try to get the answer mentally!

1) 9 + 4+ 1

4) 23 + 19+ 7

7) 11 + 42 + 9 + 8

2) 8 + 9+ 2

5) 96 + 58+ 4

8) 34 + 25 + 4 + 6

3) 5 + 17+ 5

6) 14 + 39+ 6

9) 27 + 17 + 4 + 3

In Exercises 10 – 18, perform each multiplication problem. Apply the associative and commutative properties when performing the calculations. Try to get the answer mentally!

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10) 9 ⋅ 2 ⋅ 5

13) 9 ⋅ 10 ⋅ 2

16) 7 ⋅ 5 ⋅ 8 ⋅ 2

11) 2 ⋅ 17 ⋅ 5

14) 10 ⋅ 12 ⋅ 5

17) 2 ⋅ 6 ⋅ 3 ⋅ 5

12) 5 ⋅ 38 ⋅ 2

15) 10 ⋅ 11 ⋅ 9

18) 12 ⋅ 2 ⋅ 5 ⋅ 11

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Objective 1

Division

Understand the Meaning and Notation of Division

There are different ways to indicate division.

Here are a few.

12 12 ÷ 4 12/4 4 12 4 But what does 12 ÷ 4 or “12 divided by 4” actually mean? Why does 12 ÷ 4 = 3 ?

Recall: 4 ⋅ 3= 4 + 4 + 4 = 12 So how many 4’s does it take to make up a 12?

4 4 4 0

1

2

3 4

5

6

7 8

9 10 11 12 13 14

It takes three 4’s to make up a 12. Or we can say 4 goes into 12 three times!

Therefore 12 ÷ 4 = 3.

This is what the problem would like using long 3 division to represent 12 ÷ 4. 4 12 12 0

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So how many 4’s does it take to make up a 15?

4 4 4 R3 0

1

2

3 4

5

6

7 8

9 10 11 12 13 14 15

It takes three 4’s to make up a 15 but there are 3 units still remaining. In this case we say 4 goes into 15 three times with 3 remaining units left over!

Therefore 15 ÷ 4 = 3R3.

This is what the problem would like using long 3 R3 division to represent 12 ÷ 4.

4 15 12 3

Knowing your multiplication tables can help you with division! See if you can find a pattern in the problems below!

If 3 ⋅ 4=12 then 12 3 = 4.

If 8 ⋅ 9=72 then If 6 ⋅ 7=42 then Page 2 of 8

72

8

= 9.

43

6

= 7R1.

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Answer the following homework questions.

In Exercises 1 – 9, perform each division problem. Try to do each problem mentally using your multiplication tables. Remember that if 9 ⋅ 8 = 72 then

72

9

= 8.

1) 20 ÷ 4

4) 108 ÷ 12

7) 15 ÷ 4

2) 64 ÷ 8

5) 132 ÷ 11

8) 23 ÷ 6

3) 55 ÷ 11

6) 65 ÷ 5

9) 17 ÷ 3

In Exercises 10 – 15, rewrite each division problem as an equivalent multiplication problem.

Sample: 24 ÷ 8 = 3 is equivalent to 8 ⋅ 3 = 24

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10) 36 ÷ 12 = 3

12) 54 ÷ 9 = 6

14) 100 ÷ 25 = 4

11) 42 ÷ 7 = 6

13) 63 ÷ 7 = 9

15) 150 ÷ 10 = 15

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Objective 2

Understand how to Perform Long Division

Consider 131

÷

4

4 131

How many times does 4 go into 1?

0 4 131

How many times does 4 go into 13?

Subtract.

03 4 131 12 11

Subtract.

Three

3 × 4 goes here.

Bring down the 1.

How many times does 4 go into 11?

032 4 131 12 11 8 3

Zero

Two

2 × 4 goes here.

There are 3 units left over.

Therefore, 131

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÷

4 = 32R3.

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Objective 3

Understand Division with Zero

Recall that if 12 3 = 4, then 3 ⋅ 4=12.

7 If 92 = 8, then 9 ⋅ 8=72.

If 20 5 = 4, then 5 ⋅ 4=20.

Do you notice the pattern between division and multiplication?

With this in mind, consider 50 .

If 50 = 0, then 5 ⋅ 0=0. This is true!

Similarly, if 80 = 0, then 8 ⋅ 0=0. Again true!

Conclusion: Zero divided by any number (except zero), is always zero!

But what about

5

0

0r 00 ?

To evaluate 50 , we ask ourselves “0 times what 5 0 ⋅ ?=5 number equals 5?”

0= ?

Since 0 times any number is always 0, there is 5 no answer. In math, we say that 0 is undefined.

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To evaluate 00 , we ask ourselves “0 times what number equals 0?”

0 0 ⋅ ?=0 0= ?

Since 0 times any number is always 0, any number will work! There is no defined answer! In 0 math, we say that 0 is undefined. In other words, we cannot divide by zero!

Conclusion: Any number divided by zero is always undefined!

Answer the following homework questions.

In Exercises 16 – 21, perform each division problem using the long division.

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16) 84 ÷ 4

18) 41 ÷ 4

20) 155 ÷ 6

17) 96 ÷ 3

19) 23 ÷ 6

21) 191 ÷ 8

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In Exercises 22 – 27, write in the correct number to make the equation true.

22) 54 ÷ ___= 9 23)

___ ÷ 9= 8

24) 24 ÷ ___= 3 25)

___ ÷ 8= 4

26) 88 ÷ ___= 11 27) ___ ÷ 12 = 7

In Exercises 28 – 31, write in the correct number to make the equation true.

28) 54 ÷ ___= Undefined 29)

___ ÷ 9= 0

30) 0 ÷ ___= 0 31) ___ ÷ 0= Undefined

Objective 4 Write a mathematical expression using words.

Definition

The quotient of two numbers a and b is written a ÷ b.

The word quotient indicates division.

Example 1: Using the word quotient, write “56 ÷ 8” as a word statement and find the value of the quotient.

We first begin our sentence by defining the mathematical operation first and then define the numbers. Notice how the word “and” is used.

The word statement is written as: “The quotient of fifty-six and eight.”

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The value of the quotient is 7.

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Example 2: Using the word quotient, write “231 ÷ 7” as a word statement, and find the value of the quotient.

Answer the following homework questions.

32) The word “quotient” is used to represent ______________.

33) Write “the quotient of 54 and 6” using math symbols.

34) Write “the quotient of x and y” using math symbols.

35) Using the word quotient, write “732 ÷ 6” as a word statement and find the value of the quotient.

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Translating between Math and Word Statements Objective 1

Learn the Word Statement for Arithmetic Operations.

The word sum indicates _________.

The word difference indicates _________.

The word product indicates _________.

The word quotient indicates _________.

Objective 2

Translate math to word statements.

Example 1: Translate the given math statements to word statements. a) 5 ⋅ 7 b) 6 ÷ x c) a + b d) 31 − 14 The math symbol that is represented by the word “is”, is the equal sign written “=”.

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Example 2: Translate the given math statements to word statements. a) 42 ÷ 6=a b) 8 ⋅ y=56 c) 48 − x=29 d) b + 15 =47 Objective 3

Translate word to math statements.

Example 3: Translate the given word statements to math statements. e) The sum of two and six is eight. f) The quotient of x and y is 4. g) The difference of nine and four is five. h) The product of p and q is 16.

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Answer the following homework questions.

In Exercises 1 – 4, write each math statement as a word statement.

1)

20 4

=5

2) 7(8)=56 3) 36 + 48 = k 4) 94 − 47 = p

In Exercises 5 – 8, write each word statement as a math statement.

5) The product of 8 and x is 32.

6) The difference of y and 12 is 7.

7) The quotient of 28 and x is 7.

8) The sum of x and y is 5.

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Objective 1

Exponents

Understand the Relationship between Multiplication and Exponents

Recall that 3 ⋅ 4represents an addition problem.

3⋅ 4 = 3 +3 +3 +3.

But what about the expression 3 ⋅3 ⋅3 ⋅3?

This is where the exponent is used!

Exponents are used to represent repetitive multiplication of a quantity.

Using an exponent of 4, we write 3 ⋅ 3 ⋅ 3 ⋅ 3 as 34 where 3 is called the base and 4 is the exponent.

Base

34

Suppose we are given which is equal to 8.

If we are given

23 . This represents 2 ⋅ 2 ⋅ 2

x3 , this represents

So what do you think Page 1 of 5

Exponent

x ⋅ x ⋅ x.

x3 ⋅ x4 is equal to?

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Well since x3 = x ⋅ x ⋅ x and x4 = x ⋅ x ⋅ x ⋅ x we get the following result. x3 ⋅ x4 = x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x = x 7 Notice how we can just add the exponents. x3 ⋅ x4 = x3+4= x 7 When multiplying two quantities that have the same base, we can add the exponents.

xa ⋅ xb = xa

+b

Suppose we are given x + x + x + x . Notice that we have four x’s being added together.

Recall that multiplication is used to represent repetitive addition of the same quantity.

Using the Commutative Property for multiplication we can now show that 4 ⋅ x = x ⋅ 4 = x + x + x + x = 4x

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Notice that and

4x = x + x + x + x x4 = x ⋅ x ⋅ x ⋅ x .

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Note:

x3 is said “x raised to the third power” or

“x cubed”. x2 is said “x raised to the second power” or “x squared”.

Answer the following homework questions.

In Exercises 1 – 9, write each quantity in expanded form.

Recall: 4x=x+x+x+x and x4 = x ⋅ x ⋅ x ⋅ x .

1) 5 2

4) a4

7) ab2

2) y2

5) 3a

8) x 2y2

3) 4y

6) 2x

9) p3q 4

In Exercises 10 – 18, add or multiply as indicated, if possible.

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10) 3x+4x

13) a4 ⋅ a3 ⋅ a2

16) 2h + 2b

11) y3 ⋅ y3

14) w ⋅ w 2

17) h2 + b2

12) 3y+3y

15) w + 2w

18) 2c ⋅ c 2

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Objective 2

Understand the Zero Exponent

What is the value of 2 ?

We can arrive at a general conclusion by noticing the following pattern.

4 2 = 2 ⋅ 2 ⋅ 2 ⋅ 2 = 16 Here the numbers Notice that the 23 = 2 ⋅ 2 ⋅ 2 = 8 are being divided 2 exponents are by the base 2 as ⋅ 2 2 4 = = 2 decreasing by 1! you move 1 downward!

2 =2 2= 1 =1 20 = 0

Let’s now try this with a base of 3.

4 3 = 3 ⋅ 3 ⋅ 3 ⋅ 3= 81 Notice that the 33 = 3 ⋅ 3 ⋅ 3 = 27 2 exponents are 3 ⋅3 = 9 decreasing by 1! 3 = 3 =3 31 = 1 =1 30 =

Here the numbers are being divided by the base 3 as you move downward!

Regardless of what base you use (except for 0), you will always get the zero power to equal 1!

A base of 0 does not work because we can never 0 divide by zero! 0 is undefined!

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Answer the following homework questions.

In Exercises 19 – 27, find the value of each expression!

Note: In every problem you must first evaluate the quantity with the exponent before performing the arithmetic operation!

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19) 2 3 + 32

22) 34 ⋅ 1 12

25) 2 3 ⋅ 32

20) 32 − 12 0

23) 0 10 ÷ 7 2

26) 11 2 − 8 2

21) 43 ÷ 2 3

24) 102 ÷ 03

27) 5 0 + 40

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Objective 1

Order of Operations

Understand the Four Steps of Order of Operations

Problems often have parenthesis, exponents, and arithmetic operations that we need to perform in a specific order. WE always work these problems following the four steps of Order of Operations.

Step 1: Perform all the operations within a parenthesis or other grouping symbols.

Step 2: Simplify any expressions with exponents.

Step 3: Multiply or divide working left to right, whichever comes first.

Step 4: Add or subtract working left to right, whichever comes first.

Objective 2

Page 1 of 3

Use the Order of Operations

Example 1: Evaluate each expression below following the Order of Operations. a) 8 − 5 + 1 b) 8 − ( 5 + 1 ) c) 8 + 5 − 1

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Example 2: Evaluate each expression.

10 2 3 2 − − 6 4 3 ( ) a) 3 − 5 ÷ 5 b)

Example 3: Evaluate.

7 2 − ( 42 − 5 ) − 6  + 10

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Objective 3

Understand when Parenthesis are needed to define a negative Base.

2

When evaluating the expression -3 , we must pay close attention to what the base is. In the 2 expression -3 , the base is positive 3. This is 2 2 because -3 = -1 ⋅ 3 .

Note: -32 is said “Negative one times three squared”.

To correctly evaluate -32 , we must follow the Order of Operations and evaluate the exponent before we multiply by -1.

Note: 2 2 9 9 -3 = -1 ⋅ 3 = -1 ⋅ = - Negative x Positive =Negative If the base is to be -3 , then parenthesis must be used to indicate this.

2 ( -3 ) = ( -3 )( -3 ) = 9 Answer the following homework questions.

In Exercises 1 – 9, evaluate each expression.

1) 42 − ( 13 − 10 )

Page 3 of 3

2

-5

4) 3 − 2 2 ÷ 4 ⋅ 2

7)

2) 3 + 4  17 − 2 ( 5 − 1 ) 

5) 2 3 + 33 ÷ 9 − 2

8) ( -8 )

3) 5  36 ÷ 2 ( 5 2 − 42 ) 

6) 48 ÷ 2 3 ⋅ 9 − 2 3

9)

2

2

-12

0

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Objective 1

Area and Volume

Calculate the Area of a Figure

Recall that the perimeter of a figure is a one-dimensional length and is express as ft, cm, in., etc.

But area involves 2 dimensions. For rectangles we measure its length and width and then multiply the two together. Since we are multiplying these two lengths together, area is a two-dimensional quantity and is expressed as ft 2, cm 2, in 2 , etc.

The formula for the area of a rectangle is: A=length ⋅ width –orA= l ⋅ w Example 1: Find the area of the given 5 cm rectangle.

2 cm

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2 x ⋅ x = x Note: Recall that . cm 2 .

Similarly, cm ⋅ cm =

The term cm 2 is said “centimeters squared” or “square centimeters”.

A parallelogram is a quadrilateral, where opposite sides are both parallel and have the same length.

The formula for the area of a parallelogram is: A=base ⋅ height –or- A= b⋅ h height base

Example 2: Find the area of the given parallelogram.

4 in.

7 in.

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A triangle is a three sided polygon.

The formula for the area of a triangle is: 1 A=(b⋅ h) ÷ 2 –or- A= 2(b ⋅ h) height

base

Example 3: Find the area of the given triangle.

4 cm 6 cm

Example 4: Find the area of the given triangle.

7m 9m

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A trapezoid is a quadrilateral where exactly one pair of opposite sides are parallel. Unlike a parallelogram, opposite sides do not necessarily have the same length.

The formula for the area of a trapezoid is: A= [h ⋅ (b1+ b2)] ÷ 2 –or- A= 21 ⋅ h ⋅ (b1+ b2) b1

height

b2

Example 5: Find the area of the given 20 ft trapezoid.

40 ft

160 ft

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Objective 2

Calculate the Area of a Composite Figure

In many cases we need to partition our figure so that it consists of familiar shapes such as parallelograms, rectangles, trapezoids, or triangles. The total area is the sum of the individual areas.

Example 6: Find the area of the figure below.

4 in.

3 in.

6 in.

7 in.

2 in.

12 in.

Partition the figure into two rectangles. Notice that you only need the lengths related to the dimensions of each individual rectangle.

4 in.

I

3 in.

II

12 in.

4 in.

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I

3 in.

II

12 in.

2 in.

2 in.

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4 in.

I

3 in.

2 in.

II

12 in.

A1= l ⋅ w=4 in. ⋅ 3 in. = 12 in2 A2= l ⋅ w=12 in. ⋅ 2 in. = 24 in2 ATotal= 12 in2 + 24 in2 = 36 in2 Example 7: Find the area of the figure below.

3m

4m

2m 2m

8m

Partition the figure into three separate shapes; a triangle, parallelogram, and a rectangle. Find their individual areas and add them together to get the total area.

Page 6 of 11

I

II

III

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I

II

III

Page 7 of 11

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Objective 3

Calculate the Volume of a Rectangular Prism Recall that the area of a figure is a twodimensional quantity and is express as ft2, cm2, in2, etc.

But volume involves 3 dimensions. For rectangles we measure its length, width, and height then find the product of these three lengths. Since we are multiplying these three lengths together, volume is a threedimensional quantity and is expressed as ft3, cm3, in3, etc.

The formula for the volume of a rectangular prism is: V=length ⋅ width ⋅ height –orA= l ⋅ w ⋅ h height width

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length

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3 x ⋅ x ⋅ x = x Note: Recall that .

3 ft ft ft ft ⋅ ⋅ = Similarly, .

Example 8: Find the volume of the figure below.

10 cm

6 cm 14 cm

Example 9: Find the volume of the figure below.

6 in.

3 in.

7 in.

6 in.

3 in.

2 in.

13 in.

To find the volume of this particular figure, we will partition the figure using a horizontal cut.

We could also use a vertical cut if preferred.

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6 in.

3 in.

I

6 in.

7 in.

3 in.

II

2 in.

13 in.

I

6 in.

3 in.

2 in.

3 in.

II

13 in.

V1= l ⋅ w ⋅ h = V2= l ⋅ w ⋅ h = VTotal=

Page 10 of 11

2 in.

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Answer the following homework questions.

In Exercises 1 – 3, fill in the blank to make the statement true.

1) Perimeter has ________ dimension.

2) Area has ________ dimensions.

3) Volume has ________ dimensions.

Note: In Exercises 4 and 5, you will need to find missing side lengths.

4) Find the perimeter and area of the given figure.

6 ft

All angles are right angles.

4 ft

8 ft 5 ft 3 ft 19 ft

5) Find the volume of the given figure.

4 mm 6 mm 4 mm 2 mm Page 11 of 11

2 mm

15 mm

9 mm

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Objective 1

Inequalities

Understand the Meaning of Inequalities -7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

We see that as you move to the right on the number line, the numbers get larger. Notice that -3 is to the right of -4. This means that “-3 is greater than -4”. The mathematical symbol for “is greater than” is “>”. Notice that this symbol looks like the head of an arrow that points to the right.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

Similarly, we see that as you move to the left on the number line, the numbers get smaller. Notice that -6 is to the left of -5. This means that “-6 is less than -5”. The mathematical symbol for “is less than” is “<”.

Notice that this symbol looks like the head of an arrow that points to the left.

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-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

It would be a true math statement if we wrote “-3>-4” since -3 is to the right of -4. It would be false to write “-3<-4”.

Similarly, it would be a true math statement if we wrote “-6<-5” since -6 is to the left of -5. It would be false to write “-6>-5”.

Example 1: Answer True or False. a) 0>-1 e) -37>-36 b) -5>-4 f) 36<37 c) 3<6 g) -20<-19 d) 5>-5 h) 20<19 Example 2: Translate each word statement into a math statement. a) Four is less than seven. b) Negative seven is greater than negative eight.

Page 2 of 4

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Note: The word statement “5 greater than 4” represents the math statement “5+4”.

However the word statement “5 is greater than 4” represents “5>4”. Notice how using the word “is” makes a difference!

Example 3: Translate each math statement into a word statement. a) 7<10 b) -16 >-17 Example 4: Translate each word statement into a math statement. a) The sum of negative eight and ten is less than three. b) The difference of negative seventeen and five is greater than negative twenty three.

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Answer the following homework questions.

In Exercises 1 – 15, fill in the blank with either “>” or “<” to make the statement true.

1) -5 ___ -6

6) 0 ___ -1

11) -8 ___ -7

2) -1 ___ 0

7) 1 ___ 0

12) -21 ___ 0

3) -8 ___ -9

8) 56 ___ 65

13) 21 ___ 0

4) 4 ___ 7

9) -64 ___ -63

14) 0 ___ -32

5) 8 ___ -8

10) 19 ___ 20

15) 0 ___ 32

In Exercises 16 – 19, translate each math statement into a word statement.

16) 17>4

18) 0 >-1

17) -8<-5

19) 5 >0

In Exercises 20 – 22, translate each word statement into a math statement.

20) The quotient of twenty and five is less than five.

21) The product of three and four is greater than eleven.

22) The difference of negative two and six is less than negative seven.

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Objective 1

Absolute Value

Understand the Meaning of Absolute Value Absolute Value is a mathematical representation of distance on the number line.

The mathematical symbol for absolute value is

.

The absolute value of three, written 3 represents its distance from 0 on the number line.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

On the number line above, notice that -3 is a distance of three units from zero on the number line. Similarly, 3 is also a distance of three units from zero on the number line. In both cases, their absolute values will both equal positive 3. Remember, that we always express distance as a positive quantity. You would never say that the distance from your home to the library is negative five miles, would you?

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-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

To mathematically represent the distance between -3 and 0 on the number line, we would write - 3 which is said, “the absolute value of negative three”.

Finally, we can state that - 3 = 3.

Similarly, we can state that 3 = 3. Notice that both answers are positive! In fact, absolute value answers will always be positive since they represent distances!

Example 1: Evaluate each absolute value. a) -8 e) - 30 b) 8 f) 30 c) 0 g) - 1 d) 14 h) - 100

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Note: Absolute values should be treated as a grouping symbol in the order of operations.

Therefore, you must evaluate the absolute value before moving on with the problem.

Secondly, you must simplify the expression inside the absolute value before you can evaluate it.

Example 2: Find the value of each expression below.

30 -3 − 4 − 8 5 4 + 3 ⋅ a) b) c) 6 20 + 3 20 + 3 23 d) -4 - 2 − 3 − 5 -4 - 4 − 3 − 5 -4 − 7 − 5 -4 ⋅ 7 − 5 -28 − 5 -33 2

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2 -8 -4 3 − +6 ( ) e) 2

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Answer the following homework questions.

In Exercises 1 – 6, find the value of each expression.

Page 4 of 7

1) 10 − -5 − 2

4) -9 − -4 − 7 − 8

2) -4 − 62

5) -2 2 − 32

3) 5 2 -42 + 3

6) -4 -32 − ( 2 − 3)2

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Objective 2

Understand the Opposite of a Number -7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

On the number line, numbers that are the same distance from zero on opposite sides are called “opposite numbers”. -3 and 3 are opposites of each other since they are both a distance of three from zero.

The sum of two opposite numbers is always zero! This is why sometimes they are called “Additive Inverses” of each other.

-3+3 = 0

Example 3: Find the opposite of the given quantity. Note: You must first evaluate the absolute value before you find the opposite.

Page 5 of 7

a) -12

b) 12

c) -8

d) 10

e) -2

f) 3

g) x

h) -x

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When you look at a number line, you may realize that the opposite of any negative number is always positive!

Most calculators have an opposite key on the key pad. The calculator key should have a “ ± ” symbol on it or it could have a “ (−) ” symbol. In either case, pressing this key will always give you the opposite of what displayed on the screen. Give it a try!

Later we will find that multiplying a number by -1 is how we calculate the opposite of a number. Actually, when you press the “ ± ” key on the calculator, it simply multiplies whatever is on the screen by -1.

So if -3 is displayed on your calculator screen and then you press the “ ± ”, positive three will be displayed. This means that -1(-3)= 3 This is why a “negative ⋅ negative = positive”!

Page 6 of 7

Algebra2go®

Answer the following homework questions.

7) What number is its own opposite?

8) What is the opposite of the absolute value of negative 2?

9) What is the opposite of negative 2?

10) What is the opposite of positive 2?

11) Why do all negative numbers have opposites that are positive?

12) How do you tell which calculator key is the opposite key? What does this key do to the number that is displayed on the screen?

In Exercises 13 – 16, find the opposite of the given quantity.

Note: You must first evaluate the absolute value before you find the opposite.

Page 7 of 7

13) (-7 − 3)

15) -42 − 10

14) (-4 − 1 10 )

16) -32 − 2 2

Algebra2goÂŽ

Objective 1

Adding Negative Numbers

Understand how to add a Negative Number Remember that when finding the sum of four and three, start at four and move right three unit to get seven.

4+3 4+3=7 -7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

Now consider 4+(-3). Notice that in this case we are adding a negative three! Applying the Commutative Property for addition we get the following: 4+(-3)= -3+4 Since -3+4=1, by the Commutative Property it must be true that 4+(-3)= 1 .

4+(-3)

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

Notice that 4+(-3) is the same as performing 4 â&#x2C6;&#x2019; 3. We can now make a general conclusion.

Page 1 of 4

Algebra2go速

Conclusion: Adding a negative number is the same as subtracting its opposite.

Rewriting the addition of a negative quantity to subtracting its opposite, is a necessary skill for algebra! Although, with problems that have numeric values such as 5+(-3), you may be able to get the answer without rewriting it.

Example 1: Rewrite the following addition problems as equivalent subtraction problems.

Next, find the value of the expression if possible. a) a+(-b) b) x+(-y) c) -3+(-2) d) 5+(-8) Example 2: Apply the Commutative Property for addition to each expression and evaluate. a) 4+(-7) b) 8+(-12) Page 2 of 4

Algebra2go®

Example 3: Evaluate each expression. a) (3 − 12)+(-7+5) b) 5+(-2) − 8+(-6)

Example 4: Write an expression for each word statement. Next, evaluate the expression. a) Find the sum of -8, -10 and 3.

b) Find the sum of 3, -7, 6, and -9.

Page 3 of 4

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 15, use the number line to evaluate each expression.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

1) -5 + (-8)

6) 0 + (-6)

11) 5 − 4 +(-6)

2) -2 + (-7)

7) 0 + (-8)

12) 3 − 8 +(-2)

3) 7 + (-12)

8) 5 + (-5)

13) 4 − 6 +(-3)

4) -1 + (-5)

9) 3 + (-3)

14) -2 − 2 +(-2)

5) 6 + (-10)

10) -3 + (-3)

15) -3 − 2 +(-1)

In Exercises 16 – 18, write an expression for each word statement.

Next, evaluate the expression.

16) Find the sum of -3, -5, and -5.

17) Find the sum of -20, 30, and -40.

18) Find the sum of -32, 27, and -46.

Page 4 of 4

Algebra2go®

Objective 1

Subtracting Negative Numbers

Understand how to Subtract a Negative Number

It can be difficult to fully understand what happens when you subtract a negative number. But with enough practice it will eventually become easy.

Let first talk about subtraction and the phrase “take away”. Think about the phrase “4 take away 4” which means “4 − 4”. In both cases it makes sense that the result is 0. Next, look at the number line below which visually shows “4 take away 4” is 0.

“4 take away 4”

-7 -6 -5 -4 -3 -2 -1

Page 1 of 5

0

1

2

3 4

5

6

7

Now let’s think about the phrase “-4 take away -4” which means “-4 − (-4)”. In both cases it makes sense that the result is 0.

So how should this be represented on the number line?

Algebra2go®

If “-4 take away -4” which means “-4 − (-4)” is equal to 0, then here is how we represent this on the number line.

“-4 take away -4”

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

Notice how the number line above shows that “-4 take away -4” or “-4 − (-4)” is the same as performing -4+4! In all three cases the result is zero! We can therefore conclude that -4 − (-4) = -4+4.

Conclusion: Subtracting a negative number is the same as adding its opposite.

Rewriting the subtraction of a negative quantity to adding its opposite, is a necessary skill for algebra!

Remember that rewriting the addition of a negative quantity to subtracting its opposite, is also a necessary skill for algebra!

Page 2 of 5

Algebra2go®

Example 1: Rewrite the following subtraction problems as equivalent addition problems.

Next, find the value of the expression if possible. a) a − (-b) b) x − (-y) c) -3 − (-2) d) 5 − (-8) Example 2: Evaluate each expression. a) 5 − (-5) e) -2 − (-4)+(-6)

Page 3 of 5

b) -5 − (-5)

f) -10+(-20) − (-30)

c) 0 − (-8)

g) 17 − (-4) − 15

d) -4 − (-6)

h) 11 − (-13) − (-6)

Algebra2go®

Example 3: Evaluate each expression. a) (-4 − 10) − (-8+2) b) -4 − (-5)+ 9 − (-7)

Example 4: Write an expression for each word statement. Next, evaluate the expression. a) Subtract negative twelve from the product of three and four.

b) Subtract negative three from the sum of eight and negative seven.

Page 4 of 5

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 15, use the number line to evaluate each expression.

-7 -6 -5 -4 -3 -2 -1

0

1

2

3 4

5

6

7

1) -5 − (-8)

6) 0 − (-6)

11) 5 − 4 − (-6)

2) -2 − (-7)

7) 0 − (-4)

12) 3 − 8 − (-2)

3) -7 − (-12)

8) 2 − (-2)

13) 4 +(-10) − (-3)

4) -1 − (-5)

9) 3 − (-3)

14) -2 +(-5) − (-13)

5) 2 − (-3)

10) -3 − (-3)

15) -3 +(-3) − (-6)

In Exercises 16 – 18, write an expression for each word statement.

Next, evaluate the expression.

16) Find the difference of negative fifteen and negative thirty-one.

17) Find the difference of negative thirty-one and negative fifteen.

18) Subtract negative twenty-one from the sum of negative nineteen and seven.

Page 5 of 5

Algebra2go®

Multiplication with Negative Numbers

Objective 1

Understand why a “Negative times a Positive” or “Positive times a Negative” is Negative Remember that multiplication represents repetitive addition of a number.

Recall: 3 ⋅ 4 = 3 + 3 + 3 + 3 = A “positive times a positive” will always represent a positive number since we are summing positive quantities.

Page 1 of 6

But a “negative times a positive” implies that we are summing negative quantities. We therefore will always get a negative result in these cases. −3 ⋅ 4 = ( −3 ) + ( −3 ) + ( −3 ) + ( −3 ) = −5 ⋅ 3 = ( −5 ) + ( −5 ) + ( −5 ) = −1 ⋅ 6 = ( −1 ) + ( −1 ) + ( −1 ) + ( −1 ) + ( −1 ) + ( −1 ) = By the Commutative Property we can state −3 ⋅ 4 = 4 ⋅ ( −3 ) .

So we can conclude that a “positive times a negative” is also negative!

Algebra2go®

Example 1: Rewrite the following multiplication problems as equivalent addition problems. Next, find the value of the sum. a) 5 ⋅ 4 = 5+5+5+5= b) -3 ⋅ 6 = c) -6 ⋅ 5 = d) 4⋅(-2) = But what about a negative times a negative?

Notice that multiplying -1 to a number always results in the opposite of the number!

-1 ⋅ 2 = (-1)+(-1)= -2 -1 ⋅ 3 = (-1)+(-1)+(-1)= -3 -1 ⋅ 4 = (-1)+(-1)+(-1)+(-1)= -4 So what happens if we multiply -1 to a negative number?

Since the opposite of any negative number is always positive, the result must be positive.

-1 ⋅ (-2) = 2 -1 ⋅ (-3) = 3 -1 ⋅ (-4) = 4 Page 2 of 6

Algebra2go®

We can now make a general conclusion that negative times a negative will be positive!

To summarize things, we will look at a pattern that occurs in the columns below.

1⋅2=2 2⋅2=4 3⋅2=6 4⋅2=8

(-1) ⋅ 2 = -2 ( -2) ⋅ 2 = -4

( -3) ⋅ 2 = -6 ( -4) ⋅ 2 = -8

1 ⋅ (-2)= -2 (-1) ⋅(-2) = 2 2 ⋅ (-2)= -4 (-2) ⋅(-2) = 4

3 ⋅ (-2)= -6 (-3) ⋅(-2) = 6

4 ⋅ (-2)= -8 (-4) ⋅(-2) = 8

When multiplying two numbers with the same sign, the product will be positive.

When multiplying two numbers with different signs, the product will be negative.

Now let’s think about the product of three negative numbers.

(-2)⋅(-2)⋅(-2) Working left to right, we get the following: (-2)⋅(-2)⋅(-2) = 4⋅(-2) = -8 Page 3 of 6

Algebra2go®

Now let’s think about the product of four negative numbers.

(-2)⋅(-2)⋅(-2)⋅(-2) Working left to right, we get the following: (-2)⋅(-2)⋅(-2) ⋅(-2) = 4⋅(-2)⋅(-2) = -8 ⋅(-2)= 16 We can now state the following conclusion.

When multiplying an odd number of negative quantities, the product will be negative.

When multiplying an even number of negative quantities, the product will be positive.

Answer the following homework questions.

In Exercises 1 – 15, find each product.

1) -8 ⋅(-7)

2) 5 ⋅(-9)

7) -2 ⋅ 14

4) 0 ⋅(-5)

9) 6 ⋅(-6)

3) -11 ⋅ 12

5) 6 ⋅(-3)

Page 4 of 6

6) 12 ⋅(-8)

8) -2 ⋅(-16)

10) -1 ⋅ 0

11) -5 ⋅ (-4) ⋅ (-3)

12) -2 ⋅ (-3) ⋅ 8

13) 4 ⋅ (-8) ⋅ 10

14) 2 ⋅(-3)⋅(-1)⋅(-4)

15) -5 ⋅(-2)⋅(-3)⋅(-6)

Algebra2go®

Objective 2 Understand

Negative Numbers with Exponents It is important to understand the difference 2 2 − 3 between the two expressions −3 and ( ) .

The expression −32 is read “negative one times three squared”.

Therefore −32 is equivalent to −1 ⋅ 32 .

Following order of operations and evaluating the exponent first before multiplication, we find that −32 is equal to −1 ⋅ 9 or − 9 .

However, the expression ( −3 ) is read “negative three squared”. Notice how a set of parenthesis is used to define the negative base.

2

Therefore ( −3 ) is equivalent to ( −3 )( −3 ) .

In this case, we see that two negatives are being multiplied together. Therefore ( −3 )2 is equal to ( −3 )( −3 ) or 9.

2

Page 5 of 6

Algebra2go®

Answer the following homework questions.

In Exercises 16 – 30, find the value of each expression.

Note: Be sure to follow the rules of Order of Operations!

16) 2 3

21) ( −3 )

17) −2 3

22) ( −3 )

18) ( −2 )

3

19) −2 4 20) ( −2 )

27) 4 − ( −3 )

3

2

23) −34

28) 4 − 32

24) −33

29) − 102 − ( −4)2

25) ( −1 )

4

26) 1 − 2 2

4

99

30) − ( −2 )2 − ( −3 )3

In Exercises 31 – 36, find the value of each expression.

Note: Be sure to follow the rules of Order of Operations!

31) − −2

2

32) −32 − −2

Page 6 of 6

3

33) − 2 2 − 32

35) −42 − −42

34) − ( −4)2 − −2 3

36) −62 − 62

Algebra2go®

Division with Negative Numbers

Objective 1

Perform Division with Negative Numbers Remember that multiplication represents repetitive addition of a number.

Recall 4 ⋅ 3 = 4 + 4 + 4 = 12 This means that “4 goes into 12” three times!

4 4 4 0

1

2

3 4

5

6

7 8

9 10 11 12 13 14

We can see that it takes three 4’s to make 12.

3.

12 , this means 12 ÷ 4 = Since 4 ⋅ 3 = 12 12 Note: 4 = 3 since 4 ⋅ 3 = Do you see a pattern in the above note? ( −4) + ( −4) + ( −4) =−12 Recall: −4 ⋅ 3 = We can see that it takes three -4’s to make -12.

3. − 12 , this means − 12 ÷(− 4) = Since − 4 ⋅ 3 =

Note: −−12 4 = 3 since − 4 ⋅ 3 =− 12 Do you see a pattern in the above note? Notice that negative divided by negative is positive!

Page 1 of 4

Algebra2go®

Understanding our multiplication tables can help us with mentally determining basic division problems.

Notice that since 7 ⋅ 3 = 3 ⋅ 7 =21 we can conclude that 21 3= 21 7 = d an .

3 7 Similarly, since -7 ⋅ 3= 7 ⋅(-3)= -21 we can conclude that -21 -3 -21 -7 an = = d .

3 7 We can now make a general conclusion regarding division with integers.

When dividing two numbers with the same sign, the quotient will be positive.

When dividing two numbers with different signs, the quotient will be negative.

Page 2 of 4

Algebra2go®

Example 1: Find each quotient and then rewrite it as an equivalent multiplication problem by filling in the blank.

15 = a) 5 -42 = b) 7

45 = c) -9

-54 = d) -6

since 7 ⋅( ___) -42 =

since -9 ⋅ ___ 45 =

= since -6 ⋅ ___ -54

0 = e) -12

= since -12 ⋅ ___ 0

-84 = f) 4

= since 4 ⋅( ___) -84

-128 = g) 8

since 8 ⋅( ___) -128 =

162 = h) -9

= since -9 ⋅ ___ 162

216 = i) -12 Page 3 of 4

since 5 ⋅ ____ 15 =

= since -12 ⋅ ___ 216

Algebra2go®

Answer the following homework questions.

In Exercises 1 – 6, find each quotient.

÷ (-7) 132 ÷ (-11)

÷ (-8) -54 ÷ 9

÷ (-4) ÷ 5 -56 ÷ (-7) ÷ 8

1) 28

3) -32

5) 40

2)

4)

6)

In Exercises 7 – 10, write each word statement as a mathematical expression then find the value of the expression.

7) The quotient of -30 and 5.

8) Subtract -3 from the quotient of 27 and -9.

9) The quotient of -3 squared and -9.

10) The product of -1 and -4 squared, divided by -2.

In Exercises 11 – 19, evaluate each expression.

20 4(-6) 11) -4 14) -3 17) (-2)2 + 20 ÷ 4 0 7(-6) 2 12) -8 15) -3(-2) 18) -3 + 24 ÷( −8) -6 2 2 2 13) 0 16) (-3) + 21 ÷ 7 19) -2 −(-8) ÷(-4)

Page 4 of 4

Algebra2go: Working with Fractions and Mixed Numbers

Algebra2goÂŽ

Review of Fractions

1. Understand Fractions on a Number Line Fractions are used to represent quantities between the whole numbers on a number line. A ruler or tape measure is basically a number line that has both whole numbers and fractions represented on their labels. When measuring objects with a ruler, an essential skill is to be able to read off measurements of length that involve mixed number representations such as 1 3 inches. To develop this skill, we will first look at some number lines with different 8 fractional representations.

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

These number lines can be used to help us visually understand the meaning of equivalent fractions and the concept of basic mathematical operations with fractions. Additionally, they can help us master the skill of reading off measures using a variety of measuring scales.

Letâ&#x20AC;&#x2122;s first discuss the meaning of equivalent fractions. Equivalent fractions have the same 3 6 location on the number line. Since and have the same location on the number line, they 8

16

6

are considered to be equivalent fractions. You may recall that the fraction can be reduced 16 by dividing both the numerator and denominator by the common factor of 2. Doing so results 3 in the fraction .

8

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Page 1 of 10

Algebra2go速

Algebra2go: Working with Fractions and Mixed Numbers

Suppose you are asked to evaluate the expression 1 + 1 + 1 . You may recall, in order to add 3

4

6

fractions, they must all be written with the same denominator. These fractions can each be rewritten as an equivalent fraction with a denominator of 12.

Using our fraction number lines, we can see that 1 is equivalent to 4 , 1 is equivalent to 3 , 12 4

3

12

and 1 is equivalent to 2 .

6

12

1 6

1 4

1 3

1

0 We can now evaluate the expression by replacing each fraction with its equivalent fraction.

1 1 1 4 3 2 4+3+ 2 9 3 + + = + + = = = 3 4 6 12 12 12 12 12 4 This addition process can also be demonstrated on a number line as follows.

+

4 12

+

3 12

=

2 12

9 12

=

3 4

1

0

We can also use shaded areas to create a visual representation of this arithmetic operation.

+ 1 3

+ 1 4

+ 4 12

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1 6

+ 3 12

=

= 2 12

9 12

3 4

Page 2 of 10

Algebra2go: Working with Fractions and Mixed Numbers

Algebra2go®

Use the fraction number line diagram to answer the following questions.

For Exercises 1 – 6, represent each fraction as an equivalent fraction with the indicated denominator.

1)

3)

5)

3 , 16 4

2)

4 ,5 10

4)

2 , 12 3

6)

1 , 12 4 12 ,4 16 1 , 10 2

For Exercises 9 – 12, use the fraction number line diagram to re-write each fraction with the same denominator. Then evaluate the expression.

9)

3 2 − 5 10

10)

11 3 − 16 8

11)

2 1 3 − + 3 2 4

12)

5 2 − 6 3

1 be written as a fraction with a 3 denominator of 16? Why or why not?

13) Can

0 on the number line?

6 8 8) Where is on the number line?

8

7) Where is

14) How do we represent the whole number 1 as a fraction with a denominator of 12?

2. Adding and Subtracting Unlike Fractions Now, let’s look at the arithmetic approach of finding the value of the expression 1 + 1 + 1 .

3

4

6

Recall that we need to make all the fractions have the same denominator before we add the fractions. Therefore, we first need to determine what the least common denominator (LCD) is.

Remember, the LCD is the smallest number that all the denominators divide evenly into.

The smallest number that a 3, 4, and 6 divide evenly into is 12. Therefore, 12 is our LCD. We now multiply each fraction by a factor of 1 to get the equivalent fraction having the LCD. Remember, when multiplying fractions, multiply straight across the top, and straight across the bottom.

1 = 3

1 = 31

()

1 4 =   3 4

4 12

1 = 4

1 = 4 1

()

13 =   43

3 12

1 = 6

1 = 6 1

()

12 =   62

2 12

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1 1 1 + + 3 4 6 4 3 2 + + 12 12 12

LCD = 12

Each fraction is now re-written with a denominator of 12.

4 + 3 + 2 Add the numerators of the fractions in the previous step. The denominator remains unchanged.

12 9 Reduce the fraction by dividing both the 12 numerator and denominator by 3.

3 4

Page 3 of 10

Algebra2go: Working with Fractions and Mixed Numbers

Example 1: Evaluate the expression

Algebra2go®

5 3 1 3 − + − .

8 4 2 16

We first determine that the LCD = 16 since it is the smallest number that 8, 4, 2, and 16 divide evenly into. We now multiply each fraction by a factor of 1 to get its equivalent fraction having the LCD of 16.

5 = 8

5 = 8 1

()

5 2 =   8 2

10 16

5 3 1 3 − + − 8 4 2 16

3 = 4

3 = 4 1

()

34 =   44

12 16

10 12 8 3 − + − 16 16 16 16

1 = 2 3 = 16

()

1 = 2 1

LCD = 16

18 =   28

10 − 12 + 8 − 3 16

8 16

()

24) Where is the fraction

7 on the number 3

line?

For Exercises 25 – 28, find the value of each expression.

2 1 3 25) − + 3 6 2

17)

1 2 , 9 3

18)

2 2 , 7 21

19)

5 1 3 , , 6 4 8

20)

1 2 2 , , 2 3 5

26)

3 2 1 − + 4 3 6

21)

3 4 7 , , 8 5 10

22)

9 3 6 , , 14 4 7

27)

9 1 3 − − 7 2 4

3 on the number 2

28)

13 1 3 − − 8 5 4

23) Where is the fraction

Add or subtract as indicated the numerators of the fractions in the previous step. The denominator remains unchanged.

3 16

1 18 3 =  = 1 2 2  8  16

For Exercises 15 – 22, identify the LCD and then re-write each fraction as an equivalent fraction having the LCD. 3 5 2 5 15) , 16) , 4 8 3 6

Each fraction is now re-written with a denominator of 16.

line?

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Page 4 of 10

Algebra2go: Working with Fractions and Mixed Numbers

Algebra2go®

3. Adding and Subtracting Mixed Numbers

Many calculations in Health Care careers require calculations that involve mixed numbers. A mixed number is simply the sum of a whole number and a fraction.

2

2

+ = 2 3 2= 3

6+2 3

6 2 + = 3 3

=

8 3

8 8 . Recall that the improper fraction represents a number 3 3 that is greater than 1. We can also verify this by performing the division problem 8 ÷ 3 which 8 is represented by the fraction .

3 2 R2 2 8 3 8 2 R2 = = −6 3 3 2 Notice that we find that

2

23 =

2

Note: When we add mixed numbers, we can add the whole number parts and the fractional parts separately. Using these two results, we then write our answer in mixed number format.

Example 2: Find the value of

2

1

2 3 + 1 4 . Write your answer in mixed number format.

( 2 + 1) +  3 + 4  2 3 + 1= 4 2

 2 1  Add the whole number parts and fractional parts separately.   3  Both fractions from the previous step are 8 +  +  re-written as equivalent fractions with an  12 12  LCD of 12.  11  Add both fractions in the previous step. +    12 

1

=

3

=

3 =

11

3 12

Write the final answer in mixed number format.

When we subtract mixed numbers, we can subtract the whole number parts and the fractional parts separately. Using these two results, we again write our answer in mixed number format.

Note: These problems can be approached by first changing the mixed numbers into improper fractions and then performing the operation. This approach will be demonstrated later in this section.

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Page 5 of 10

Algebra2go: Working with Fractions and Mixed Numbers

Example 3: Find the value of

5

46 5

46

−

Algebra2go®

2

2 3 . Write your answer in mixed number format.

2 − 2= 3

( 4 − 2) +  56 − 23 

=

2

=

2 =

Subtract the whole number parts and fractional parts separately.

 5 4  Notice that 23 was re-written as an +  −  equivalent fraction with an LCD of 6. 6 6 1 +   Subtract both fractions in the previous step. 6 1

26

Write the final answer in mixed number format.

Note: In some cases, when finding the sum of mixed numbers, we need to carry over a whole number from the fractional part. The following example demonstrates the process.

Example 4: Find the value of

7

28

+

3

1 4 . Write your answer in mixed number format.

  + +  + 1 2 ( ) 2 8 + 1= 4 8 4 7

3

7

=

3

=

3

=

3

=

4 =

3

7 6 + +  8 8  13  +   8

Add the whole number parts and fractional parts separately.

Notice that

3 was re-written as an 4

equivalent fraction with an LCD of 8.

To get this result, we added the fractional parts.

Notice we have an improper fraction which will be re-written as a sum in the next step.

 5 we write the improper fraction as a + 1+  Here sum of a whole number and fraction.  8  5  Here we added the whole numbers +   parts. 8 5

48

Write the final answer in mixed number format.

Note: In some cases, when finding the difference of mixed numbers, we need to borrow from the whole number part of the mixed number to avoid a negative result in the fractional part. The following example demonstrates the process.

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Page 6 of 10

Algebra2go: Working with Fractions and Mixed Numbers 2

53

Example 5: Find the value of

−

Algebra2go®

3

2 4 . Write your answer in mixed number format.

(5 − 2) +  3 − 4  5 3 − 2= 4 2

=

3

=

2

=

2

=

2

=

2

=

Example 6: Evaluate 1

4 10

+

 2 3  Subtract the whole number parts and fractional parts separately.   9  From the previous step, both fractions are re-written as 8 +  −  equivalent fractions with an LCD of 12. Notice that performing  12 12  the subtraction operation would give us a negative result.

8 9  In this step, we borrowed a 1 from the whole  + 1 + −  number part to avoid the negative result.

12 12  

3

1

4 10

9  Here we represent the number 1 with an  12 8 + + −  equivalent fraction having an LCD of 12.  12 12 12   20 9  Here we followed the rule for order of operations + −  and worked left to right adding the first two fractions   12 12  in the previous step.  11  +   Subtract both fractions in the previous step.  12 

11

2 12

+

1

Write the final answer in mixed number format.

5

2 2 − 3 6 . Write your answer in mixed number format.

− 3 ( 4 + 2 − 3) +  10 + 2 − 6  2 2= 6 1

5

=

3

=

3

=

2

=

2

=

2

=

2 =

www.Algebra2go.com

23

2 30

1 5  As indicated, add and subtract the whole number parts 1 and fractional parts separately.    3 15 25  From the previous step, all fractions are re-written + + −  as equivalent fractions with an LCD of 30.  30 30 30  we followed the rule for order of operations and worked left to right  18 25  Here the first two fractions in the previous step. We can see that + −  adding  30 30  performing the subtraction in this step would give us a negative result.

18 25  Here we borrowed a 1 from the whole number  + 1 + −  part to avoid the negative result.

30 30    30 18 25  Here we represent the number 1 with an +  + −  equivalent fraction having an LCD of 30.  30 30 30   48 25  Again we follow the rule for order of operations and work left to right +  −  adding the first two fractions in the previous step.  30 30   23  +   Subtract both fractions in the previous step.  30 

Write the final answer in mixed number format.

Page 7 of 10

Algebra2go: Working with Fractions and Mixed Numbers

Algebra2go®

As was mentioned earlier, we can first change the mixed numbers to improper fractions before we add or subtract. In some cases this is easier, and in some cases it may be more difficult. In Example 7, we will repeat the problem given in Example 5 using this approach.

Example 7: Find the value of

2

−

53

3

2 4 by first writing the mixed numbers as

improper fractions. Write your answer in mixed number format.

2

3

17 11 Both mixed numbers are re-written − 3 4 as improper fractions.

68 33 From the previous step, both fractions are re-written as = − 12 12 equivalent fractions with an LCD of 12.

68 − 33 Here we write the difference of the two = fractions in the previous step.

12 35 Here we calculated the difference of 68 and 33. = 12 The denominator remains unchanged. get this result, we perform long division and = 2R11 To divide 12 into 35.

53 − 24 =

11

=

2 12

For Exercises 29 – 38, evaluate each expression.

29)

2 3 + 14

1

1

30)

3 8 − 14

7

3

31)

4 5 3 5 + 16

32)

1 43 −

4 27

33)

6 8 + 5 10

34)

8 16 − 4 8

35)

7 2 + 6 10

37)

55 + 46

38)

23

7

7

1

9

4

2

5

−

7

18

3

5

3 11 36) 11 − 5 4 16

The result is finally written in mixed number format.

For Exercises 39 – 44, evaluate each expression by first changing the mixed numbers to improper fractions. Write your final answer in mixed number format.

63

+

91

www.Algebra2go.com

3

310 + 2 8

41)

68 + 45

43)

6 − 310 + 2 5

44)

8 10 − 5 + 7

2

−

1

39)

3

1

7

5

2

40)

46 − 23

42)

2 3 − 14

1

1

4

3

4

Page 8 of 10

Algebra2go: Working with Fractions and Mixed Numbers

Algebra2go®

Review Exercises

For Exercises 45 – 56, evaluate the expression.

45) 10 ⋅ 9 ⋅ 7

57) In your own words, write down the rules for order of operations.

58) When was the last time you had to perform a multiplication problem for a household task? Write down the event in your own words.

46) 10 ⋅ 6 ⋅ 8 47) 6 − 14 + 3

59) When was the last time you had to perform a division problem for a household task? Write down the event in your own words.

48) 17 − 31 + 8 49) 13 + 7 ⋅ 6

60) Besides math instructors, what type of careers use mathematics on a daily basis? Write down at least five careers.

50) 11 − 2 ⋅ 9 51) 16 ÷ 4 ⋅ 2 52) 32 ÷ 2 ⋅ 4 53) 4 + 32 ⋅ 2 54) 40 − 20 ÷ 22 55) (11 − 7 )

2

÷ 23

56) ( 6 − 2 ) − ( 2 ⋅ 3) 3

www.Algebra2go.com

2

Page 9 of 10

Algebra2go: Working with Fractions and Mixed Numbers 1 2

0

2 8

1 10

0

1 12

1 16

2 10

2 12

2 16

3 5

3 16

www.Algebra2go.com

4 10

3 10

3 12

4 16

5 16

6 16

7 16

6 12

8 16

5 6

6 8

5 8

6 10

5 10

5 12

4 12

1

4 6

4 8

3 8

1

4 5

3 6

2 6

1 8

0

3 4

2 5

1 6

0

1

2 4

1 5

0

0

2 3

1 4

0

0

1

1 3

0

Algebra2go速

7 10

7 12

9 16

8 12

10 16

11 16

7 8

8 10

9 12

12 16

1

1

9 10

10 12

13 16

1

11 12

14 16

15 16

1

1

Page 10 of 10

Algebra2go®

Reducing Fractions

Objective 1

Write a Fraction in Lowest Terms (Reducing) Note: A fraction is written in lowest terms or reduced when the numerator and denominator have no common factors other than 1.

3 Let’s begin with the fraction 8 . In this case, both the numerator 3 and denominator 8 have no common factors other than 1. Therefore, this fraction is in lowest terms.

6 Now let’s look at 8 . Here, the numerator and denominator have a common factor of 2.

To reduce this fraction we divide out the common terms between the numerator and denominator. 6 = 6 ÷ 2 = 3

8

8÷2

4

Notice that dividing both the numerator and denominator by the same factor results in an equivalent fraction!

In some cases, we may have to divide out common factors more than once to reduce the fraction to lowest terms.

Page 1 of 5

Algebra2go®

28 Let’s take a look at the fraction 42 . It may not be obvious that 28 and 42 have a common factor of 14. Since they are both even numbers we can begin by dividing out the common factor of 2. 28 = 28 ÷ 2 = 14

42

42 ÷ 2

21

Since 14 and 21 both have a common factor 14 of 7, the fraction 21 is not written in lowest terms. Therefore we divide out the common factor of 7.

14 14 ÷7 2 = =

21

21 ÷ 7

3

Since 2 and 3 have no common factors other 28 than 1, we can now state that 42 is now written 2 in lowest terms as the equivalent fraction 3 .

Page 2 of 5

Since we separately divided out a common factor of 2 and 7, this means that 2 ⋅ 7 or 14 is a common factor of 28 and 42. Therefore, 28 the original fraction 42 could have been reduced in one step using the common factor of 28 = 28 ÷ 14 = 2 14.

42

42 ÷ 14

3

Algebra2go®

Recall that -21 ÷ 3=-7. When writing this -21 quotient using a fraction bar, we have 3 .

Dividing out the common factor of 7, we get the following result.

-21 -21 ÷= 3 -= 7 -7 = 3÷3 3 1

Suppose we were given 21 ÷ (-3) which also equals -7. Using a fraction bar we have the equation 21 -3 = -7 .

-21 21 -7 We can see that 3= -= . Therefore, 3 whenever we have one negative sign in either the numerator or denominator (not both), we can move it to the front of the fraction to indicate a negative answer.

-21 = 21 = 21 = − -3 3 −7 3

Answer the following homework questions.

In Exercises 1 – 6, write each fraction in lowest terms.

42 6 6 1) 20 3) 51 5) 28 8 80 6 2) 52 4) 54 6) 24 Page 3 of 5

Algebra2go®

Let’s now reduce a fraction without writing a division symbol. Suppose we are asked to 8 reduce 10 . Since 8 and 10 are both divisible by 2, we will divide out this common factor using the following notation.

Here it is understood that you are dividing out a common factor of 2.

8= 8=4 4 10 10 5 5

Below is the same process without writing down a division symbol.

8 8= ÷2 4 = 10 10 ÷ 2 5

Objective 2 Reducing

Fractions that have Variables 8x3 fraction x 2 .

Let’s begin with the Remember that exponents are used to represent repeated multiplication. Therefore 8xx = 8 ⋅ xx ⋅⋅ xx ⋅ x .

Since xx = 1 , we can cancel variable terms as 1 1 3 follows. 8 x= x 8= x 8x 8 ⋅ x ⋅ x ⋅= 3

2

x2

x ⋅ x1

1

1

Notice that when cancelling variable terms, they get replaced with 1’s.

Page 4 of 5

Algebra2go®

20n 2 6n 5 .

Suppose we were given Again remember that exponents are used to represent repeated multiplication. Therefore we use the following notation to reduce the fraction to lowest terms. 10 1

1

10 10 20n 20 = ⋅n⋅n = = 5 6n 6 ⋅ n ⋅ n ⋅ n ⋅ n ⋅ n 3 ⋅ n ⋅ n ⋅ n 3n3 2

3

1

1

Answer the following homework questions.

In Exercises 7 – 15, write each fraction in lowest terms.

15 3a 2 xy 7) 20 10) 3b 13) 8 xyz

Page 5 of 5

80 8) 24

4 xy 11) − 4 y

2 3 14) 3aabb

12 9) 52

5a 12) −5ab

7 2b 3c 5 15) 63aab 4c7

Algebra2go®

Multiplying Fractions

Objective 1

Perform Multiplication with Fractions Recall that multiplication represents repeated addition of the same quantity.

1⋅ = 1+1+1+1+1+1 = 2 6 2 2 2 2 2 2

3

We can also write the 6 as an improper 6 fraction 1 and multiply. We will reduce by dividing out a common factor of 2.

3 1 ⋅ 6 = 1 ⋅ 6 = 1 ⋅6 = 6 = 6 = 3 = 1 2 1 2 1 2⋅1 2 2 1

3

Notice how we multiply 21 to 61 . We multiply straight across the numerators and straight across the denominators.

Whenever we are multiplying fractions together we can use a technique called “crosscancelling”, but it is very important that you remember that this technique can only be used when multiplying fractions together!

3

Page 1 of 3

1 ⋅6 = 1 ⋅6 = 3 = 2 1 21 1 1

3

Here it is understood that you are dividing out a common factor of 2 before multiplying.

Algebra2go®

Dividing out common factors before multiplying fractions together is very useful, especially when you are working with large numbers. Sometimes you will have to divide out factors multiple times. Notice how the following challenging problem is worked.

8

1

2

5

12

2

72 ⋅ 55 ⋅ 14 = 72 ⋅ 55 ⋅ 14 = 8 ⋅ 1 ⋅ 2 = 35 108 110 35 108 110 5 ⋅ 12 ⋅ 2 2

1

8= ⋅1⋅2 ⋅1⋅1 2 2= 5 ⋅ 12 ⋅ 2 5 ⋅ 3 ⋅ 1 15 3

1

There are different ways of approaching the problem above. But no matter how many steps it takes you, we should all end up with the same answer!

Answer the following homework questions.

In Exercises 1 – 9, multiply by first dividing out common factors.

8 1) ⋅ 3 6 2) ⋅ 8

3) Page 2 of 3

1 2 4 3 40 3

⋅ 109

4) 5) 6)

⋅

7)

⋅

8)

24 9 6 8 15 5 10 30 16 x 20 5 12 y

⋅

9)

⋅

6 x 2 21a 7 a 2 12 x x y y x ab c c c a b

⋅

⋅ ⋅

Algebra2go®

Recall that exponents are used to represent repeated multiplications.

Example 1: Simplify each expression. a)

 1 −   2

2

−9

 1  1  1   −  −  −  −  2  2  2  −

1 8

−

1   3

2

b)

9 3  3  1

1   2

2

 1  1      2  2 

1

9⋅ 9 1

1 4

⋅8

+  2  3

2

⋅9

⋅8 +  23  23 ⋅9

⋅8

+

1 9

⋅9

Answer the following homework questions.

In Exercises 10 – 15, simplify each expression. 1 10)   2

11) Page 3 of 3

2

⋅8

2 7  −   3 6

12)

3   2

3

⋅

5 13) 16   4

2

8 9

⋅

7 25

14)

3⋅

15)

1 −  2

7   3 2

2

⋅ 215

⋅ 65 ⋅ 16

Algebra2go®

Division with Fractions

Objective 1

Perform Division with Fractions Suppose we are given the problem 4 ÷ 31 .

This problem is asking you how many one-thirds will go into a 4?

Consider the number lines below. Notice that 12 it takes “twelve-thirds” or 3 to make a 4.

0

1

2

3

4

0 1 2 3 4 5 6 7 8 9 10 11 12 3 3 3 3 3 3 3 3 3 3 3 3 3

We can demonstrate this visually using the following diagram.

1 1 3 3 1 3

1 1 3 3 1 3

1 1 3 3 1 3

1 1 3 3 1 3

We can see there are 12 one-thirds in 4 -wholes, where each whole contains 3 one-thirds.

Page 1 of 3

We can arithmetically calculate 4 ÷ 31 by multiplying 4 by the reciprocal of 31 . The reciprocal of 31 represents the number of one-thirds in 1 whole. The reciprocal of 31 is 3 since there are 3 one-thirds in 1whole.

Algebra2go®

When dividing a quantity by a fraction, multiply the quantity by the reciprocal of the fraction. = ÷ 31 4 4= ⋅ 3 12

Change division of the fraction to multiplication by its reciprocal.

Example 1: Divide.

3 ÷ 7 14

a) − 6

−

6 7

⋅ 143

b) − 4 ÷

−

4 1

8 5

⋅ 85

3 4

÷2

3 4

÷ 12

c)

Answer the following homework questions.

In Exercises 1 – 9, perform the indicated operations.

2 3 40 25 1) 3 ÷ 10 4) ÷ 7) 2) 3) Page 2 of 3

4 3 ÷ 5 10 64 8 ÷ 9 27

5) 6)

69 46 17 17 ÷ 30 30 16 x 4 x ÷ 10 5

6x2 5a 2

÷ 7a

18 x 2

8) 9)

2 2 x ÷   ÷ 3 5 y 3c c 3c ÷ ÷ 4a a 4a

Algebra2go®

Example 2: Divide. a)

7 6 4 5

xy 2 z y z

b)

xy 2 z

7 4 ÷ 6 5

÷

c)

3a 5b3 a3 10b 2

y z

Answer the following homework questions.

In Exercises 10 – 15, perform the indicated operations.

Page 3 of 3

10)

1 ÷ 4 2

11)

1 2

÷4

÷  − 32  +

12)

4 9

13)

7 10

÷ − 1 4

2 5

4 3

14)

15)

3÷ 6⋅

3÷

1 2

3   4

2

÷6

Algebra2go®

More with Fractions Objective 1

Add and Subtract Fractions with Different Denominators Remember: In order to add or subtract fractions the denominators must be the same.

2+1−5 Let’s begin by working the problem 3 4 6 .

In order to perform the indicated operations of addition and subtraction, we must rewrite each fraction as equivalent fractions having the same denominator.

We begin by first finding the Least Common Denominator (LCD) of all three fractions. The LCD can simply be thought of as the smallest number that all your denominators divide evenly into.

2+1−5 3 4 6 Here our denominators our 3, 4, and 6. The smallest number that 3, 4, and 6 divide evenly into is 12. Therefore 12 is the LCD.

Page 1 of 5

Note: The LCD is never smaller than the largest denominator. In fact, it is always a multiple of the largest denominator.

Algebra2go®

Another method of finding the LCD is to find the Least Common Multiple (LCM) of the denominators. A simple way of doing this is to make a list of multiples of the denominators to find the lowest common multiple. This quantity will be the LCD. 2+1−5 For the problem 3 4 6 , we will make list of multiples for 3, 4, and 6 starting with the largest denominator.

6: 6, 12, 18, 24, 30, 36, …

4: 4, 8, 12, 16, 20, 24, 30, …

3: 3, 6, 9, 12, 15, 18, 21, 24, …

Notice that 12 is the lowest common multiple and therefore 12 is the LCD.

Page 2 of 5

Note: When the denominators involve very large numbers, making a list of common multiples can be very time consuming. In these cases, using prime factorization to find the LCD may be a better approach. This method will be covered in a later section.

Algebra2go®

2 1 5 For the problem 3 + 4 − 6 we have the LCD=12.

To rewrite each fraction as an equivalent fraction with a denominator of 12, we must multiply each fraction by an appropriate factor of 1.

2+1−5 3 4 6 2 4 +1 3 −5 2 3 4 4 3 6 2 8 + 3 − 10 12 12 12 8 + 3 − 10 12 1 12

( ) ( ) ( )

Page 3 of 5

Algebra2go®

Example 1: Perform the indicated operations.

2 4 3 5 1 3 5 a) − 6 + 10 − 5 b) − 8 + ( − 4 ) − 32 Multiples of Denominators 10: 10, 20, 30, 40, 50, 60, 70, …

6: 6, 12, 18, 24, 30, 36, 42, …

5: 5, 10, 15, 20, 25, 30, 35, …

LCD=30

( )

( ) ( )

5 + 3 3 − 4 10 −5 6 5 10 3 3 10

9 − 40 − 25 + 30 30 30

− 25 + 9 − 40 30 − 56 30 28

− 56 30 15

28 − 15 Page 4 of 5

( )( )

1 −1 − 5 − 83 + − 4 4 32 1 − 5 − 83 + 16 32 Multiples of Denominators.

32: 32, 64, 96, 128, …

16: 16, 32, 48, 64, 80, 96, …

8: 8, 16, 24, 32, 40, 48, 56, 64, …

LCD=32

( )

( )

1 2 − 5 − 83 4 + 4 16 2 32 12 + 2 − 5 − 32 32 32 −12 + 2 − 5 32 −15 32 15 − 32

Algebra2go®

Answer the following homework questions.

In Exercises 1 – 12, perform the indicated operations.

3+2 3 − −1 3 2−2− 2 1) 4 5 5) 12 9) 7 9 21 2 2 2 3 2 1 3 2 3 − − 2) 4 + 5 − 10 6) 81 − − 4 10) − 2 3 3 5 1 3 2 1 2 6 3) 9 − − 6 7) − 8 − 6 + 3 11) 25 − 15 4 1 3 + 5 3+ 1 4) 3 + t 8) 12) 40 36 5 2h (LCD=3t) (LCD=10h)

( ) ( )

( )

2

Page 5 of 5

( )

( )

Algebra2goÂŽ

Clearing Fractions (Kung Fu Fractions) Objective 1

Recognize when a Whole Number times a Fraction is a Whole Number Consider the product 12 ( 4 3 ). Will the result be a whole number or a fraction?

Writing 12 as a fraction and multiplying will get us the result.

4 4 12 4= 16 16 = = 12= â&#x2039;&#x2026; â&#x2039;&#x2026; ( 43 ) 12 1 3 1 3 1 1

Notice that the result above was a whole number. This was because the denominator of the fraction divided evenly into the whole number.

When finding the product of a whole number and fraction, if the denominator of the fraction divides evenly into the whole number, the result will be a whole number. If the denominator does not divide evenly into the whole number, the result will be a fraction.

Consider the product 8 ( 2 ) . Will the result be a whole number or a fraction?

5

Page 1 of 4

Algebra2go®

With 8 ( 2 ) we say to ourselves, “2 goes into 8 four times, and four times 5 is 20”. Therefore 8 ( 52 ) = 20 . We could use the following short hand notation. Can you see how this works?

4 ( 52 ) = 20 1 5

8

With lots of mental practice we can read off answers very quickly. In this case we say we used “Kung Fu math” to clear the fraction.

Example 1: Use “Kung Fu math” to find each product.

4 5 7 4 a) 9( 3 ) b) 14( 7 ) c) 10 ( 2 ) d)18 ( 6 ) 3) 7 ( Consider the product 2 . Will the result

Page 2 of 4

be a whole number or a fraction?

In this case, the result will be a fraction since the denominator does not divide evenly into the whole number. We must proceed as 1 follows. 7= or 10 ⋅ 23 21 ( 23 ) 71= 2 2

Algebra2go®

Now consider 8 ( 4 + 2 − 8 ) . Notice that the denominators of the fractions within the parenthesis divide evenly into the whole number 8 . In this case it may be easier to distribute the 8 rather than simplifying the expression within the parenthesis using an LCD. This process is demonstrated below.

3 5

8(

11

3 + 5 − 11 4 2 8

)

) 8( 8( ) 8( ) 8( ) 6 + 20 − 11 15 2

3 + 5 − 11 4 2 8 4 1 3 + 5 − 11 2 4 8 1

Page 3 of 4

1

1

The more you practice this “math Kung Fu” technique, the faster you will get. A “Kung Fu math” black belt will solve the problem in two steps! 8 ( 43 + 52 − 11 8 ) = 6 + 20 − 11 = 15

Algebra2go®

Example 2: Use “Kung Fu math” to simplify the expressions to a whole number.

15 − 1 − 3 ) ( 16 a) b) 12 ( 2 − 53 + 63 ) 8 2 4

Answer the following homework questions.

In Exercises 1 – 12, simplify each expression.

Page 4 of 4

5)

5 ( 101 − 151 )

9)

63( 72 − 29 − 212 )

24( 58 )

6)

18 ( 59 − 61 )

10)

30( 1521 − 3 + 115 )

3)

5 ( 21 )

7)

7 − 11 36( 12 18 )

11)

20( 56 − 5 + 8)

4)

3( 76 )

8)

13 − 27 ) 100( 20 50

12)

36( 5 + 126 − 8)

1)

16( 43 )

2)

Algebra2go®

Complex Fractions Objective 1

Learn how to simplify Complex Fractions using the Clearing Fractions Technique 3+1 4 3 Consider the complex fraction 5 3 .

6−2 While simplifying this complex fraction looks a bit complicated, it can be simplified rather easily using the clearing fractions technique.

Using the LCD for all four fractions, we can clear away all four fractions! This can be done by multiplying the LCD to the top and bottom of the complex fraction.

This technique is demonstrated below.

3 1 4+ 3 5−3 6 2

LCD=12

12 ( 43 + 31 ) 12 ( 56 − 23 ) Page 1 of 3

12 ( 43 ) + 12 ( 31 ) 9 + 4 13 = = = 3 5 − − 8 12 ( 6 ) − 12 ( 2 ) 10 18

−

13 8

Algebra2go®

Again, the more you practice the clearing fractions technique, the faster you will get at simplifying the complex fraction expressions.

Example 1: Use the clearing fractions technique to simplify the complex fraction.

2 a) 31 5

LCD=15

15 ( 23 ) 15 ( 51 )

b)

2 + 83 − 61 5 − 12

1

LCD=24

24( 2 + 83 − 61 ) 5 −1 24( 12 ) 24( 2 ) + 24( 83 ) − 24( 61 ) 5 − 24 1 24( 12 ) ( )

Page 2 of 3

Algebra2go®

Part b) in Example 1 can be done very quickly once you master this technique. Here’s what the work of a “math Kung Fu” black belt would look like. See if you can follow the work.

2 + 83 − 61 5 − 12

LCD=24

1

48 + 9 − 4 10 − 24 53 − 14 Answer the following homework questions.

In Exercises 1 – 6, simplify each complex fraction.

1)

2) Page 3 of 3

8 7 6 5 3 8 2 9

3

3) 11 6

4)

5) 1

−1

2

2 1

3− 3 + 2 7 4 + 10 5

2

3

−3+

5 6

2− 3 + 4

1+ 4

6)

1 1 + − 3 6 2 5 + − 9 6

2

1

Algebra2go®

Combining Like Terms Objective 1 Learn how to Combine and Identify Like Terms

Terms like 3x and 5x are considered to be like terms because the variable parts are identically the same. The variable part is “x”.

To find the sum 3x+5x, we simply add the numeric parts. The variable parts remained unchanged. We can use the Distributive Property to demonstrate the process.

3x+5x = x(3+5) = x(8) = 8x Suppose we have 4y2 − y2 or 4y2 − 1y2 .

Since the variable parts are exactly alike, we can combine these two terms as follows.

2 4y2 − y= y2 ( 4 − 1= ) y2 ( 3= ) 3y2 The terms 8b2 and 2b are not like terms since the variable parts are not identically the same. Similarly, 5 x 2 and 3y2 are not like terms. Note the following.

8b2 − 2b = Cannot Combine 5x 2 + 3y2 = Cannot Combine

Page 1 of 5

Algebra2go®

Example 1: Combine like terms if possible.

3 3 3 a) 2x+7x f) 8n + 6n − 3n

Page 2 of 5

2+5 b) x x

8 + 6 − 3 g) b2 b2 b2

c) 3a2 − 7a2

h) 3ab2 − 7ab2 + ab2

d) 8a2 − 11b2

2 3 2 3 i) 14x y − 10x y

3 2 2 3 25 p q − 15 p q e)

j) a7b5 c 4 − a7b5 c 4

Algebra2go®

Objective 2 Combine Like Terms within an Expression

Suppose we are asked to simplify the expression 12x − 7 − 6x + 3 by combing like terms.

Remember that the rules for Order of Operations state we must work left to right when we have additions and subtractions. But we are unable do this with the expression 12x − 7 − 6x + 3 since the first two terms 12x and 7 are not like terms!

However, if rewrite the subtractions as “adding negative quantities”, we can then rearrange terms within the expression.

Remember, subtracting a negative number is the same as adding its opposite! Applying this technique to an expression is demonstrated below.

12x − 7 − 6x + 3 12x + (-7) + (-6x) + 3 Page 3 of 5

Algebra2go®

Having the expression 12x + (-7) + (-6x) + 3 , we can now rearrange the expression as follows.

12x + (-7) + (-6x) + 3 12x + (-6x) + (-7) + 3 Note: You can rearrange terms within an expression when all the terms are being added.

Since all the terms are now being added, we can now combine like terms as shown below.

12x + (-6x) + (-7) + 3 6x + (-4) 6x − 4 Once you completely understand the process of combining like terms within an expression, you can start using “Kung Fu math” to quickly simplify an expression. Below is how a black belt in “Kung Fu math” would work the problem. Can you explain how this student did not follow the rules of Order of Operations and get the right answer?

12x − 7 − 6x + 3 Page 4 of 5

6x − 4

Algebra2go®

Example 2: Simplify each expression by combing like terms. a) 8a − 4a − 6 − 2 f) -a − b − 3a + 5 − 4b

b) 3x − 5 − x − 7

g) -8 − 7 − 5 − 2

c) -5+a2 − 10 − 4a2

h) - ( -8p ) − 3q − 4p + 6q

d) -8 x − 7 − ( -5 x ) − ( -2 )

i) -10y3 − 10x 2 − 10y2

e) 7 p3 − q 2 − 10p3+5q 2 j) a2b2 − b2 c 2 + a2 c 2

Page 5 of 5

Algebra2go®

Objective 1

The Distributive Property and Expressions Understand how to use the Distributive Property to Clear Parenthesis The Distributive Property The Distributive Property states that multiplication can be distributed across addition and subtraction.

X(a+b) = ax + bx a(x − y+z) = ax − ay+az -a(x − y+z) = -ax + ay − az

Page 1 of 9

Consider the expression 3(x+2). While the rules of Order of Operations state we must first work on the expression within the parenthesis, this cannot be done. The expression x+2 cannot be simplified since x and 2 are not like terms.

However, we can remove the parenthesis by distributing the 3 using multiplication to each term within the parenthesis.

3(x+2) 3⋅x+3⋅2 3x+6

Algebra2go®

Suppose we need to remove the parenthesis from the expression 4(a + b − c). To do this, we will distribute the 4 using multiplication to each term within the parenthesis.

4(a + b − c) 4⋅a + 4⋅b − 4⋅c 4a + 4b − 4c When distributing a positive quantity across addition and subtraction within a parenthesis, the operations remain unchanged.

But what happens when we distribute a negative quantity? You will notice that the operations will change. Can you explain why?

-4(a+b − c)

Page 2 of 9

-4(a+b − c) (-4)(a) + (-4)(b) − (-4)(c) (-4a) +(-4b) − (-4c) -4a − 4b+4c

Algebra2go®

When distributing a negative quantity across addition, you end up adding a negative quantity. This is why addition changes to subtraction! Review how to add negative numbers.

When distributing a negative quantity across subtraction, you end up subtracting a negative quantity. This is why subtraction changes to addition!

Review how to subtract negative numbers.

Most students do not write out all the steps when distributing a negative quantity across addition and subtraction. Most students choose to use “Kung Fu math”. Try it on the following example.

Example 1: Apply the Distributive Property to remove the parenthesis. Use “Kung Fu math”. a) -2(x − 4) c) -5(-8 − t) e) -2(x − y+3) -2x+2y − 6 -2 x + 8 40 + 5 t b) -5(a − 2) Page 3 of 9

d) -10(-3 − p)

f) -4(a − b + 3)

Algebra2go®

The expression -(x − y+z) implies that a "- 1 " is being multiplied to the parenthesis. So writing down -(x − y+z) is the same as writing down -1(x − y+z) . The process of distributing a negative is shown below.

-(x − y+z)

-x + y − z

In cases where we have an addition or subtraction symbol in front of a parenthesis, we must develop techniques to remove the parenthesis.

When there is an addition operation in front of a parenthesis, we can simply remove the parenthesis.

5 + (x + y − z) = 5 + x + y − z

But if the first term in the parenthesis is negative, we must subtract its opposite! Once again, remember that adding a negative number results in subtracting its opposite!

5 + (-x + y − z) = 5 − x + y − z

Page 4 of 9

Algebra2go®

What happens when there is a subtraction operation in front of a parenthesis? Consider the following expression.

5 −(x − y+z)

In this case we can treat the subtraction symbol as an addition of a "- 1 " and write the equivalent expression 5 + (-1)(x − y+z).

This approach is demonstrated below.

5 −(x − y+z) Rewrite the subtraction 5 + (-1)(x − y+z) symbol as adding -1.

Distribute the -1 into 5 +( - x + y − z) the parenthesis.

5 − x + y − z Remove the parenthesis.

When there is a quantity following the subtraction symbol, we use a similar approach.

Note that you cannot do 10 − 8 because the 8 is being multiplied to the parenthesis. You must perform multiplication before subtraction!

10 − 8(x − y+z) Rewrite subtract 8 10 + (-8)( x − y+z) as adding -8.

Now distribute the -8 10 + (-8x + 8y − 8z) into the parenthesis.

10 − 8x + 8y − 8z Remove the parenthesis.

Page 5 of 9

Algebra2go®

Once you have practiced enough you will be able to correctly remove parenthesis without writing down all the steps.

Example 2: Simplify the expression by removing the parenthesis and combining like terms. a) 3x+y − (x + 2y) b) a+2b + (-a + b)

In some cases we have remove multiple sets of parenthesis before we can combine like terms. See the example below.

-3(2 x+y) − 4(-3x − 2y)

-3(2 x+y) − 4(-3x − 2y) -6x − 3y + 12 x + 8y -6x − 3y + 12 x + 8y 6x + 5y

Page 6 of 9

Here we use the Distributive Property to remove the parenthesis.

Here we identify and combine like terms.

Algebra2go®

Example 3: Simplify the expression. a) 3(a − 2) e) 6(2 x + 1) − 2

b) -(a − 2)

c)

3 −(a − 2)

d) 3 − 6(a − 2b)

Page 7 of 9

f)

-2(3y − 3)+4y

g) 2(x + 1)+4(x − 1)

h) -3(2x + 5 ) − 4(x − 2)

Algebra2go®

Objective 2

Find the Value of an Expression given the value of the Variable Term or Terms We use variables to represent unknown quantities. In the expression x+2, the symbol x is the variable term. We cannot solve for x, as x+2 is not an equation, it is an expression.

Equations have equal signs and expression do not!

We could find the value of the expression if we are given a number to represent the variable. In this case, we say we are evaluating the expression.

Example 4: Evaluate the following expressions given x=12. a) 3x − 8 b) -x ÷ 4 c) -x 2 − 44

3(12) − 8 36 − 8 28

Page 8 of 9

Algebra2go®

Some expressions can have more than one variable. In these cases, you must be given the value of both variables to “find the value” or “evaluate” the expression.

Example 5: Evaluate the following expressions given that x=3 and y=-2.

2 x a) 3x + 2y b) x 2 − y2 c) 2

y

Page 9 of 9

Algebra2go®

Solving Equations Objective 1

Use Properties of Equality to Solve Equations Recall that an expression, like x+3, can only be evaluated if given the value of the variable x. Expressions cannot be solved!

For example, if x=5 then the value of x+3 is 8.

But what about x+3=8? This is an equation because it contains an equal sign. In the case, we will be asked to solve the equation for the unknown value of x. By inspection the solution to the equation x+3=8 is x=5. This is because 5+3=8.

Suppose we are asked to solve equation was x − 5=2? By inspection the solution to the equation is x=7. This is because 7 − 5=2.

So how do we solve equations algebraically?

We can use “Properties of Equality” which state we can add, subtract, multiply, and divide a number to both sides of an equation without changing the solution.

Page 1 of 8

Algebra2go®

The “Properties of Equality” will be demonstrated in the following four examples.

When solving an equation, our goal is to get the variable isolated on one side of the equation.

Example 1: Solve the equation x − 5 = 2 for x. Use the Addition Property of Equality.

In this example we must add 5 to both sides of the equation to isolate the variable.

Vertical Method

x−5 = 2

x−5 = 2 +5 +5 x +0 = 7 x= 7

Horizontal Method

x−5+5 = 2+5 x−5+5 = 2+5 x +0 = 7 x= 7

Either method can be used to solve the equation for x. Remember to circle or box your final answer. Verify that your solution is correct by going back to the original equation and replacing the variable with your solution.

Check that both sides of the equation are in fact equal.

Page 2 of 8

Algebra2go®

Example 2: Solve the equation x + 8 = 6 for x.

Use the Subtraction Property of Equality.

In this example we must subtract 8 to both sides of the equation to isolate the variable.

Vertical Method

x+8 = 6

x+ 8 = 6 −8 −8 x +0 = -2 x = -2

Horizontal Method

x+8 −8 = 6−8 x+8 −8 = 6−8 x + 0 = -2 x = -2 1

Example 3: Solve the equation 3 x = 5 for x.

Use the Multiplication Property of Equality.

In this example we must multiply 3 to both sides of the equation to isolate the variable.

1 = 3x

Page 3 of 8

5 both sides by 3( 31 )( x) = 3(5) Multiply 3 to clear the fraction.

1 the fraction on the left 3( 31 )( x) = 3(5) Clear side and multiply on the right.

1 1( x) = 15 Multiply 1(x) to get x. x = 15

Algebra2go速

1

The equation 3 x = 5 is equivalent to the x= equation 3 5 and can be solved using the same technique. x = 3 3 ( 3x ) = 1 3( 3x ) = 1

5 both sides by 3(5) Multiply 3 to clear the fraction.

Clear the fraction on the left 3(5) side and multiply on the right.

x = 15 Example 4: Solve the equation 12 x = 5 for x.

Use the Division Property of Equality.

In this example we must divide 12 to both sides of the equation to isolate the variable.

12 x = 5 12 x = 5

12 1 12 x = 12 1 1x = Page 4 of 8

12 5 12 5 12 x= 5 12

Divide both sides by 12 to get a 1 in front of the x.

Reduce the fraction to 1x.

Write 1x as x.

Algebra2go®

Consider the equation below. x+2 = 5 4 3 6 Notice that we have fractions on both sides of the equation. To clear the fractions or “Kung Fu” them, we can multiply both sides of the equation by the LCD of all three fractions!

Before we move forward, let’s review how to clear fraction.

12 ( 4x )

12 ( 23 )

12 ( 56 )

12 (

)

12 ( 56 )

3( x )

4( 2 )

3x

8

2(5 ) 10

12 ( 3

x 4 1

)

4

2 3 1

2

1

These calculations will be used in the following example.

Page 5 of 8

Algebra2go®

Example 5: Solve the equation. x+2 = 5 4 3 6

LCD=12

12 ( 4x + 23 ) = 12 ( 56 )

12 ( 4x ) + 12 ( 23 ) = 12 ( 56 ) 3x + 8

Apply the distributive property to the left side of the equation.

= 10

3x + 8 = 10 −8 −8 3x + 0 =2 3x = 2

3x = 2 3 3 x= 2 3 Page 6 of 8

Multiply both sides by 12.

Clear the fractions.

Subtract 8 from both sides of the equation to isolate the variable term on the left side of the equation.

Divide both sides of the equation by 3 to isolate the variable on the left side.

Algebra2go®

Sometimes we need to clear parenthesis and combine like terms before we can use the properties of equalities. The example below demonstrates how to deal with these types of equations.

Example 6: Solve the equation.

-(x + 4) + 4(2 x − 3) = 12 -(x + 4) + 4(2 x − 3) = 12 12 -x − 4 + 8x − 12 = -x − 4 + 8x − 12 = 12 7 x − 16 + 16 7x − 0 7x 7 x

Page 7 of 8

Apply the distributive property to the clear the fractions.

12 = + 16 = 12 = 12 7 = 12 7

Identify and combine like terms.

Add 16 to both sides of the equation to isolate the variable term on the left side of the equation.

Divide both sides of the equation by 7 to isolate the variable on the left side.

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 15, solve each equation for the unknown.

1) x + 4 = 12

6) -7w + 4 + 8w = 9 − 12

2) p − 8 = 13

7)

3) 3t − 5 = 7

8)

4) 2m + 4 = -16

9) 5b − 10 + 3 ( -2 − b ) = 12

5) 3 + 4r = 12 11) 12) 13) 14) 15)

Page 8 of 8

-6 − 4s − 11 + 8s =6 − 18 8 − ( y + 4) − 8 + 2y = -18

10) - ( a + 4) − 2 ( -2a + 4) = 12

1 x= 2 4 3 k+ 1 = 3 2 4 8 3 - 5 C = 15 3= t 5 −1 4 2 6 1 d − 2 d+ 1 = −2 2 4 3 5 3

Algebra2go®

More Equations Objective 1

Solve Equations with Variable Terms on both Sides of the Equation Some equations like 5x − 3 = 4 +8x have variable terms on both sides of the equations.

As always, we must isolate the variable term to one side. It does not matter which side. Below the equation is solved in two different ways.

5 x − 3 = 4 + 8x

5 x − 3 = 4 + 8x −8x − 8x -3x − 3 =4 +3 +3 -3x = 7 -3x = 7 -3 -3 7 x =-

3

5 x − 3 = 4 + 8x −5 x −5 x -3= 4 + 3x −4 −4 -7 = 3x -7 = 3x 3 3 7 - =x

3

-or-

x =Page 1 of 4

7

3

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 10, solve each equation for the unknown.

6) -7h + 4 + 8h = 8 − 6h

1) 6= x − 13

-= 4 3r + 8

7)

p 2 p + 15 3) =

8)

2) 4)

-2k + 3= -15 − 4k

5) x + 2 = 8 − 5 x

Objective 2

9) 10)

-2 ( -b − 5 ) = 12 + 3 ( -2 − b ) 5 ( 4 + 2e ) = 12e − 2 ( -2e + 4)

Solve Equations using a Reciprocal 5 20 Some equations like 6 x = 3 can be solved very quickly by multiplying both sides of the equation by the reciprocal of the fraction in front of the variable. Note that multiplying reciprocals together will always result in 1.

5 6

x = 20 3

6 20 x = ) 5( 3 )

( 6 5 x = 6 20 5(6) 5( 3 )

6 5 5 6

Page 2 of 4

-w − 11 =6 + w -1 + y = 7 − 2 ( y + 4 )

1

1

2

4

1

1

1

1

x =8

Multiply both sides by the reciprocal of

5 which is 6 .

5 6

Clear the fraction on the left side and reduce the fraction on the right side.

Algebra2go®

Using a reciprocal will not work efficiently 5 1 x 20 when given an equation like 6= 3 − 4 . You would have to distribute on the right side of the equation. You will likely end up with a fraction on the right side of the equation.

5= 20 − 1 6x 3 4

( ) ( ) 6= 5 x 6 20 − 6 1 ( 5 6) 5 ( 3 ) 5 ( 4)

6 = 5 x 6 20 − 1 5 6 5 3 4

x =

8

3 − 10

3 = x 80 − 10 10

x=

77 10 or

LCD=10

7 7 10

1 − x 20 Try working the problem 56= 3 4 using the clearing fraction technique with an LCD of 12.

Using “Kung FU math” you will immediately get 10x = 80 − 3 . We can quickly solve this to get the solution!

Page 3 of 4

Algebra2go速

Answer the following homework questions.

In Exercises 11 - 20, solve each equation for the unknown. Begin each problem by multiplying both sides of the equation with the reciprocal of the quantity in front of the variable.

12 5 15 16) 2m = 5 11) 8 p = 4 3 t= 9 =-8 -6 w 17) 12) 20 3 10 3 y= 63 18) 21b = - 2 11 13) 32 4 19) 8a = - 5 14) 5 x = 7 8 3 15) - 5 r = 10 20) - 3 e = -8

6

Page 4 of 4

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go速

1.1 Digits and Place Value and Introduction to Decimals 1. Understand Digits and Place Value with Decimals

Digits are mathematical symbols that are arranged in a specific order to represent numeric values. There are ten different digits in our number system. They are listed below.

0 1 2 3 4 5 6 7 8 9 We use these ten digits (or ten symbols) to create numbers by placing them in a specific order.

It is the position of each digit within a number that determines its place value. One digit alone can also represent a number. A single digit that represents a number is said to be in the ones place value position.

To assist us in determining place value, we use commas to separate periods of a number, and also use a decimal point to define the location of the ones place. The ones place is just to the left of the decimal point. When writing down whole numbers we normally do not write down the decimal point. In this case it is understood that the digit furthest to the right, or rightmost place, is in the ones place.

We will now look at a whole number with four full periods. The name of each period as well as the place value of each digit is labeled. Can you see a pattern in the diagram below?

4 9 7, 5 4 8, 6 0 1, 3 7 2. billions period

millions period

thousands period

ones period

Next we have a number that has digits to the right of the decimal point. Be sure to again look for a pattern by imagining the ones place as the middle of the number.

7 3 6, 4 8 5, 0 3 7. 9 2 0 8 1 9 5 6 Can you see the pattern that is mirrored about the ones place? Once we learn how to identify the place value of digits, we then can learn how to read and write numbers properly. www.Algebra2go.com

Page 1 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

Example 1: Write down the place value of the digit 4 in the following numbers. Use the place value diagrams on the previous page to help find the answer. a) 114,235

The four is in the one-thousands place.

b) 2,297,465

The four is in the hundreds place.

c) 0.0004

The four is in the ten-thousandths place.

d) 10.259843

The four is in the hundred-thousandths place.

e) 0.1030804

The four is in the ten-millionths place.

f) 4,250,006,258

The four is in the one-billions place.

2. Review How to Read and Write Whole Numbers Knowing place values as well as knowing how the periods of a number are ordered, enables us to read and write whole numbers correctly.

When writing whole numbers using words, we always include the period(s) in our word statement with exception of the ones period. The diagram below will help us write the number 2,015,325 using words. Pay close attention to the numbers within each period and how the commas are used in the word statement below.

2, 0 1 5, 3 2 5. millions period

thousands period

ones period

Two million , fifteen thousand , three hundred twenty-five .

In the sentence above, notice how the commas break up the sentence to define the periods.

Note that the ones period is excluded. Also, notice that we do not use the word “and” when writing down whole numbers using words. The word “and” is used to connect the decimal (or fractional) parts to the whole number. This will be addressed later in this section.

Now let’s write the number 11,982,050,307 using words.

1 1, 9 8 2, 0 5 0, 3 0 7. billions period

millions period

thousands period

ones period

Eleven billion, nine hundred eighty-two million, fifty thousand, three hundred seven.

Once again, notice how the commas break up the sentence to define the periods. Also, notice that the ones period is again excluded from the word statement. www.Algebra2go.com

Page 2 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

3. Understand How to Read and Write Decimal Numbers Less Than 1.

Next we will learn how to correctly read and write decimal numbers less than 1. Let’s begin with 0.053 which represents a number less than 1.

To write a decimal number less than 1 using words, we first need to define the place value of the digit furthest to the right. In the number 0.053, the digit 3 is in the rightmost place. Using our place value pattern, we can see that the digit 3 is in the one-thousandths place.

0.0 5 3 Next, we write down the number to the right of the decimal point. In this case we have the number 53. Because the 53 terminates in the one-thousandths place, it means we have “fifty-three one-thousandths”. So we write this number using words as follows.

Fifty-three one-thousandths.

53 1,000 and both are written using words as “fifty-three one-thousandths”. Notice that the numerator 53 is represented by the numeric value to the right of the decimal point. of the fraction 1,000

Recall that a decimal number represents a fraction. Therefore we can express 0.053 as

Note: In many cases it is acceptable to write down “fifty-three thousandths” rather than “fifty-three one-thousandths”. Check with your instructor to see if this is acceptable.

Now let’s try the number 0.01089 which is again a number less than 1.

0.0 1 0 8 9 In this case, the number to the right of the decimal point is 1089 and it terminates in the hundred-thousandths place. Notice that the digit 9 is in the rightmost place. This means we have “one thousand eighty-nine hundred-thousandths”.

1 ,089 Therefore we can express 0.01089 as and write the number using words as follows.

100,000

One thousand eighty-nine hundred-thousandths. www.Algebra2go.com

Page 3 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

Example 2: Write each of the following numbers using words. Also represent each decimal as a fraction for parts e) – h). a) 52,003

e) 0.9

b) 907,000

f) 0.085

c) 84,000,250

g) 0.0030

d) 108,581,609,004

h) 0.00000406

Notice in parts a) – d), the numbers given are whole numbers.

Remember, when writing whole numbers using words, we always include the period(s) in our word statement with exception of the ones period.

a) 5 2, 0 0 3.

Notice we have 52 in the thousands period, and 3 in the ones period.

Fifty-two thousand, three. b) 907,000.

Notice we have 907 in the thousands period.

Nine hundred seven thousand. c) 84,000,250.

Here we have 84 in the millions period, and 250 in the ones period.

Eighty-four million, two hundred fifty. d) 108,581,609,004.

Here we have 108 in the billions period, 581 in the millions period, 609 in the thousands period, and 4 in the ones period.

One hundred eight billion, five hundred eighty-one million, six hundred nine thousand, four.

Notice in parts e) – h), the numbers are less than 1.

To write a decimal number less than 1 using words, we first write down the numeric value to the right of the decimal point, followed by the place value of the rightmost digit.

Here we have the number 9 to the right of the decimal point and

e) 0.9 it is in tenths place.

Nine tenths.

Here we have the number 85 to the right of the decimal point.

f) 0.085 The 5 is the rightmost digit and it is in the one-thousandths place.

Eighty-five one-thousandths.

Here we have the number 30 to the right of the decimal point.

g) 0.0030 The 0 is the rightmost digit and it is in the ten-thousandths place.

Thirty ten-thousandths.

Here we have the number 406 to the right of the decimal point.

h) 0.00000406 The 6 is the rightmost digit and it is in the hundred-millionths place.

Four hundred six hundred-millionths. www.Algebra2go.com

Page 4 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

Answer the following questions.

1) Write down the place value of the digit 7 in the following numbers.

3) Write each of the following numbers using words.

a) 947,025

e) 0.007000

a) 500,009

b) 306.007

f) 0.065070

b) 0.0018

c) 580.85670

g) 9.871324

c) 456,800

d) 657,289,634 h) 6.0578238

d) 0.00507 e) 13,000,060,105

2) Using the number below, identify the digit in the given place value.

f) 0.08060 4) Write each of the following numbers using digits.

20,546,318.72968467 a) one-millions

a) Seventy-five one-thousandths.

b) ten-millionths

b) One hundred eight million.

c) one-thousandths

c) Sixteen ten-millionths.

d) hundred-thousands

d) Thirty-three thousand.

e) hundreds

e) Four million, six-hundred seventy-five.

f) hundredths g) one-millionths

f) Ninety million, two thousand, one hundred four.

h) ten-thousands

4. Understand How to Read and Write Numbers The number 125.87 has a whole number part and a decimal (or fractional) part. The whole number part represents a quantity that is greater than 1, and the decimal part represents a quantity that is less than 1.

The whole part of the number 125.87 is 125 and is read “one hundred twenty-five”. The decimal part of the number is .87 and is read “eighty-seven hundredths”. The decimal point is used to connect the whole number part to the decimal (or fractional) part by addition. This means that the number 125.87 actually represents a mixed number!

=

125.87 125

+

.87

=

125 + 87 = 100

87 125 100

Recall that the mixed number format represents a sum of a whole number part and a fractional part.

To write the number 125.87 using words, we first write down the whole number part. Next, we use the word “and” to connect the whole number part to the decimal (or fractional) part.

125.87 One hundred twenty-five and eighty-seven hundredths. www.Algebra2go.com

Page 5 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

Suppose we are given the number 1,002.0050 which again has both a whole number part and a decimal (or fractional) part. The whole number part is 1,002 and is written “one thousand, two”. The decimal part of the number is .0050 and is written “fifty ten-thousandths”.

As before, to write the number 1,002.0050 using words, we first write down the whole number part. Next, we use the word “and” to connect the decimal (or fractional) part.

1, 002.0050 One thousand, two and fifty ten-thousandths.

When we need to write out a check, we must always indicate the dollar amount in two forms.

First we write the number using digits, and second we write the number using words.

Example 3: In the appropriate space, write in the dollar amount of the check using words.

The dollar amount of the check is 1,834.18 which has both a whole number part and a decimal (or fractional) part. To fill in the indicated dollar amount using words, we write the following words on the dollar amount line in the check above.

1,834.18 One thousand, eight hundred thirty-four and eighteen hundredths

It is also acceptable to write the decimal part as a fraction.

1,834.18 One thousand, eight hundred thirty-four and

www.Algebra2go.com

18 100 Page 6 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

Now we will look at how to write a number using digits given a word statement. We will begin with a whole number. The word statement we will work with is written below.

Fifty billion, three thousand, twenty-one.

Notice that a millions period is not present in the word statement above. When writing the number using digits, the millions period must be included. To represent the millions period in this case, we place three 0’s within this period. The result is represented in the diagram below.

Fifty billion, three thousand, twenty-one.

5 0, 0 0 0, 0 0 3, 0 2 1. billions period

millions period

thousands period

ones period

Observe the three 0’s in the millions period. These zeros are required in order to represent the number “fifty billion, three thousand, twenty one” correctly.

Additionally, notice that there are always three digits between any two commas.

In the case of the ones period, always remember that it must contain three digits before you begin entering digits in the thousands period.

Next we will deal with a number that contains both a whole number part and a decimal part.

Suppose we are asked to write “three hundred two thousand, twenty and two hundred one tenthousandths” using digits. The diagram below represents the result.

Three hundred two thousand, twenty and two hundred one ten-thousandths.

3 0 2, 0 2 0.0 2 0 1 thousands period

ones period

Notice that the digits 3, 0, and 2, are in the thousands period. This represents “three hundred two thousand”. In the ones period are the digits 0, 2, and 0, which represent twenty.

To the right of the decimal point are the digits 0, 2, 0, and 1. Because the digit 1 is the rightmost digit and is located in the ten-thousandths place, the decimal part .0201 represents “two-hundred one ten-thousandths”. We can also say that 201 terminates in the ten-thousandths place.

www.Algebra2go.com

Page 7 of 8

Algebra2go: Review of Place Value and Introduction to Decimals

Algebra2go®

Example 4: Write each of the following numbers using digits. a) Three hundred one.

301

Here we have 301 in the ones period.

b) One thousand and fifty-four hundredths.

1,000.54

Here we have 1 in the thousands period, three 0’s in the ones period, and to the right of the decimal point, 54 terminates in the hundredths place.

c) Two thousand, thirteen and eighty-seven one-thousandths.

2,013.087

Here we have 2 in the thousands period, 13 in the ones period, and to the right of the decimal point, 87 terminates in the one-thousandths place.

d) Six hundred ninety-three billion, nine thousand and six one-millionths.

693,000,009,000.000006

Here we have 693 in the billions period, three 0’s in the millions period, 9 in the thousands period, three 0’s in the ones period, and to the right of the decimal point, 6 is in the one-millionths place.

For Exercises 5 – 10, write each of the numbers using words.

For Exercises 11 – 16, write the number using digits.

5) 687.05

11) Three and five hundredths.

6) 1, 000.001

12) Sixteen ten-thousandths.

7) 32,870, 051.369

13) Four million and one onethousandths.

8) 50, 000, 090.0030 9) 304, 000, 000, 000 10) 0.000050801

14) Two hundred-thousandths.

15) Nine thousand and nine hundred hundred-thousandths.

16) Thirty-two thousand, eight hundred one-millionths.

Review Exercises Evaluate the expression.

17) 9 − 5 + 4 18) 4 − 8 + 2 19) 3 − −2 − 4 20) 7 − 11 − 22

For Exercises 25 – 28, find the value of each expression if x = 3 and y = −2 .

25) 3 − x − y 26) x 2 + y 2

21) − 32 22) ( − 3)

2

23) − ( − 5 )

2

24) − − 6

2

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27)

3 + y3 5 − x2

28)

x 2y − 2y x

Page 8 of 8

Algebra2go速

Objective 1

Operations with Decimals

Perform Addition and Subtraction with Decimals

The technique for performing addition and subtraction with decimals requires that we arrange our numbers in columns of common place value. Because decimal values have a labeled decimal point, we can use it as a guide.

Example 1: Calculate the sum of 82.3, 0.54, and 32.

We will use the vertical format to get the result. Be sure to line up the numbers in columns according to place value.

8 2..3 0 0.. 5 4 + 3 2.0 . 0 1 1 4. 8 4

Page 1 of 6

Note: In this problem we must line up the numbers by place value. The decimal point is used as a guide. Notice that we wrote 82.3 as 82.30 and 32 as 32.00. This was to match the place value of 0.54.

Note: When subtracting two decimal numbers, we also line up the numbers by place value.

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 15, add or subtract as indicated. Be sure to follow the rules for Order of Operations.

1) 0.2+0.3

6) 0.2 − (1.5 − 3)

11) -(1 − 2.7) − (1+5.4)

2) 0.0008+0.7

7) -1.2 +(2 − 0.5)

12) 1.4 − 2.04 − 5

3) 1.8 − 0.5

8) 1.05 − (-3.2)

13) -0.008 − 5.4+10

4) 2.1 − 0.0004

9) 2.17 +(-5 − 1.4)

14) -1.009 − 0.2

5) 0.5 − 0.019

Objective 2

10) -(5 − 2.9)+2.7

15) 2 − (4.1 − 60.8)

Perform Multiplication and Division with Decimals

Suppose we want to calculate the product of 0.7 and 0.4. It may be easier to perform this calculation by first converting the decimals to fractions.

7 4 28 0.28 = = ⋅ 10 10 100 41 241 8 18 = 1 = ⋅ ⋅ 100 10 100 10 4338 = 4.338 1000

0.7 ⋅ 0.4=

2.41 ⋅ 1.8=2 Page 2 of 6

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To calculate the product of 2.41 and 1.8 using the vertical format, we do not have to line up the decimals! We line up the numbers on the right. The calculation is shown below.

2.4 1 Now add the 1.8 × results.

First multiply 241 by 8.

3

241 8 × 19 2 8

Next, multiply 241 by 10.

24 1 × 10 24 1 0

2 41 × 18 1928 + 2410 43 3 8

At this point, our result is 4338. But this is not the final answer. We need to know where to place the decimal point. To figure out where, we count the number of decimal places in the two numbers we multiplied together.

We count from the right as shown below.

2.4 1 1.8 We count a total of three place values. ×

Page 3 of 6

Finally, we place the decimal three places from the right in our result. 4338 4.3 3 8

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Next, we will perform long division with 1.32 decimal numbers. Consider 1.36 á 0.4 or 0.6 .

0.4 1 .36

Our divisor is 0.4 and our dividend is 1.36.

When performing long division, we want our dividend to be a whole number.

If we look at our problem in fractional form, we can see that multiplying both numerator and denominator by 10 will make our dividend a whole number. Note that multiplying a number by 10 moves the decimal point one place value to the right.

1.36 10 = 13.6 0.4 ( 10 ) 4 This process is replicated in long division notation as follows.

0.4 1 .36 = 4 1 3.6

Now we place our decimal above the long division symbol and perform long division.

.

4 1 3.6 Page 4 of 6

Algebra2go®

After making the dividend a whole number, we perform long division just as if we are using whole numbers. Note the we do not move the decimal when we get our final answer.

.

4 1 3.6

How many times does 4 go into 1?

. .

0 4 13 6

Zero

How many times does 4 go into 13?

Subtract.

. .

03 4 13 6 12 16

3 × 4 goes here.

Bring down the 6.

How many times does 4 go into 16?

. .

Subtract.

03 4 4 13 6 12 16 16 0

Three

Four

4 × 4 goes here.

There are 0 units left over.

Finally, 1.36 Page 5 of 6

÷

0.4 equals 3.4.

Algebra2go®

Answer the following homework questions.

16) Find the product of 3.5 and 0.4.

17) Find the product of 2.09 and 8.1.

18) Find the quotient of 15.2 and 0.8.

19) Find the quotient of 10.5 and 0.05 In Exercises 20 – 25, multiply and divide as indicated.

Page 6 of 6

20) 3.76 ( 0.4)

22) 1.44 ÷ 1.2

24) 13.87 ÷ 7.3

21) 0.25 ⋅ 0.2

23) 13.2 ÷ 0.11

25) 29.25 ÷ 4.5

Algebra2go速

Objective 1

More on Decimals

Learn how to Round Decimal Numbers

The technique for rounding decimal numbers is slightly different than that for rounding whole numbers.

Sometimes we need to add zeros to the decimal to represent place values. Such is the case when we need to round the whole number 3 to the nearest hundredth. In this case, we would have to write in the decimal point to the right of the whole number and add two zeros.

8 = 8.00

The rightmost 0 is in the hundredths place.

The two quantities above are equivalent, but the quantity on the right of the equal sign is expressed as a quantity rounded to the nearest hundredth.

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Algebra2go速

Rounding Decimal Numbers First locate the digit that is one place to the right of the place value you want to round to.* If that digit is less than 5, remove it as well as all other digits to the right.

If the digit is greater than or equal to 5, add 1 to the digit to its left and remove it as well as all other digits to the right.

If the number has no digit in the place value you want to round to, add zeros to the right of your number up to and including the place value you want to round to. *

Example 1: Round the following numbers to the nearest one-thousandth.

The 2 is in the a) 0.532439 one-thousandths The digit to the place.

right is less than 5.

0.532439

Remains unchanged.

0.532 Page 2 of 4

Remove these numbers.

Algebra2go®

b) 50.6

There is no digit in the one-thousandths place.

Include these digits.

50.6_ _ 50.600

c) 5.21684

The 6 is in the one-thousandths place.

Add two zeros to include the one-thousandths place.

The digit to the right is greater than or equal to 5.

5.21684 Add 1.

5.217

Remove these two digits from the rounded number.

Answer the following homework questions.

In Exercises 1 – 6, round each number to the nearest one-thousandth.

1) 1.2597

3) 0.0055468

5) 123

2) 15.1

4) 0.9999999

6) 0.3333333

In Exercises 6 – 11, round each number to the nearest tenth.

Page 3 of 4

6) 52

8) 18,576.3265

10) 9.99

7) 6.785

9) 0.00005

11) 0.5000

Algebra2goÂŽ

Objective 2

Write Fractions as Decimal Equivalents 1

The fraction 2 is equivalent to the quotient 1 á 2 . Performing this division problem by hand or with a calculator will result with 0.5 or 5 5 1 10 . Notice that 10= 2= 0.5 (five-tenths).

Example 2: Write each fraction as a decimal rounded to the nearest hundredth.

3 = 35 d) 2 16 a) 12 c) 58 13 b) 43 16 0.923 Answer the following homework questions.

In Exercises 12 - 19, convert each fraction to a decimal rounded to the nearest ten-thousandth.

3 2 12) 5 14) 3 5 5 13) 6 15) 16

Page 4 of 4

1 16) 8 2 17) 8

3 18) 8 4 19) 8

Algebra2go®

Circles, Cylinders, and Spheres Objective 1

Find the Circumference and Area of a Circle In math, there are numbers that come up so often, we give them their own symbol. One such special number is pi.

The symbol for pi is π .

π ≈ 3.14

The number π shows up as the ratio of a circle’s circumference to its diameter. The circumference is the distance around the circle and the diameter is the distance across the circle through the center. Another dimension we will often mention is the radius of a circle. The radius is half the length of the diameter. This means the diameter is twice the length of the radius.

Circumference

Page 1 of 5

Algebra2go®

The formula for the circumference of a circle is C = 2π r .

The formula for the area of a circle is A = πr 2 .

Example 1: Find the circumference and the area of the circle.

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Algebra2goÂŽ

Objective 2

Find the Volume of a Sphere A sphere is a 3-dimensional object like a soccer ball. The radius of a sphere is the distance from the center to the outer edge.

The formula for the volume of a sphere is 3 V= 4 Ď&#x20AC; r .

3 Example 2: Find the volume of the sphere.

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Algebra2goÂŽ

Example 3: Find the volume of the hemisphere.

Objective 3 Find the Volume of a Right Circular Cylinder

A right circular cylinder is a threedimensional object where the two ends are circles. A soup can is a right circular cylinder with a height of h.

The formula for the volume of a cylinder is V = Ď&#x20AC; r 2h .

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Algebra2go速

Example 4: Find the volume of the cylinder.

Page 5 of 5

Algebra2go®

Solving Equations with Decimals Objective 1

Solving Equations by Clearing Decimals Suppose we are asked to solve the equation 0.10x + 0.05x = 2.1 .

We could rewrite the decimal values as fractions and then clear them using the LCD. This approach is demonstrated below.

0.1x + 0.05x = 2.1 1 x + 5 x = 2.1 LCD=100 10 100 1 x + ( 100 ) 5 x = ( 100 ) 2.1 ( 100 ) 10 100 10x

+

5x = 210 15 x = 210 x = 14

If we think of the decimals as fractions with denominators of powers of 10, we can then simply clear or “Kung FU” the decimals just as we would fractions. Compare the technique below with the solution above.

0.1x + 0.05x = 2.1 LCD=100 100 ( 0.1 ) x + 100 ( 0.05 ) x = 100 ( 2.1 )

Page 1 of 4

10x + 5 x = 210 15 x = 210 x = 14

Algebra2go®

Example 1: Solve the equation.

0.25 t − 0.88 = 0.03t LCD=100 100 ( 0.25 ) t − 100 ( 0.88 ) = 100 ( 0.03 ) t

Example 2: Solve the equation.

LCD=10 5 t + 0.6 = t + 1 10 ( 5 t ) − 10 ( 0.6 ) = 10 ( t ) + 10 ( 1 )

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Example 3: Solve the equation.

LCD=100 0.02 + 0.5 a = -0.3 100 ( 0.02 ) +100 ( 0.5 a) = 100 ( 0.03 )

Example 4: Solve the equation.

LCD=10 0.5x + 0.1( x + 30 ) = 4.8 10 ( 0.5 ) x + 10 ( 0.1 )( x + 30 ) = 10 ( 4.8 )

Page 3 of 4

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 10, solve each equation for the unknown.

9.3 1) 0.2x + 5.7 =

6) 1.4x + 0.73 = 1.8x + 1.61

2.6 2) 0.5m − 4.9 =

7) 1.3c + 0.67 = 1.37c + 2.31

0.2 3) 0.5 − 0.4p =

8) 0.1 ( y + 4) − 2 + 2.4y = -5

0.15 4) 0.5 − 0.2k =

9) 0.5b − 1 + 0.3 ( -2 − b ) = 1.2

6.3a+5.4 5) 5.7a + 1.2 =

Page 4 of 4

10) -0.01 ( a + 4) − 0.02 ( -2a + 4) = 0.12

Algebra2go®

Square Roots Objective 1

Understand the meaning of a Square Root What does it mean to square a number?

If we square 8, we have 8 ⋅ 8 = 8 2 = 64.

If we square root 64, we get 8. The mathematical symbol for square root is We also call it a radical.

Note: It is note the long division symbol

. .

The mathematically statement 64 is asking us “what is the square root of 64”. In other words, “what positive number do you square to get 64”.

There are actually two integers you can square to get 64. These are -8 and 8. But the square root function only gives the “principal root”. Which means the square root of a number is always positive.

Finally, we can make the statement 64 = 8 .

Page 1 of 6

Algebra2go速

The square of an integer is known as a perfect square. There are several perfect squares you should already know. These are 144, 121, 100, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0.

Notice the following.

144 = 12 121 = 11 100 = 10 81 = 9 64 = 8 49 = 7 36 = 6

25 16 9 4 1 0

=5 =4 =3 =2 =1 =0

Most square roots we need a calculator to calculate like for 2 . We can only approximate it answer. For example, 2 = 1.414 rounded to the nearest one-thousandth. We will only be working with square roots of perfect squares in this section.

Page 2 of 6

Algebra2go®

Objective 2

Evaluate Expressions with Square Roots Step 1 of the rules for order of operations states to “Perform all the operations within a parenthesis or other grouping symbols”. The square root symbol, or radical, is considered a grouping symbol. Therefore, if there is an expression under a square root, you must first simplify the expression beneath the radical symbol before taking the square root.

Example 1: Evaluate each expression following the rules for Order of Operations. a) 20 − 42 c) 3 16 e) -2 36 −( 25 − 3) 20 − 16 4

2

b) 6 + 13 2

Page 3 of 6

3 ⋅ 16

3⋅4

12

d) 5 25

2

f) ( 1 − 121 ) − 2 81

Algebra2go®

Answer the following homework questions.

In Exercises 1 - 12, evaluate each expression following the rules for order of operations.

1)

1+ 0

7)

81 ÷

144 − 2 36

8)

( 36 ÷ 6 )2

49 + -6

9)

-2 2 + 20

81 − 100

10)

2) 3)

-

4)

5) 3 4 6)

Objective 3

4+ 4+ 4

12)

61 − ( 3 − 8 )

2

1 4 64 16

Understand the Pythagorean Theorem for Right Triangles A right triangle is a triangle that has a right  angle of 90 degrees ( 90 ).

These side lengths are called the legs of the triangle.

Page 4 of 6

11)

9

This side length is known as the hypotenuse. It is the side opposite the 90o or right angle. It is also always the longest side length of the triangle.

This squared symbol denotes the location of a right angle within the triangle.

Algebra2goÂŽ

The Pythagorean Theorem states,

a2 +b2 =c 2

where a and b represent the lengths of the legs of the triangle (in no particular order) and c represents the length of the hypotenuse.

The side lengths of any right triangle must satisfy the theorem. If the side lengths do not, then it is not a right triangle!

Example 2: Show that the triangle below is a right triangle using the Pythagorean Theorem.

Let a=5, b=12, and c=13.

12 f t

13 f t

5 ft Page 5 of 6

a2 +b2 =c 2 5 2 +12 2 =132 25+144=169 169 = 169 ď ?

The Pythagorean Theorem is satisfied, therefore this is a right triangle. The right angle is located opposite the hypotenuse which is 13 ft in length.

Algebra2go®

Example 3: Find the length of the hypotenuse in the triangle.

Let a=3 and b=4.

c

a2 +b2 =c 2 32 +42 =c 2 9+16=c 2 25 = c 2

3 yd

4 yd

2 Since = 25 c= or c 2 25 , we as ourselves “what number do we square to get 25?” Since the square root of 25 is 5 ( 25 = 5 ), this means that c must equal 5.

Therefore the length of the hypotenuse in the triangle above must be 5 yards.

Answer the following homework question.

In Exercise 13-14, find the missing side length in each right triangle.

13)

Page 6 of 6

c 8 in.

14)

6 in.

13 cm

b

5 cm

Algebra2go®

Ratios

Objective 1

Understand the meaning of a Ratio What is a ratio? We use ratios to compare two quantities. For example, the ratio of 3 to 4 can be written as 43 or 3:4 .

We often see ratios used in recipes. For example, suppose your recipe requires 4 cups of powdered mix to create a serving for 12 people.

4 In this case you would have a ratio of 12 .

If we reduce this ratio we get 31 which means that our recipe requires 1 cup of powdered mix to create a serving for 1 person.

Note: When we write a ratio, we do not include the units. When we do include the units, we call this a rate. Rates will be covered in the following section.

6 5

12 25

Example 1: Write the ratio of to as a reduced ratio comparing two whole numbers.

Page 1 of 4

6 1 5 5 6 ÷ 12 6 ⋅ 25 6 ⋅ 25 5 = 5 = = = 25 5 12 5 12 2 12 1 2 25 6 6 5 25 Here we are using the 5 5 30 5 = = = 2 clearing fractions technique 12 12 12 using the LCD of 25.

25 2 25 25

( ) ( )

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Example 2: Write each ratio as a reduced ratio comparing two whole numbers. a) 0.4 to 4 b) 4.8 to 0.8 c) 0.12 to 0.4

0.4 4

LCD=10

10 ( 0.4) 10 ( 4)

4 40 1 4 40 10 1 10 Example 3: Write the ratio as a reduced ratio comparing two whole numbers.

8 5 to 0.3

Page 2 of 4

Algebra2goÂŽ

Sometimes a ratio can provide us with useful information in everyday situations and also provide us with some statistical information.

Example 4: Candice drove her hybrid vehicle 480 miles on 10 gallons of gas. What is the ratio of miles to gallons for Candiceâ&#x20AC;&#x2122;s hybrid?

Example 5: At a certain high school there are 425 female students and 375 male students. a) What is the ratio of female students to male students? b) Based on your reduced ratio in part a), theoretically in a classroom of 32 students, how many should be female? c) What is the ratio of female students to the total student population.

Page 3 of 4

Algebra2go速

Answer the following homework questions.

In Exercises 1 - 12, write each ratio as a reduced ratio comparing two whole numbers.

2) 75 to 50

7 18 9 to 21 10 to 15 54 27 6)

10) 0.04 to 12

3) 0.5 : 5

7) 1.2 to 3.4

11)

4) 3.5 : 0.7

8) 0.204 to 0.6

12)

1) 7 to 8

Page 4 of 4

5)

9) 2.1 to 0.03

2 21 3 43

3 to 4 to

2 23

Algebra2go速

Rates and Unit Price

Objective 1

Understand the meaning of Rates and Unit Price Remember that a ratio compares two quantities. For example, the ratio of 3 to 4 can 3 be written as 4.

A rate is compares two quantities and we write in the units. The two quantities have different units.

110 miles 2 hours

76 miles 2 gallons

For example or are considered to be rates since we included the units.

A Unit rate is a rate with a denominator of 1.

110 miles For example, if we reduce the rate 2 hours to 55 miles 1 hour , we call this a unit rate and write 55 miles/hour or 55 mph.

76 miles 2 gallons

Similarly, if we reduce to also call this a unit rate and write 38 miles/gallon or 38 mpg.

Page 1 of 4

38 miles 1 gallon ,

we

Algebra2goÂŽ

Example 1: A car travels at a rate of 200 miles per 4 hours. What is the cars unit rate in miles per hour?

200 miles 50 miles 50 miles/hour = 50 mph = = 4 hours 1 hour

Example 2: Charlieâ&#x20AC;&#x2122;s scooter can travel 270 miles in 6 gallons of gas. What is the scooters unit rate in miles per gallon? miles = 6 gallons

miles = gallon

miles/gallon =

mpg

Example 3: In 3 hours, a family traveled 255 kilometers (km). What is the familyâ&#x20AC;&#x2122;s unit rate in km per hour? km = 3 hours

Page 2 of 4

km = hour

km/h*

* The

km/h abbreviation for km per hour is most commonly used.

Algebra2goÂŽ

Example 4: A local farmerâ&#x20AC;&#x2122;s market sells wheat germ at a rate of $2.72 per 16 ounces. What is the unit price in cents per ounce?

2.72 dollars 272 cents = 16= ounces 16 ounce

cents/oz

Unit price between different brands of product can be used to determine the better buy. Most stores display unit price on the price tags now days. Next time you are at the store, take a look at the unit prices. It may save you some money!

Example 5: A home improvement store sells Brand A light bulbs for $4.36 for 4 bulbs.

Brand B cost $6.48 for 6 bulbs. Which is the better buy?

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Algebra2goÂŽ

Example 6: A department store sells Brand A socks for $6.42 for 6 pairs. Brand B cost $9.72 for 9 pairs. Which is the better buy?

When you receive a paycheck for working part-time, the amount you receive is based on the hours you work and your hourly wage. It is always important that you check to see if you are paid correctly.

Example 7: Candiceâ&#x20AC;&#x2122;s time sheet is shown below.

If she was paid $391.50 for working these hours, what is her hourly wage?

Monday 5.0 hours

Page 4 of 4

Tuesday 4.0 hours

Thursday 4.0 hours

Friday 8.0 hours

Sunday 8.0 hours

Algebra2go速

Proportions

Objective 1

Solve Proportion Problems A proportion is an equation with two ratios! x =2 The equation 4 3 represents a proportion and it can be solved using a technique call crossmultiplying. It can also be solved by first using an LCD to clear the fractions.

Sometimes it may be easier to clear the fractions, but most students tend to like to cross-multiply.

Here we will solve the proportion problem using the cross-multiplication technique.

x =2 4 3

x =2 4 3 3x = 8 x = 83 or 2 2 3 Page 1 of 4

Algebra2go®

Now we will solve the proportion problem using the clearing fraction or “Kung Fu” fraction technique. x =2 LCD=12

12 (

4

)

3

( )

x = 2 12 3 4 3x = 8

x = 83 or 2 2 3 Remember that you can only cross-multiply across an equals (=) sign and you should only do this when you have a proportion. Do not attempt to cross-multiply on the equation below. It is not a proportion. Use the clearing fractions technique in this case.

Page 2 of 4

x=3 + 1 2 4 5 19 x= 10

LCD=20

Algebra2go®

Example 1: Solve the proportion.

1 2

x

4 = 2 3

In this problem we will cross-multiply twice to reach the solution.

1 2

x

4 =

1⋅2 = 2 3 Remember, this is called “cross-cancelling” and can only be performed across a multiplication operation!

1

1⋅2 = 21 3

2 3

Remember, we can only perform “cross-multiplication” across an equals sign!

4⋅ x 4⋅ x 4

1 = x 3 1 = 4x 3 1

Here we rewrite 4x as a fraction by placing it over 1. This way we can again cross-multiply.

1 = 12x 12 12 1 = x or x = 1 12 12 Page 3 of 4

Algebra2go速

Example 2: Solve each proportion problem. x =5 a) 4 8

2 =4 b) 5 x

25 = x c) 100 4

Answer the following homework questions.

In Exercises 1 - 9, solve each proportion problem. Try not to use a calculator.

4 1) x = 29 2) 3x = 10 11

1 3) 0.5 1.2 = x

Page 4 of 4

x = 0.15 4) 1.4 3 x = 0.5 5) 0.3 2.5 x = 0.05 6) 12 0.12

7)

6 5 1 = 2 2 3

x

2

8) 3 = 8 x 1 2

9)

1 1 2 = 3 x 14

Algebra2go®

Applications of Proportions

Objective 1

Set up and Solve Proportion Problems Proportions can be used to solve many different types of problems. Be sure to read the problem carefully and try to estimate what a reasonable answer is.

Remember, a proportion is an equation of two ratios and we should always write in our units when we set up the problem.

Additionally, make sure the units mirror each other on both sides of the equation.

Example 1: Suppose Tache’s car can travel 72 miles on 3 gallons of gas. How many miles can Tache’s car travel on 12 gallons?

We first begin by setting up the proportion.

Notice how the units mirror each other on both sides of the equation.

miles = gallons

Page 1 of 7

miles gallons

Algebra2goÂŽ

Since we are being asked to find out how many miles Tacheâ&#x20AC;&#x2122;s car can travel, we let our variable x represent these unknown miles.

Because these unknown miles correspond with 12 gallons, we can set up our first ratio on the left side of the proportion.

x miles = 12 gallons

miles gallons

On the right hand side of the equation we will write in our given ratio. Notice that the problem tells us that the car can travel 72 miles on 3 gallons of gas. This is our given ratio. Writing these quantities on the right side of the equation completes the setup of our proportion.

x miles = 72 miles 12 gallons 3 gallons

We now solve the proportion for x.

x = 72 12 3

Page 2 of 7

Algebra2goÂŽ

We begin by first cross-multiplying.

x = 72 3 12

x = 72 3 12 3x = 864 3x = 864 3 3 x = 288 Recall that we let represent the unknown quantity in units of miles.

We finally answer the question with a complete sentence!

Answer: Tacheâ&#x20AC;&#x2122;s car can travel 288 miles on 12 gallons of gas.

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Algebra2go速

Example 2: A recipe calls for 5 cups of sugar for 30 servings. How many cups of sugar are needed to make 8 servings?

x cups = 30 servings

Page 4 of 7

cups servings

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Example 3: On a travel map, 1 inch represents 75 kilometers. If the measured distance between two cities is 4.5 inches on the map, what is the actual distance between the two cities in kilometers?

4.5 inches = 1 inches x kilometers 75 kilometers

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Algebra2goÂŽ

Example 4: At 2 PM. Mariaâ&#x20AC;&#x2122;s shadow is 8 ft long and she is 5 ft tall. If at this same time, the flag pole casts a shadow that is 8.4 feet long, how tall is the flag pole?

Page 6 of 7

Algebra2go速

Example 5: The standard wide screen wide ratio is 16:9, width to height. A new movie theatre is being constructed with 4 different screen sizes.

Fill in the missing dimensions in the table below. Use proportions to find the missing lengths.

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Algebra2go速

Percent

Objective 1

Understand the Meaning of Percent A percent represents a part per hundred. For example, 3% represents 3 per hundred.

Percentages can also be represented as an equivalent decimal or fraction. For example, 45 45% is equivalent to 0.45 or 100 .

Objective 2 Learn how

to Rewrite a Percentage as Decimal or Fractional Equivalents To rewrite a percent as a decimal equivalent simply move the decimal point two places to the left.

Note: When writing whole percents, there is an implied decimal point that is generally not written in. For example, 45% has an implied decimal point just to the right of the 5. If we wanted to write in the decimal point, we would write 45.% which is the same as writing 45%.

To write a percent as a fraction, we first write the percent as a decimal and then write the fractional equivalent.

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Algebra2go速

Example 1: Write each percent as a decimal and a fraction. Reduce the fraction if possible.

1 4 25 a) 4% = 0.04 = 100 = b) 17% c) 35% d) 100% e) 125% f) 0.4% g) 0.03% h) 1.25% Objective 3 Learn how

to Rewrite a Decimal as a Percent To write a decimal as a percent, simply move the decimal point two places to the right.

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Example 2: Write each decimal as a percent. a) 0.5 b) 0.31 c) 1.49 d) 0.007 e) 0.375 f) 0.0067

Algebra2goÂŽ

To represent a fraction as a percent, first change the fraction to a decimal and then convert the decimal to a percent. In most cases you will have to round off the percent to a specified place value.

For example, supposed we are asked to write 3 the fraction 16 as a percent rounded to the nearest tenth.

We first convert the fraction to a decimal by dividing the numerator by the denominator.

3 á 16 = 0.1875

We do not want to round off this result. We are being asked to round off to the nearest tenth only after we have written our percent.

Now we convert 0.1875 to a percent by moving the decimal point two places to the right.

18.75% Finally we round our percent to the nearest tenth.

18.8% Page 3 of 4

Algebra2go速

Example 3: Write each fraction as a percent rounded to the nearest tenth.

3 a) 4 = 0.75 = 75% = 75.0% 1 b) 2 = 0.5 = 50% = 50.0% 5 c) 16 7 d) 8 2 e) 3 3 f) 7 1 g) 10 1 h) 6 Sometimes we need to write numbers as percents. With whole numbers, it may be helpful to write in the decimal point to the right of the number. With mixed numbers, first convert the fractional part to a decimal.

Page 4 of 4

Example 4: Write each number as a percent rounded to the nearest hundredth. a) 2 = 2. = 2.00% b) 10 1 12 c) 16 = 12.0625 5 d) 3 8

Algebra2go®

Percent Problems

Objective 1

Solve a Basic Percent Problem Basic percent problems are problems that are generally given as a simple word statement.

For example, suppose you were asked the question, “what number is 50% of 20?”

To solve this basic percent problem, we want to translate the given word statement into a mathematical equation using a variable to represent the unknown quantity. Then we solve the equation and answer the question.

Let’s now translate the question “what number is 50% of 20?” into an equation.

What number is 50% of

⋅

20?

X = 0.50 20 Notice that the unknown number is represented by x, the “is” by an equals sign, 50% by 0.50, “of” by a multiplication symbol, and the 20 represents itself.

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Algebra2go®

The equation x = 0.50 ⋅ 20 tells us that x=10. So to properly answer the question, what number is 50% of 20, we should write: 10 is 50% of 20.

Example 1: What is 38% of 90?

Suppose we are asked to the question, what percent of 48 is 12? Again, we want to translate the given word statement into a mathematical equation using a variable to represent the unknown quantity.

What percent of 48 is 12?

⋅

X 48 = 12 Here we have the equation x ⋅ 48 = 12 or 1 48x = 12. Solving for x we get, x= 4 or x=0.25.

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Algebra2goÂŽ

Because x in this problem represents a percent, we convert the decimal number to a percent by moving the decimal point two places to the right. This gives us x=25%.

To properly answer the question, we write the following: 25% of 48 is 12.

Example 2: What percent of 32 is 20?

Example 3: 29 is 4% of what number?

29 = 0.04

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â&#x2039;&#x2026;

x

Algebra2goÂŽ

Example 4: 30% of 60 is what number?

0.30

â&#x2039;&#x2026; 60

=

x

Example 5: What percent of 91 is 52? Round your final answer to the nearest tenth of a percent.

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Algebra2go®

Applications of Percent

Objective 1

Solve an Applied Percent Problem Applied percent problems can generally be solved using the following structured format.

(A percent) of (a total) is (a portion). (

%

) ⋅ (Total) = (Portion)

Let’s now use this structured format to solve the following problem.

A basketball player makes 72 out of 90 free throws. What percent of free throws does the basketball player make?

In this problem, the basket player attempts 90 total free throws. Therefore, 90 is our total.

The player makes 72 of the 90 free throws.

Therefore, 72 is our portion.

We can now set up an equation using our structured format where we let the variable x represent the unknown percent.

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Algebra2go®

(

%

) ⋅ (Total) = (Portion)

⋅

X 90 = 72 Here we have the equation x ⋅ 90 = 72 or 72 90x = 72. Solving for x we get, x= 90 or x=0.8 as a decimal.

Because x represents a percent, we convert the decimal number to a percent by moving the decimal point two places to the right. This gives us x=80%.

To properly answer the question, we can write the following: The basketball player makes 80% of attempted free throws.

Example 1: If 40 students enrolled in a music class but only 34 completed the course, what percent of students completed the course?

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Algebra2go®

Example 2: Watermelon is 91% water. How many pounds of a 12 pound watermelon is water? ( % ) (Total) = (Portion)

⋅

0.91

⋅

12

=

x

Example 3: If Thao deposits 30% of her paycheck into a savings account and the amount she deposits is $255, what was the total amount of her paycheck?

(

%

) ⋅ (Total) = (Portion) 0.30

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⋅

x

=

255

Algebra2go®

In the Health Care Career field, many calculations involve “mixtures or solutions” that are made of water and some other ingredient.

For example, a liquid solution that is marked 75% sodium means that the solution is 75% water and 25% sodium.

Similarly, a 1% iodine solution means that the solution is 99% water and 1% iodine.

Example 4: How much sodium is in a 60 milliliter bottle labeled 75% sodium?

(

%

) ⋅ (Total) = (Portion) 0.75

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⋅

60

=

x

Algebra2go®

Example 5: If 46% of a solution is water and there is 184 milliliters of water in the solution, how many milliliters is the total solution?

(

%

) ⋅ (Total) = (Portion)

Example 6: How much hydrochloric acid (HCl) is in a 70 milliliter bottle that is labeled 34% HCl? ( % ) (Total) = (Portion)

⋅

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Algebra2go®

Sales Tax and Commission

Objective 1

Solve an Applied Sales Tax Problem Applied sales tax problems are very similar to applied percent problems and also generally can be solved using the following structured format.

(A percent) of (a total) is (a portion).

(

(

) ⋅ (Total) = (Portion)

%

% Sales Tax

) ⋅(

)=(

Purchase Price

Amount of Sales Tax

)

We refer to the purchase price as the “Total”.

This can also be thought of as the price shown on the price tag. The amount of sales tax you pay at the register is a portion of the purchase price and is calculated once you are at the cash register. Finally, the “Total Amount” you pay at the register is the sum of the purchase price and the amount of sales tax.

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(

Total Amount

)= (

)+ (

Purchase Price

Amount of Sales Tax

)

Algebra2go®

Example 1: Suppose the sales tax rate in a certain county is 7%. If the price of a calculator is $34.95, what is the amount of sales tax?

What is the total price paid at the register?

( (

Amount of Sales Tax Amount of Sales Tax

(

Page 2 of 5

Total Amount

) =( )⋅( ) ) = ( 0.07 ) ⋅(34.95 ) % Sales Tax

)= (

)+ (

Purchase Price

Purchase Price

Amount of Sales Tax

)

Algebra2goÂŽ

Example 2: If the purchase price of a television is $276 and the amount of sales tax is $13.80, what is the percent sales tax?

(

Amount of Sales Tax

) =(

) â&#x2039;&#x2026;(

% Sales Tax

Purchase Price

)

Example 3: If the purchase price of a stereo system is $625 and the amount of sales tax is $37.50, what is the percent sales tax?

Page 3 of 5

Algebra2go®

Objective 2

Solve a Commission Rate Problem Some careers pay their sales personnel a portion of their total sales. This portion is called the amount of commission. The amount of commission is actually a percentage of the total sales. This percentage is called the commission rate.

The way we calculate the amount of commission is very similar to the way we calculate the amount of sales tax.

(

) =(

Amount of Commission

% Commission

) ⋅( ) Total Sales

Example 4: A real estate agent has a commission rate of 3.5% and sells a property for $240,000. What is the amount of her commission?

(

) =(

Amount of Commission

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% Commission

) ⋅( ) Total Sales

Algebra2go速

Example 5: If the commission for selling a car is $516 and the car sold for $12,000, what is the commission rate?

Example 6: If the commission rate for selling an appliance is 2.5% and the appliance sold for $945, what is the amount of commission?

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Algebra2go®

Percent Increase or Decrease and Discount

Objective 1

Solve a Percent Increase Problem A percent increase refers to an amount that is a portion of a given total quantity. For example, if you receive a percent raise on your salary, the amount of the raise is a portion of your total salary. As in the previous sections, we can use the following structured format to calculate the amount of increase on any total quantity.

(Portion) = (

(

Amount of Increase

)=(

%

) ⋅ (Total)

% Increase

) ⋅(

Total Quantity

)

Example 1: An employee earning $60,000 a year receives a 15% raise. What is the amount of the raise? What is the new salary?

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Algebra2go速

Example 2: The annual tuition at a college is presently $4,500. Next year the tuition will increase by 18%. What is the amount of the tuition increase? What will the annual tuition be next year?

Example 3: If an employee making $8.20 per hour receives a raise of $1.23 per hour, what is the percent increase?

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Algebra2go®

Objective 2

Solve a Percent Decrease Problem A percent decrease refers to an amount that is again a portion of a given total quantity. As with percent increase problems, we can again use a structured format to solve these types of problems.

(Portion) = ( % ) (Total)

⋅

(

Amount of Decrease

)=(

% Decrease

) ⋅(

Total Quantity

)

Example 4: A large university currently has 40,000 students. Next year, it is expected that the university will see a 3.2% decrease in its current student population. How many students are expected to attend the university next year?

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Algebra2go®

Example 5: One year ago, Duyen bought a 16GB USB drive for $48. Today she bought the same USB drive for $12. What is the percent decrease in the price of the USB drive?

(

Objective 3

Page 4 of 6

Amount of Decrease

)=(

% Decrease

) ⋅(

Total Quantity

)

Solve a Discount Problem When department stores have clearance sales, they regularly discount the purchase price of much of their merchandise to attract potential shoppers. Stores will often place signs throughout the store stating things like “15% Off” or in some cases “50% Off”. These signs indicate that there is a percent decrease in the purchase price of the item. The new discounted price is referred to as the sale price.

Algebra2go®

When we buy things from a store and there is a percent decrease in the purchase price, we refer to this as a discount. The amount of the discount is always a portion of the purchase price and can be calculated using the following structured format.

(Portion) = ( % ) (Total)

⋅

(

Amount of Discount

)=(

% Discount

) ⋅(

Purchase Price

)

Example 6: During a clearance sale, a pair of shoes that usually sells for $39.00 is marked “15% Off”. What is the discount amount and what is the sale price?

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Algebra2go®

Example 7: The regular price of a pair of designer jeans costs $80. If the jeans are marked on sale for “15% Off”, what is the total cost of the jeans if the sales tax rate is 8%?

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Algebra2go®

Objective 1

Simple Interest

Solve a Simple Interest Problem When we take a loan from a bank we must pay an interest amount on the loan amount.

The interest amount is a portion of the total loan amount.

Credit cards can sometimes have high interest rates sometimes upwards of 20%! In these cases the amount of interest you must pay can be very large.

Simple Interest is calculated based on the loan amount. With simple interest problems we refer to the loan amount as the principal.

The formula for calculating simple interest is given below.

⋅

⋅ ⋅

Interest = Principal Rate Time I = P R T “I” represents the amount of interest.

“P” represents the amount borrowed.

“R” represents the annual interest rate.

“T” represents the time in years.

⋅

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Algebra2go®

Note: In finance calculations such as simple interest calculations, it is assumed that 1 year = 360 days or equivalently 1 month = 30 days.

Therefore, if you took out a loan for 1 month or 30 days(any month of the year), 30 the time in years of your loan would be 360 1 or 12 of a year.

If you took out a loan for 6 months or 180 days, the time in years of your loan 180 1 would be 360 or 2 year.

Example 1: A student takes out a loan for $700 at an annual interest rate of 14%. How much does the student pay in interest if the student pays off the loan on 90 days?

I = P R T 90 I = 700 0.14 360 I = 700 0.14 0.25 I = 24.5 Therefore the student must pay $24.50 in interest to pay off the loan in 90 days.

⋅ ⋅ ⋅

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⋅ ⋅ ⋅

Algebra2go®

Example 2: A student takes out an emergency loan for $600 to buy school supplies. If the annual interest rate is 6%, how much must the student pay to completely pay off the loan in 6 months?

I = P R T 180 360 I = 600 0.06 I = 600 0.06 0.5 I = 18 After 6 months, the student owes $18 in interest. Therefore to completely pay off the loan, the student must pay $618.

⋅ ⋅ ⋅

⋅ ⋅ ⋅

We can also use the simple interest formula to calculate the amount of interest we earn when depositing money into a savings account. Although it should be noted that most banks today do not use the simple interest formula when calculating interest owed or interest gained. A compound interest formula is generally used.

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Algebra2go速

Example 3: How much interest is gained after 1 year if $10,000 is put into a savings account at an annual interest rate of 7%?

Example 4: Everlyn deposited $8,500 into a savings account for 30 months at 5% interest.

How much money is in the account after this time?

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Algebra2go速

Algebra2go: Understanding the Metric System

The Metric System

1. Understand the Basic Units of Length used in Health Care Careers The metric system is the most commonly used system of measurement in the Health Care career field. A meter is the basic unit of length in the metric system and 1 meter (abbreviated 1 m) is approximately 3 inches longer than 1 yard. Both larger and smaller units of measure are expressed using a prefix on the word meter. Below are diagrams of meter sticks whose lengths are scaled with markings of unit measures that are less than 1 meter. The abbreviations for the prefix on the word meter are shown in the parenthesis.

1 meter = 10 decimeters (dm) 0.5

0 dm

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

7.5

8

8.5

9

9.5

10

1 meter = 100 centimeters (cm) 5

0 cm 0

10 10

15

20 20

25

30 30

35

40 40

45

50 50

55

60 60

65

70 70

75

80 80

85

90 90

95

100 100

1 meter = 1,000 millimeters (mm) 50

0 mm

100

150

200

250

300

350

400

450

500

550

600

650

700

750

800

850

900

950

1000

Use the above meter sticks and previous content to answer the following questions.

1) Write down the abbreviations for meters, decimeters, centimeters, and millimeters.

5) A length of 3 dm is equal to a length of how many millimeters?

2) How does the abbreviation for miles differ from the abbreviation for meters?

6) A length of 70 cm is equal to a length of how many decimeters?

3) Order the following lengths of measure from smallest to largest.

1 dm, 1 mm, 1 m, 1 cm 4) Order the following lengths of measure from largest to smallest.

1 450 mm, 40 cm, m, 6 dm, 1 yd 2

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1 m is equal to a length of 4 how many millimeters?

7) A length of

1 m is equal to a length of 5 how many decimeters?

8) A length of

Page 1

Algebra2go: Understanding the Metric System

Algebra2goÂŽ

Many of the things we use on a daily basis can help us estimate lengths of other objects. For example, the length of a dollar bill is 6.14 inches and its width is 2.61 inches. In metric units its length is approximately 16 cm and its width approximately 7 cm. A mechanical pencil is slightly smaller than the length of a dollar bill, and therefore its length can be approximated as 15 cm.

Here are some other items that we may use on a daily basis that can help us estimate metric lengths.

The diameter of a quarter is about 25 mm.

A USB plug is about 1 cm wide.

A dime is about 1 mm thick.

The diameter of a DVD is about 14 cm.

Using an incorrect metric prefix to represent measurements of quantities such as length, mass (weight), or drug dosages, can result in dangerous errors. In order to determine if a given or calculated quantity makes sense, we need to develop good estimation skills. Letâ&#x20AC;&#x2122;s now begin developing our estimation skills by doing problems that require us to write in an appropriate metric unit.

Fill in the blank with the appropriate metric unit. Choose m, dm, cm, or mm.

9) The width of the palm of your hand is 10 ____ .

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10) The height of a soda can is 12 ____.

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Algebra2go速

Algebra2go: Understanding the Metric System

11) The length of a key is 50 ____.

15) A cellular phone is approximately 120 ____ in length.

12) The diameter of a nickel is about 2 ____.

16) A plastic fork is approximately 16 ____ in length.

13) The length of your index finger is about 7 ____.

17) Your teacher is about 1.7 ____ tall.

14) The average length of an adult female femur bone is about 45 ____.

18) A CD-ROM disk has a thickness of 1.2 ____ .

2. Converting between Metric Units using Powers of 10 Looking at the meter stick diagrams below, we notice that the prefix deci is used to represent unit 1 lengths that are of a meter and therefore 10 dm = 1 m. Similarly, we see that centi is used to 10 1 of a meter. Therefore, 100 cm = 1 m. Finally, we see that milli is represent unit lengths that are 100 1 of a meter and therefore 1,000 mm = 1 m. used to represent unit lengths that are 1, 000 You may have noticed that one-half of a meter, or 0.5 m, is equal to 5 dm. The diagrams below show us that 5 dm is equal to 50 cm, and that 50 cm is equal to 500 mm. Summarizing this, we get the relationship 0.5 m = 5 dm = 50 cm = 500 mm. Do you see a pattern?

1 meter = 10 decimeters (dm) 0.5

0 dm

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

7.5

8

8.5

9

9.5

10

1 meter = 100 centimeters (cm) 5

0 cm

10

15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

90

95

100

1 meter = 1,000 millimeters (mm) 50

0 mm

100

www.Algebra2go.com

150

200

250

300

350

400

450

500

550

600

650

700

750

800

850

900

950

1000

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Algebra2goÂŽ

Algebra2go: Understanding the Metric System

Letâ&#x20AC;&#x2122;s now complete a table to demonstrate the pattern. Fill in the blank cells.

Equivalent Lengths

0.5 m 0.7 m 0.3 m 0.6 m

5 dm 7 dm 3.dm 5 dm

50 cm 70 cm 35 cm 50 cm

500 mm 700 mm 350 mm 625 mm

When moving across each row to the right, we see that the numbers are multiplied by a factor of 10.

When moving across each row to the left, the numbers are divided by a factor of 10. Remember that multiplying a number by 10 moves the decimal point to the right one place value. Dividing a number by 10 moves the decimal point to the left one place value.

Now letâ&#x20AC;&#x2122;s apply what we have learned to the following questions.

Convert each measure to the indicated unit.

19) 25 cm to mm

23) 7.6 dm to cm

27) 1.5 m to cm

20) 35 dm to mm

24) 278 mm to cm

28) 17.5 dm to m

21) 0.2 m to cm

25) 3.2 dm to m

29) 1,578 mm to m

22) 0.7 cm to mm

26) 5 mm to dm

30) 349 cm to m

3. Understand Units of Length Greater Than 1 Meter Up to this point we have mainly dealt with lengths that measure less than 1 meter. What about lengths that are more than 1 meter? In this case we again use a prefix on the word meter to represent measures of length that are greater than 1 meter.

The prefix deka is used to represent a length that is 10 meters.

The prefix hecto is used to represent a length that is 100 meters.

The prefix kilo is used to represent a length that is 1,000 meters.

1 dekameter = 10 meters 1 hectometer = 100 meters 1 kilometer = 1,000 meters

Again you may notice that there is a pattern involving powers of 10. Notice that the prefix deka is used to represent unit lengths that are 10 times that of a meter. Therefore, 1 dekameter = 10 m.

Next, we see that hecto is used to represent unit lengths that are 100 times that of a meter.

Therefore, 1 hectometer = 100 m. Finally, we see that kilo is used to represent unit lengths that are 1,000 times that of a meter and therefore 1 kilometer = 1,000 m.

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Algebra2goÂŽ

Algebra2go: Understanding the Metric System

Letâ&#x20AC;&#x2122;s again complete a table to demonstrate the pattern. Fill in the blank cells. The abbreviations for the prefix on the word meter are shown in the parenthesis.

Equivalent Lengths

kilometers (km)

hectometers (hm)

dekameters (dam)

meters (m)

2 km 0.865 km 4.675 km 1.009 km

20 hm 8.65 hm 46.75 hm 10.09 hm

200 dam 86.5 dam 467.5 dam 100.9 dam

2,000 m 865 m 4,675 m 1,009 m

Again we can see that when moving across each row to the right, the numbers are multiplied by a factor of 10. When moving across each row to the left, the numbers are divided by a factor of 10.

Remember that multiplying a number by 10 moves the decimal point to the right one place value.

Dividing a number by 10 moves the decimal point to the left one place value.

Letâ&#x20AC;&#x2122;s now continue to develop our estimation skills by doing problems that require us to write in an appropriate metric unit.

Fill in the blank with the appropriate metric unit. Choose m, dam, hm, or km.

31) The length of a car is about 5 ____.

35) The height of the Statue of Liberty is about 1 ____.

32) The height the Empire State Building is about 45____.

36) The traveled length across the Golden Gate Bridge is about 2 ____.

33) The radius of the earth is approximately 6,000 ____.

37) A Boeing 747 Jumbo Jet is approximately 6.4 ____ in length.

34) The distance from New York to Los Angeles is about 4,000 ____.

38) The length of a NFL football field is approximately 9.1 ____ in length.

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Algebra2go: Understanding the Metric System

Algebra2go®

The following diagram organizes the unit measures covered in this section in order from largest to smallest. Notice how the powers of ten are used to move from one unit measure to the next. Each arrow represents the movement of the decimal point one time. Use this diagram to answer the following questions.

km

hm

dam

m

dm

cm

mm

Convert each measure to the indicated unit by moving the decimal point appropriately. Write down the number of times you moved the decimal point and the direction you moved it.

39) 3.8 hm to dm

43) 80.04 cm to hm

47) 3,498 dm to km

40) 2,385 mm to hm

44) 31.08 m to hm

48) 0.164 km to cm

41) 0.7 cm to m

45) 19 hm to m

49) 0.028 dam to dm

42) 0.91 dam to dm

46) 31 dam to cm

50) 1,578 mm to dm

Review Exercises Evaluate the expression.

2 5 51) − + 8 3 6 52)

− 3 + 8  2  3

Fill in the blank with the appropriate metric unit.

3

÷

9

Simplify the expression as much as possible.

57) The length of a paper clip is approximately 3.2 ____.

58) The diameter of a nickel is approximately 21____.

53) −16 − 12 x − 5 + 3x Answer True or False.

54) 8 − 5 ( 2 x − 3) + 6 x 55) −2 −3 − 8 + 7

59) Dividing a number by 1,000 is the 1 same as multiplying by .

1, 000

56) 6 − −5 + 12 + 11

60) 220 cm is 20 cm more than 2 dm.

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Page 6

Algebra2go: Conversion Calculations and Dosage Calculations

Conversions with Lengths, Weight (Mass), and Volume

1. Learn How to Perform Conversions using One Conversion Factor A conversion factor is a fraction or ratio involving two equivalent quantities that are expressed in different units. For example, if you wish to convert inches to centimeters, you will need to use a conversion factor to perform the calculation. Recall that 2.54 cm = 1 in..  2.54 cm   1 in.  This gives us the conversion factor   which is equivalent to   . Next, we  1 in.   2.54 cm  will need to develop a sense of how to choose the appropriate conversion factor or factors, for a given conversion calculation.

Example 1: How many centimeters are in 10 inches?

In this example we are being asked to convert 10 in. to cm. We begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.  10 in.     1  Next, we will multiply are given quantity using the appropriate conversion factor to get the desired result in centimeters.  10 in.   2.54 cm      1   1 in.  Given Quantity

Conversion Factor

Notice that the denominator of our conversion factor contains units of inches. This allows us to divide out the units of inches leaving the desired units of centimeters. Here is what our completed conversion calculation will look like.  10 in.   2.54 cm     = 25.4 cm  1   1 in. 

Based on our conversion calculation, we can now answer the question by stating there are 25.4 centimeters in 10 inches.

When performing conversion calculations it is always important to show how you reached your solution. This will allow someone else to easily verify that your calculation is correct by checking your work.

Note: In Example 1 above, the conversion factor was written for converting inches to centimeters. In the next example we will convert from centimeters to inches.

Notice how the conversion factor differs in Example 2.

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Page 1

Algebra2go®

Algebra2go: Conversion Calculations and Dosage Calculations

Example 2: How many inches are in 35 centimeters?

In this example we are being asked to convert 35 cm to inches. Again, we begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.  35 cm     1  Next, we will multiply our given quantity using the appropriate conversion factor to get the desired result in inches.  35 cm   1 in.      1   2.54 cm  Given Quantity

Conversion Factor

In this example, notice that the denominator of our conversion factor contains units of centimeters. This allows us to divide out the units of centimeters leaving the desired units of inches. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.  35 cm   1 in.     = 13.780 in.  1   2.54 cm 

Based on our conversion calculation, we can now answer the question by stating there are approximately 13.780 inches in 35 centimeters.

Perform the following conversion calculations using one conversion factor.

1) How many seconds are in 17 minutes?

2) How many feet are in 40 yards?

7) If one tablet contains 150 mg of ibuprofen, how much ibuprofen is in 1 3 2 tablets?

3) Convert 10 cm to inches.

4) Convert 25 inches to centimeters.

8) Given that 1 kilogram = 2.2 pounds, how many kilograms does a 175 lb adult male weigh?

5) Convert 2 liters to quarts.

6) Convert 5,000 pounds to tons.

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Algebra2go: Conversion Calculations and Dosage Calculations

2. Learn How to Perform Conversions using Multiple Conversion Factors

Many conversion calculations require the use of more than one conversion factor to obtain the desired result. In these cases, we let the dimensions guide us through the calculations, telling us where to put the numeric values. This approach will be demonstrated in Example 3 and Example 4. We will be using the following equivalent relationships to perform these types of conversion calculations.

12 in. = 1 ft

3 ft = 1 yd

5,280 ft = 1 mile

2.54 cm = 1 in.

Other equivalent relationships can be found on the conversion handout sheet found at the end of this sections material.

Example 3: How many miles are in 500,000 inches?

As always, we begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.  500,000 in.    1   Our first conversion factor will convert inches to feet by placing units of inches in the denominator and feet in the numerator.  500,000 in.   1 ft     1    12 in.  Given Quantity

Conversion Factor

Our second conversion factor will now convert feet to miles by placing feet in the denominator and miles in the numerator.  500,000 in.  1 ft   1 mi      1   12 in.   5,280 ft  Given Quantity

Conversion Factor

Conversion Factor

Notice that the denominators of our conversion factors divide out the units in the preceding numerators. This leaves us with the desired units of miles. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.  500,000 in.   1 ft   1 mi      = 7.891 mi 1    12 in.   5,280 ft  Based on our conversion calculation, we can now answer the question by stating there are approximately 7.891 miles in 500,000 inches.

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Algebra2go: Conversion Calculations and Dosage Calculations

Example 4: How many yards are in 4,500 centimeters?

Again, we first write the given quantity as a ratio using a 1 to represent the denominator.  4,500 cm    1  

Our first conversion factor will convert centimeters to inches by placing units of centimeters in the denominator and inches in the numerator.  4,500 cm   1 in.     1    2.54 cm  Given Quantity

Conversion Factor

Our second conversion factor will now convert inches to feet by placing inches in the denominator and feet in the numerator.  4,500 cm   1 in.   1 ft      1    2.54 cm   12 in.  Given Quantity

Conversion Factor

Conversion Factor

Our third conversion factor will now convert feet to yards by placing feet in the denominator and yards in the numerator.  4,500 cm   1 in.   1 ft   1 yd       1    2.54 cm   12 in.   3 ft  Given Quantity

Conversion Factor

Conversion Factor

Conversion Factor

Again we see that the denominators of our conversion factors divide out the units in the preceding numerators. Doing so leaves us with the desired units of yards. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.  4,500 cm   1 in.   1 ft   1 yd       = 49.213 yd 1    2.54 cm   12 in.   3 ft 

Based on our conversion calculation, we can now answer the question by stating there are approximately 49.213 yards in 4,500 centimeters.

Perform the following conversion calculations using multiple conversion factors.

9) How many meters are in 1 mile?

11) Convert 3 pounds to grams.

10) How many seconds are in 1 year?

12) Convert 2 liters to ounces.

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Algebra2go: Conversion Calculations and Dosage Calculations

3. Solve Applied Problems using Conversion Calculations

Conversion calculations can be used to solve many of the applied problems seen within the Health Care career field. Probably the most important types of conversion calculations are applied problems that involve dosage calculations. With these types of problems, a given solution strength ratio or dosage strength ratio is used as a conversion factor. The following examples represent common dosage calculations using this approach.

Example 5: Suppose you found that 100 mL of a solution contains 1 gram of lidocaine. đ?&#x;?

How many mg of lidocaine are in đ?&#x;? mL of the solution? đ?&#x;?

We begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.  2.5 mL     1  Next, we multiply our given quantity by the solution strength ratio. This conversion factor will convert milliliters to grams by placing units of milliliters in the denominator and grams in the numerator. This allows us to divide out the units of mL leaving us with grams of lidocaine.  2.5 mL   1 g      1   100 mL  Given Quantity

Conversion Factor (Solution Strength)

Now we must convert the grams of lidocaine to mg. To accomplish this, we add a second conversion factor that will convert grams to milligrams.  2.5 mL   1 g   1,000 mg       1   100 mL   1 g  Given Quantity

Conversion Conversion Factor Factor (Solution Strength) (grams to milligrams)

Once again the denominators of our conversion factors divide out the units in the preceding numerators. Doing so leaves us with the desired units of milligrams of lidocaine. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.  2.5 mL  1 g   1,000 mg   = 25.000 mg of lidocaine     1  100 mL   1 g  Based on our conversion calculation, we can now answer the question by stating there are 25.000 mg of lidocaine in 2.5 mL of solution. www.Algebra2go.com

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Algebra2go: Conversion Calculations and Dosage Calculations

Example 6: Suppose you found that 1,000 mL of a solution contains 1 g of epinephrine.

How many mg of epinephrine are in 2 tbsp of solution? Assume 1 tbsp = 15 mL.

Notice that we are given a volume measured by tablespoons. Because our solution strength ratio involves volume by milliliters, we must begin our conversion calculation by first converting tablespoons to milliliters.

We start by first representing the given quantity as a ratio using a 1 to represent the denominator.  2 tbsp     1  Our first conversion factor will convert tablespoons to millimeters. Notice that by placing units of tablespoons in the denominator and milliliters in the numerator, we can divide out the units of tablespoons leaving us with milliliters of epinephrine.  2 tbsp   15 mL      1   1 tbsp  Conversion Factor (tbsp to mL)

Given Quantity

Next, we use our solution strength ratio as the second conversion factor to convert milliliters to grams. Notice that milliliters are placed in the denominator and grams in the numerator. This allows us to divide out the units of milliliters leaving us with grams of epinephrine. At this point, the calculation will give us the grams of epinephrine in 2 tbsp of solution.

1g   2 tbsp   15 mL        1   1 tbsp   1,000 mL  Given Quantity

Conversion Factor (tbsp to mL)

Conversion Factor (Solution Strength)

Because we are asked to calculate the dosage of epinephrine in mg, we need to add an additional conversion factor that will convert grams to milligrams. Doing this leaves us with the desired units of milligrams of epinephrine.

1g   1,000 mg   2 tbsp   15 mL         1   1 tbsp   1,000 mL   1 g  Given Quantity

Conversion Factor (tbsp to mL)

Conversion Factor (Solution Strength)

Conversion Factor (grams to milligrams)

Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.   1,000 mg   2 tbsp   15 mL   1 g   = 30.000 mg epinephrine     1   1 tbsp   1,000 mL   1 g  Based on our conversion calculation, we can now answer the question by stating there are approximately 30.000 mg of epinephrine in 2 tbsp of solution. www.Algebra2go.com

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Algebra2go: Conversion Calculations and Dosage Calculations

Solve the applied problems.

13) The solution strength label of a solution indicates that 100 mL contains 10 grams of magnesium sulfate. How many mL of solution will contain 350 mg of magnesium sulfate?

15) Suppose you found that 5 mL of a solution contains 0.25 grams of Amoxicillin. How many mg of Amoxicillin are in 2 tbsp of solution?

Assume 1 tbsp = 15 mL.

14) The solution strength label of a solution indicates that 2,000 mL contains 1 gram of epinephrine. How many mL of solution will contain 0.25 mg of epinephrine?

16) Suppose you found that 5 mL of a solution contains 0.1 grams of Motrin®.

How many mg of Motrin® are in 2 tsp of solution? Assume 1 tsp = 5 mL.

Review Exercises Evaluate the expression.

17)

18)

3 6 10 ⋅ ⋅ 4 7 4

21) The width of a dollar bill is approximately 6.6 ____.

3 + 5 ÷ 20 6

3

22) The diameter of a quarter is approximately 24____.

Simplify the expression as much as possible.

19)

4 y3 x2

⋅x÷

20)

5b 2 a3

÷

3

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Fill in the blank with the appropriate metric unit.

y 4

15b3 7

⋅

b 21a

Fill in the blanks with the appropriate metric prefix.

23) _____ means

1 .

1, 000

24) _____ means

1 .

100

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Algebra2go: Conversion Calculations and Dosage Calculations

Equivalent Measurement Table

LENGTH

FLUID VOLUME

12 inches = 1 foot 3 feet = 1 yard

1,000 mL = 1 L

5,280 feet = 1 mile

1,000,000 μL = 1 L 1.06 quarts ≈ 1 L

2.54 centimeter = 1 inch

1 gallon ≈ 3.79 L

AREA

WEIGHT

144 in 2 = 1 ft 2

1 pound = 16 ounces

9 ft 2 = 1 yd 2

1 Ton = 2,000 pounds

6.4516 cm 2 = 1 in 2

28.3 grams ≈ 1 ounce 2.20 pounds ≈ 1 kilogram

VOLUME 1728 in 3 = 1 ft 3

FLUID VOLUME

27 ft 3 = 1 yd 3

16.39 milliliter = 1 in 3 1 milliliter = 1 cc (1 cm3 )

FLUID VOLUME 1 cup = 8 fluid ounces

1 teaspoon ≈ 5 milliliters 1 tablespoon ≈ 15 milliliters 1 fluid ounce ≈ 29.6 milliliters

2 cups = 1 pint 2 pints = 1 quart 4 quarts = 1 gallon

TEMPERATURE 

C = = F

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5 ( F  − 32 ) 9 9  C + 32 5

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Algebra2go速

Objective 1

Conversions of Rates

Convert a Rate to an equivalent Rate of different Dimensions

In some cases we may want to know the equivalent rate in feet per second of 65 miles per hour. In cases like this we can use conversions factors to perform this calculation.

The entire calculation is done in one single problem. We can first use conversion factors to convert the numerator to the desired dimension, and then use additional conversion factors to convert the denominator to the desired dimension. This process is demonstrated in the following examples.

Example 1: Convert 65 miles per hour to feet per second. hr ft 65 mi min

(1

Page 1 of 2

hr

)(

mi

)(

min

)(

sec

)

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Example 2: Convert 1,000 cm per second to miles per hour. cm ( 1,000 1 sec ) (

in cm

)(

in )(

ft

mi ft

)(

sec min

)(

min hr

)

Example 3: Convert 65 miles per hour to kilometers per hour.

Example 4: Convert 100 kilometers per hour to centimeters per second.

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