14
Solving equations
Objectives :
This chapter:
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explains what is meant by an equation and its solution
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shows how to solve linear, simultaneous and quadratic equations
An equation states that two quantities are equal, and will always contain an unknown quantity that we wish to find. For example, in the equation 5x þ 10 ¼ 20 the unknown quantity is x. To solve an equation means to find all values of the unknown quantity that can be substituted into the equation so that the left side equals the right side. Each such value is called a solution or alternatively a root of the equation. In the example above the solution is x ¼ 2 because when x ¼ 2 is substituted both the left side and the right side equal 20. The value x ¼ 2 is said to satisfy the equation.
14.1 Solving linear equations A linear equation is one of the form ax þ b ¼ 0 where a and b are numbers and the unknown quantity is x. The number a is called the coefficient of x. The number b is called the constant term. For example, 3x þ 7 ¼ 0 is a linear equation. The coefficient of x is 3 and the constant term is 7. Similarly, ÿ2x þ 17:5 ¼ 0 is a linear equation. The coefficient of x is ÿ2 and the constant term is 17.5. Note that the unknown quantity occurs only to the first power, that is, as x, and not as x 2 , x 3 , x 1=2 etc. Linear equations may appear in other forms that may seem to be different but are nevertheless equivalent. Thus 4x þ 13 ¼ ÿ7;
3 ÿ 14x ¼ 0
and
3x þ 7 ¼ 2x ÿ 4
are all linear equations that could be written in the form ax þ b ¼ 0 if necessary.
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Solving equations
An equation such as 3x 2 þ 2 ¼ 0 is not linear because the unknown quantity occurs to the power 2. Linear equations are solved by trying to obtain the unknown quantity on its own on the left-hand side, that is, by making the unknown quantity the subject of the equation. This is done using the rules given for transposing formulae in Chapter 13. Consider the following examples.
WORKED EXAMPLE 14.1 Solution
Solve the equation x þ 10 ¼ 0. We make x the subject by subtracting 10 from both sides to give x ¼ ÿ10. Therefore x ¼ ÿ10 is the solution. It can be easily checked by substituting into the original equation: (ÿ10) þ 10 ¼ 0
as required
Note from the last worked example that the solution should be checked by substitution to ensure it satisfies the given equation. If it does not then a mistake has been made.
WORKED EXAMPLES 14.2 Solution
Solve the equation 4x þ 8 ¼ 0. In order to find the unknown quantity x we attempt to make it the subject of the equation. Subtracting 8 from both sides we find 4x þ 8 ÿ 8 ¼ 0 ÿ 8 ¼ ÿ8 That is, 4x ¼ ÿ8 Then dividing both sides by 4 gives 4x
ÿ8 ¼
4
4
so that x ¼ ÿ2. The solution of the equation 4x þ 8 ¼ 0 is x ¼ ÿ2. 14.3 Solution
Solve the equation 5x þ 17 ¼ 4x ÿ 3. First we collect all terms involving x together. This is done by subtracting 4x from both sides to remove this term from the right. This gives 5x ÿ 4x þ 17 ¼ ÿ3 That is, x þ 17 ¼ ÿ3. To make x the subject of this equation we now subtract 17 from both sides to give x ¼ ÿ3 ÿ 17 ¼ ÿ20. The solution of the
14.1
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Solving linear equations
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equation 5x þ 17 ¼ 4x ÿ 3 is x ¼ ÿ20. Note that the answer can be easily checked by substituting x ¼ ÿ20 into the original equation and verifying that the left side equals the right side. 14.4 Solution
Solve the equation
xÿ3 4
¼ 1.
We attempt to obtain x on its own. First note that if we multiply both sides of the equation by 4 this will remove the 4 in the denominator. That is, 0 1 xÿ3 A¼4 1 4 @ 4 so that xÿ3¼4 Finally adding 3 to both sides gives x ¼ 7.
Self-assessment questions 14.1 1. Explain what is meant by a root of an equation. 2. Explain what is meant by a linear equation. 3. State the rules that can be used to solve a linear equation. 4. You may think that a formula and an equation look very similar. Try to explain the distinction between a formula and an equation.
Exercise 14.1 1. Verify that the given values of x satisfy the given equations: (a) x ¼ 7 satisfies 3x þ 4 ¼ 25 (b) x ¼ ÿ5 satisfies 2x ÿ 11 ¼ ÿ21 (c) x ¼ ÿ4 satisfies ÿx ÿ 8 ¼ ÿ4 (d) x ¼ 12 satisfies 8x þ 4 ¼ 8 (e) x ¼ ÿ13 satisfies 27x þ 8 ¼ ÿ1 (f ) x ¼ 4 satisfies 3x þ 2 ¼ 7x ÿ 14
(g) 5 ÿ 2x ¼ 2 þ 3x xþ3 3x þ 2 ¼ 3 (i) þ 3x ¼ 1 (h) 2 2 3. Solve the following equations: (a) 5(x þ 2) ¼ 13 (b) 3(x ÿ 7) ¼ 2(x þ 1) (c) 5(1 ÿ 2x) ¼ 2(4 ÿ 2x)
2. Solve the following linear equations: x (a) 3x ¼ 9 (b) ¼ 9 (c) 3t þ 6 ¼ 0 3 (d) 3x ÿ 13 ¼ 2x þ 9 (e) 3x þ 17 ¼ 21
4. Solve the following equations: (a) 3t þ 7 ¼ 4t ÿ 2 (b) 3v ¼ 17 ÿ 4v (c) 3s þ 2 ¼ 14(s ÿ 1)
(f ) 4x ÿ 20 ¼ 3x þ 16
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Solving equations
5. Solve the following linear equations: (a) 5t þ 7 ¼ 22 (b) 7 ÿ 4t ¼ ÿ13 (c) 7 ÿ 4t ¼ 27 (d) 5 ¼ 14 ÿ 3t (e) 4x þ 13 ¼ ÿx þ 25 xþ3 xÿ3 1 1 (f ) ¼ (g) x þ 6 ¼ x þ 2 2 4 3 2 1 1 (h) x þ 7 ¼ x þ 5 5 3 2x þ 4 x ÿ 3 x ÿ 7 3x þ 1 (i) ¼ ( j) ¼ 5 2 8 5 6. The following equations may not appear to be linear at first sight but they can all
be rewritten in the standard form of a linear equation. Find the solution of each equation. 1 1 5 1 5 (a) ¼ 5 (b) ¼ (c) ¼ x x 2 xþ1 2 1 5 1 1 (d) ¼ (e) ¼ xÿ1 2 x 2x þ 1 3 1 1 1 (g) ¼ (f ) ¼ x 2x þ 1 x þ 1 3x þ 2 3 2 (h) ¼ x þ 1 4x þ 1
14.2 Solving simultaneous equations Sometimes equations contain more than one unknown quantity. When this happens there are usually two or more equations. For example in the two equations x þ 2y ¼ 14
3x þ y ¼ 17
the unknowns are x and y. Such equations are called simultaneous equations and to solve them we must find values of x and y that satisfy both equations at the same time. If we substitute x ¼ 4 and y ¼ 5 into either of the two equations above we see that the equation is satisfied. We shall demonstrate how simultaneous equations can be solved by removing, or eliminating, one of the unknowns. WORKED EXAMPLES 14.5
Solution
Solve the simultaneous equations x þ 3y ¼ 14
(14:1)
2x ÿ 3y ¼ ÿ8
(14:2)
Note that if these two equations are added the unknown y is removed or eliminated: x þ 3y ¼ 14 þ 2x ÿ 3y ¼ ÿ8 ____________ 3x ¼6 ____________
14.2
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Solving simultaneous equations
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so that x ¼ 2. To find y we substitute x ¼ 2 into either equation. Substituting into Equation 14.1 gives 2 þ 3y ¼ 14 Solving this linear equation will give y. We have 2 þ 3y ¼ 14 3y ¼ 14 ÿ 2 ¼ 12 12 y¼
3
¼4
Therefore the solution of the simultaneous equations is x ¼ 2 and y ¼ 4. Note that these solutions should be checked by substituting back into both given equations to check that the left-hand side equals the right-hand side. 14.6
Solve the simultaneous equations 5x þ 4y ¼ 7
(14:3)
3x ÿ y ¼ 11 Solution
(14:4)
Note that, in this example, if we multiply the first equation by 3 and the second by 5 we shall have the same coefficient of x in both equations. This gives 15x þ 12y ¼ 21
(14:5)
15x ÿ 5y ¼ 55
(14:6)
We can now eliminate x by subtracting Equation 14.6 from Equation 14.5, giving 15x þ 12y ¼ 21 ÿ 15x ÿ 5y ¼ 55 _______________ 17y ¼ ÿ34 _______________ from which y ¼ ÿ2. In order to find x we substitute our solution for y into either of the given equations. Substituting into Equation 14.3 gives 5x þ 4(ÿ2) ¼ 7
so that
5x ¼ 15
or
x¼3
The solution of the simultaneous equations is therefore x ¼ 3, y ¼ ÿ2.
Exercise 14.2 1. Verify that the given values of x and y satisfy the given simultaneous equations. (a) x ¼ 7; y ¼ 1 satisfy 2x ÿ 3y ¼ 11, 3x þ y ¼ 22 (b) x ¼ ÿ7, y ¼ 2 satisfy 2x þ y ¼ ÿ12, x ÿ 5y ¼ ÿ17 (c) x ¼ ÿ1, y ¼ ÿ1 satisfy 7x ÿ y ¼ ÿ6, x ÿ y ¼ 0
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Solving equations
2. Solve the following pairs of simultaneous (a) 3x þ y ¼ 1; 2x ÿ y ¼ 2 (c) 2x ÿ y ¼ 17; x þ 3y ¼ 12 (e) ÿx þ y ¼ ÿ10; 3x þ 7y ¼ 20
equations: (b) 4x þ 5y ¼ 21; 3x þ 5y ¼ 17 (d) ÿ2x þ y ¼ ÿ21; x þ 3y ¼ ÿ14 (f ) 4x ÿ 2y ¼ 2; 3x ÿ y ¼ 4
3. Solve the following simultaneous equations: (a) 5x þ y ¼ 36, 3x ÿ y ¼ 20 (b) x ÿ 3y ¼ ÿ13, 4x þ 2y ¼ ÿ24 (c) 3x þ y ¼ 30, ÿ5x þ 3y ¼ ÿ50 (d) 3x ÿ y ¼ ÿ5, ÿ7x þ 3y ¼ 15 (e) 11x þ 13y ¼ ÿ24, x þ y ¼ ÿ2
14.3 Solving quadratic equations A quadratic equation is an equation of the form ax 2 þ bx þ c ¼ 0 where a, b and c are numbers and x is the unknown quantity we wish to find. The number a is the coefficient of x 2 , b is the coefficient of x, and c is the constant term. Sometimes b or c may be zero, although a can never be zero. For example x 2 þ 7x þ 2 ¼ 0
3x 2 ÿ 2 ¼ 0 ÿ2x 2 þ 3x ¼ 0
8x 2 ¼ 0
are all quadratic equations.
Key point
A quadratic equation has the form ax 2 þ bx þ c ¼ 0 where a, b and c are numbers, and x represents the unknown we wish to find.
WORKED EXAMPLES 14.7
State the coefficient of x 2 and the coefficient of x in the following quadratic equations: (a) 4x 2 þ 3x ÿ 2 ¼ 0
Solution
(b) x 2 ÿ 23x þ 17 ¼ 0
(c) ÿx 2 þ 19 ¼ 0
(a) In the equation 4x 2 þ 3x ÿ 2 ¼ 0 the coefficient of x 2 is 4 and the coefficient of x is 3. (b) In the equation x 2 ÿ 23x þ 17 ¼ 0 the coefficient of x 2 is 1 and the coefficient of x is ÿ23: (c) In the equation ÿx 2 þ 19 ¼ 0 the coefficient of x 2 is ÿ1 and the coefficient of x is zero, since there is no term involving just x.
14.8
Verify that both x ¼ ÿ7 and x ¼ 5 satisfy the quadratic equation x 2 þ 2x ÿ 35 ¼ 0: