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Solving Linear Equations and Inequalities: Solving Equations of the Form ax=b and x/a=b

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

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Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the "Five-Step Method" of solving applied problems (discussed in modules ((Reference)) and ((Reference))). Objectives of this module: understand the equality property of addition and multiplication, be able to solve equations of the form ax = b and x/a = b.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

Overview

  • Equality Property of Division and Multiplication
  • Solving ax=b ax=b and x a  = b x a  = b for x x

Equality Property of Division and Multiplication

Recalling that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side suggests the equality property of division and multiplication, which states:

  1. We can obtain an equivalent equation by dividing both sides of the equation by the same nonzero number, that is, if c ≠ 0,  c ≠ 0,  then a = b a = b is equivalent to a c  =  b c a c  =  b c .
  2. We can obtain an equivalent equation by multiplying both sides of the equation by the same nonzero number, that is, if c ≠ 0, c ≠ 0, then a = b a = b is equivalent to ac=bc ac=bc .

We can use these results to isolate x, x, thus solving the equation for x x .

Example 1

Solving ax=bax=b for xx

ax = b a  is associated with   x  by multiplication. Undo the association by dividing both sides by a. ax a = b a a x a = b a 1⋅x = b a a a =1  and  1  is the multiplicative identity. 1⋅x = x ax = b a  is associated with   x  by multiplication. Undo the association by dividing both sides by a. ax a = b a a x a = b a 1⋅x = b a a a =1  and  1  is the multiplicative identity. 1⋅x = x

Example 2

Solving x a = b x a = b for x x

x = b a This equation is equivalent to the first and is solved by x. x a = b a  is associated with  x  by division. Undo the association by multiplying both sides by a. a⋅ x a = a⋅b a ⋅ x a = ab 1⋅x = ab a a =1  and  1  is the multiplicative identity. 1⋅x  =  x x = ab This equation is equivalent to the first and is solved for  x. x = b a This equation is equivalent to the first and is solved by x. x a = b a  is associated with  x  by division. Undo the association by multiplying both sides by a. a⋅ x a = a⋅b a ⋅ x a = ab 1⋅x = ab a a =1  and  1  is the multiplicative identity. 1⋅x  =  x x = ab This equation is equivalent to the first and is solved for  x.

Solving ax=b ax=b and x a  = b x a  = b for x x

Example 3

Method for Solving ax=b and  x a =b ax=b and  x a =b

To solve ax=b ax=b for x x , divide both sides of the equation by a a .
To solve x a =b x a =b for x x , multiply both sides of the equation by a a .

Sample Set A

Example 4

Solve 5x = 355x = 35 for xx.

5x = 35 5  is associated with x by multiplication. Undo the association by dividing both sides by 5. 5x 5 = 35 5 5 x 5 = 7 1⋅x = 7 5 5 =1  and 1 is multiplicative identity. 1⋅ x = x. x = 7 5x = 35 5  is associated with x by multiplication. Undo the association by dividing both sides by 5. 5x 5 = 35 5 5 x 5 = 7 1⋅x = 7 5 5 =1  and 1 is multiplicative identity. 1⋅ x = x. x = 7

Check: 5(7) = 35 Is this correct? 35 = 35 Yes, this is correct. Check: 5(7) = 35 Is this correct? 35 = 35 Yes, this is correct.

Example 5

Solve x 4  = 5 x 4  = 5 for x x .

x 4 = 5 4 is asssociated with x by division. Undo the association by  multiplying both sides by 4. 4⋅ x 4 = 4⋅5 4 ⋅ x 4 = 4⋅5 1⋅x = 20 4 4 =1 and 1 is the multiplicative identity. 1⋅x=x. x = 20 x 4 = 5 4 is asssociated with x by division. Undo the association by  multiplying both sides by 4. 4⋅ x 4 = 4⋅5 4 ⋅ x 4 = 4⋅5 1⋅x = 20 4 4 =1 and 1 is the multiplicative identity. 1⋅x=x. x = 20

Check: 20 4 = 5 Is this correct? 5 = 5 Yes, this is correct. Check: 20 4 = 5 Is this correct? 5 = 5 Yes, this is correct.

Example 6

Solve 2y9 = 32y9 = 3 for yy.

Method (1) (Use of cancelling):

2y 9 = 3 9 is associated with y by division. Undo the association by   multiplying both sides by 9. ( 9 )( 2y 9 ) = (9)(3) 2y = 27 2 is associated with y by multiplication. Undo the  association by dividing both sides by 2. 2 y 2 = 27 2 y = 27 2 2y 9 = 3 9 is associated with y by division. Undo the association by   multiplying both sides by 9. ( 9 )( 2y 9 ) = (9)(3) 2y = 27 2 is associated with y by multiplication. Undo the  association by dividing both sides by 2. 2 y 2 = 27 2 y = 27 2

Check: 2 ( 27 2 ) 9 = 3 Is this correct? 27 9 = 3 Is this correct? 3 = 3 Yes, this is correct. Check: 2 ( 27 2 ) 9 = 3 Is this correct? 27 9 = 3 Is this correct? 3 = 3 Yes, this is correct.

Method (2) (Use of reciprocals):

2y 9 = 3 Since  2y 9 = 2 9 y,  2 9  is associated with y by multiplication. Then, Since  9 2 ⋅ 2 9 =1, the multiplicative identity, we can ( 9 2 )  ( 2y 9 ) = ( 9 2 )(3) undo the associative by multiplying both sides by  9 2 . ( 9 2 ⋅ 2 9 ) y = 27 2 1⋅y = 27 2 y = 27 2 2y 9 = 3 Since  2y 9 = 2 9 y,  2 9  is associated with y by multiplication. Then, Since  9 2 ⋅ 2 9 =1, the multiplicative identity, we can ( 9 2 )  ( 2y 9 ) = ( 9 2 )(3) undo the associative by multiplying both sides by  9 2 . ( 9 2 ⋅ 2 9 ) y = 27 2 1⋅y = 27 2 y = 27 2

Example 7

Solve the literal equation 4axm = 3b4axm = 3b for xx.

4ax m = 3b m is associated with x by division. Undo the association by  multiplying both sides by m. m ( 4ax m ) = m⋅3b 4ax = 3bm 4a is associated with x by multiplication. Undo the  association by multiplying both sides by 4a. 4a x 4a = 3bm 4a x = 3bm 4a 4ax m = 3b m is associated with x by division. Undo the association by  multiplying both sides by m. m ( 4ax m ) = m⋅3b 4ax = 3bm 4a is associated with x by multiplication. Undo the  association by multiplying both sides by 4a. 4a x 4a = 3bm 4a x = 3bm 4a

Check: 4a( 3bm 4a ) m = 3b Is this correct? 4a ( 3bm 4a ) m = 3b Is this correct? 3b m m = 3b Is this correct? 3b = 3b Yes, this is correct. Check: 4a( 3bm 4a ) m = 3b Is this correct? 4a ( 3bm 4a ) m = 3b Is this correct? 3b m m = 3b Is this correct? 3b = 3b Yes, this is correct.

Practice Set A

Exercise 1

Solve 6a=42 6a=42 for a a .

Solution

a = 7 a = 7

Exercise 2

Solve −12m=16 −12m=16 for m m .

Solution

m = − 4 3 m = − 4 3

Exercise 3

Solve y 8 =−2 y 8 =−2 for y y .

Solution

y = −16 y = −16

Exercise 4

Solve 6.42x=1.09 6.42x=1.09 for x x .

Solution

x = 0.17 x = 0.17 (rounded to two decimal places)

Use a calculator to solve this equation. Round the result to two decimal places.

Exercise 5

Solve 5k 12 =2 5k 12 =2 for k k .

Solution

k=  24 5 k=  24 5

Exercise 6

Solve −ab 2c =4d −ab 2c =4d for b b .

Solution

b =  -8cd a b =  -8cd a

Exercise 7

Solve 3xy 4 =9xh 3xy 4 =9xh for y y .

Solution

y = 12h y = 12h

Exercise 8

Solve 2 k 2 mn 5pq =−6n 2 k 2 mn 5pq =−6n for m m .

Solution

m =  −15pq k 2 m =  −15pq k 2

Exercises

In the following problems, solve each of the conditional equations.

Exercise 9

3x = 42 3x = 42

Solution

x=14 x=14

Exercise 10

5y = 75 5y = 75

Exercise 11

6x = 48 6x = 48

Solution

x=8 x=8

Exercise 12

8x = 56 8x = 56

Exercise 13

4x = 56 4x = 56

Solution

x=14 x=14

Exercise 14

3x = 93 3x = 93

Exercise 15

5a = −80 5a = −80

Solution

a=−16 a=−16

Exercise 16

9m = −108 9m = −108

Exercise 17

6p = −108 6p = −108

Solution

p=−18 p=−18

Exercise 18

12q = −180 12q = −180

Exercise 19

−4a = 16 −4a = 16

Solution

a=−4 a=−4

Exercise 20

−20x = 100 −20x = 100

Exercise 21

−6x = −42 −6x = −42

Solution

x=7 x=7

Exercise 22

−8m = −40 −8m = −40

Exercise 23

−3k = 126 −3k = 126

Solution

k=−42 k=−42

Exercise 24

−9y = 126 −9y = 126

Exercise 25

x 6  = 1 x 6  = 1

Solution

x=6 x=6

Exercise 26

a 5  = 6 a 5  = 6

Exercise 27

k 7  = 6 k 7  = 6

Solution

k=42 k=42

Exercise 28

x 3  = 72 x 3  = 72

Exercise 29

x 8  = 96 x 8  = 96

Solution

x=768 x=768

Exercise 30

y −3  = −4 y −3  = −4

Exercise 31

m 7  = −8 m 7  = −8

Solution

m=−56 m=−56

Exercise 32

k 18  = 47 k 18  = 47

Exercise 33

f −62  = 103 f −62  = 103

Solution

f=−6386 f=−6386

Exercise 34

3.06m= 12.546 3.06m= 12.546

Exercise 35

5.012k = 0.30072 5.012k = 0.30072

Solution

k=0.06 k=0.06

Exercise 36

x 2.19  = 5 x 2.19  = 5

Exercise 37

y 4.11  = 2.3 y 4.11  = 2.3

Solution

y=9.453 y=9.453

Exercise 38

4y 7  = 2 4y 7  = 2

Exercise 39

3m 10  = −1 3m 10  = −1

Solution

m= −10 3 m= −10 3

Exercise 40

5k 6  = 8 5k 6  = 8

Exercise 41

8h −7  = −3 8h −7  = −3

Solution

h= 21 8 h= 21 8

Exercise 42

−16z 21  = −4 −16z 21  = −4

Exercise 43

Solve pq= 7r pq= 7r for p p .

Solution

p= 7r q p= 7r q

Exercise 44

Solve m 2 n = 2s m 2 n = 2s for n n .

Exercise 45

Solve 2.8ab = 5.6d 2.8ab = 5.6d for b b .

Solution

b= 2d a b= 2d a

Exercise 46

Solve mnp 2k  = 4k mnp 2k  = 4k for p p .

Exercise 47

Solve −8 a 2 b 3c  = −5 a 2 −8 a 2 b 3c  = −5 a 2 for b b .

Solution

b= 15c 8 b= 15c 8

Exercise 48

Solve 3pcb 2m  = 2b 3pcb 2m  = 2b for pc pc .

Exercise 49

Solve 8rst 3p  = −2prs 8rst 3p  = −2prs for t t .

Solution

t=− 3 p 2 4 t=− 3 p 2 4

Exercise 50

Solve The product of a rectangle and a star over a triangle is equal to a rhombus. for □ □ .

Exercise 51

Solve 3□Δ∇ 2∇ =Δ∇ 3□Δ∇ 2∇ =Δ∇ for □ □ .

Solution

□= 2∇ 3 □= 2∇ 3

Exercises for Review

Exercise 52

((Reference)) Simplify ( 2 x 0 y 0 z 3 z 2 ) 5 ( 2 x 0 y 0 z 3 z 2 ) 5 .

Exercise 53

((Reference)) Classify 10 x 3  − 7x 10 x 3  − 7x as a monomial, binomial, or trinomial. State its degree and write the numerical coefficient of each item.

Solution

binomial; 3rd degree; 10,−7 10,−7

Exercise 54

((Reference)) Simplify 3 a 2  − 2a + 4a(a+2) 3 a 2  − 2a + 4a(a+2) .

Exercise 55

((Reference)) Specify the domain of the equation y= 3 7+x y= 3 7+x .

Solution

all real numbers except −7 −7

Exercise 56

((Reference)) Solve the conditional equation x+6=−2 x+6=−2 .

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