UNIT
Algebra Basics
Chapters
Key Skills
Basic Algebra ...........................................340
In this chapter, we'll learn the basics of manipulating equations. This skill will help us as we move into the equation of a line and some advanced algebra in the factoring & foiling section.
Exponents & Algebra ..............................350 Applied Algebra .......................................363
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Basic Algebra The key to manipulating algebraic equations is to make careful, balanced changes. As long as you play by the rules, you can do whatever you want!
Keep it Balanced Solving basic algebra problems is just a balancing act. The test writers will give you an algebra equation, like:
2x - 7 = -21
...and then they’ll ask you to “solve for x.” Which really just means that we need to strategically add, subtract, multiply, or divide until “x” is alone on one side of the equation. But here’s the key: we have to keep the equation balanced. Anything we do to the left side of the equal sign we must also do to the right side... otherwise, we throw off the balance and change the equation: ced!
n Bala
2 x - 7 = -21 +7 +7 2 x = -14
TIP
In this example, if we add 7 to the left side of the equation and NoT to the right side, it disrupts the balance and changes the equation.
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!
nced
Bala Not
2 x - 7 = -21 +7 2 x = -21
on the right, we made the mistake of only adding 7 to one side of the equation. We can tell at a glance that something's wrong... after all, how can 2x = –21 when 2x – 7 = 21? To avoid this, remember to always make step-by-step, balanced changes to each side of the equation.
BASIC AlgEBrA
Clean Work is Happy Work Making balanced changes is the most basic algebra skill we can cover. But in the rush of the test, it s often these basic skills that suffer. It s not just "careless" students who make "careless errors" – they can result from focusing so much on the tougher topics that you forget to actually work out the simpler steps. As a result, you forget to distribute a negative or you make a basic error of substitution. To avoid these errors, get in the habit of writing out careful, step-by-step work. Follow the step-by-step instructions to complete the solution below.
If 5x – 6 = 54, then x = ? 1. Rewrite equation 2. Add to 6 both sides 3. Write the result 4. Divide both sides by 5 5. Write the result
Answers: 1) 5x – 6 = 54
2) 5x – 6 + 6 = 54 + 6
3) 5x = 60
4) 5x ÷ 5 = 60 ÷ 5
5) x = 12
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BASIC AlgEBrA
The Freedom to Tinker here s the first, key thing to understand about solving algebra problems on the SAT. As long as you follow the rule of making balanced changes to each side (as well as those pesky rules about not dividing by zero, etc.) you are free to manipulate the equation as you see fit. If you feel like adding 1,000,000 and then subtracting 1,000,000 to each side before starting the exercise above, you'd be totally free to do that! We wouldn't recommend it, but, hey, you're the boss. Experimentation is a skill that can be a huge help on tougher problems.
Solve for x in the problems below.
TIP
These problems may seem a little basic, but the skills behind them are important! Basic Algebra skills and a little balanced experimentation here and there are the keys to solving even the toughest algebra problems on the test.
Equation
x=?
1.
12x + 12 = 12
2.
3x – 7 = 14
3.
2x + 3 = 4x – 8
4.
–13x = –3x + 30
5.
15x – 10 = 12x + 11 Scratch Work
Answers: 1) 0
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2) 7
3) 11/2
4) –3
5) 7
BASIC AlgEBrA
EXAMPLE 1 What value of x satis es the e uation 13 x - 8 x = 3 + 3 ? 15
15
5
20
ON SOLUTI
TIP
When you have fractional coefficients, you add and subtract them following normal fraction rules.
Start by combining like terms on the left. Both terms have the same denominators and one x, so let’s combine and simplify:
1
combine
5 x = 15
3 + 3 5 20
2
simplify
1 x = 3
3 + 3 5 20
Now clean up the right side. We have two constants. To combine these fractions, we need a common denominator:
3
find C. .
1 x = 3
so...
1 x = 3
3 + 3 => 12 + 3 => 15 => 3 5 20 20 20 20 4 3 4
Finally, make balanced changes to isolate x. To get x by itself, we need to get rid of that fraction by multiplying both sides by 3.
4
balance
(×3) 1 x =
voila!
x =
3
3 (×3) 4 9 4
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BASIC AlgEBrA
EXAMPLE 2 If 4m + 2 = 9 what is the value of m? m–7
ON SOLUTI
TIP
It’s completely okay (and often necessary) to multiply both sides of an equation by an expression, such as (m – 7), rather than just a variable or constant. remember, you can do anything you want to the equation as long as you keep it balanced!
First, look at the left side. We have a fraction with the variable we’re looking for, m, in both the numerator and denominator. That’s messy. let’s get rid of the fraction by multiplying both sides of the equation by the denominator, then make more balanced changes until m is all by itself.
1
Multiply by (m – 7)
(m –7)
4m + 2 m–7
= 9 (m – 7)
4m + 2 = 9m – 63
2
Subtract (4m)
4m – 4m + 2 = 9m – 4m – 63 2 = 5m – 63
3
Add (63)
2 + 63 = 5m – 63 + 63 65 = 5m
4
Divide by (5) ...booyah
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5 65 = 5 5m
13 = m
BASIC AlgEBrA
Substitution: Rename and Plug-in Whenever you are given (or discover through work) another "name" for a variable, you can plug in that new name anywhere the variable appears. For example, if we have the equation:
5k + a = 26
...and we learn that k = 2, we can go back and substitute in 2 anywhere we see k, and that will help us move forward in the problem:
PORTAL Balancing and substitution form the very heart of algebra. You'll use these basic skills to solve even the toughest systems of equations problems. Don't believe us? Turn to page 494 to see for yourself!
5(2) + a = 26 10 + a = 26 10 – 10 + a = 26 – 10 a = 16
For now we are focusing on replacing a variable with a number, but this also works if you are told that k = a – 2. In this case, we just replace every k with an (a – 2):
5(a– 2) + a = 26 5a – 10 + a = 26 6a – 10 = 26 6a – 10 + 10 = 26 + 10 6a = 36 a=6
The name's k, but YOU can call me (a – 2)!
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