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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 (<link document="m21980"/>) and (<link document="m21979"/>)). 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.

Overview

  • Equality Property of Division and Multiplication
  • Solving a x = b and x a = b for 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 , then a = b is equivalent to 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 , then a = b is equivalent to a c = b c .

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

Solving a x = b for x

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

Solving x a = b 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 = a b 1 x = a b a a = 1 and 1 is the multiplicative identity . 1 x = x x = a b This equation is equivalent to the first and is solved for x .

Solving a x = b And x a = b For x

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

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

Sample set a

Solve 5 x = 35 for x .

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

C h e c k : 5 ( 7 ) = 35 Is this correct? 35 = 35 Yes, this is correct .

Solve x 4 = 5 for x .

x 4 = 5 4 is asssociated with x by division . Undo the association by multiplying b o t h 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

C h e c k : 20 4 = 5 Is this correct? 5 = 5 Yes, this is correct .

Solve 2 y 9 = 3 for y .

Method (1) (Use of cancelling):

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

C h e c k : 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):

2 y 9 = 3 Since 2 y 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 ) ( 2 y 9 ) = ( 9 2 ) ( 3 ) undo the associative by multiplying b o t h sides by 9 2 . ( 9 2 2 9 ) y = 27 2 1 y = 27 2 y = 27 2

Solve the literal equation 4 a x m = 3 b for x .

4 a x m = 3 b m is associated with x by division . Undo the association by multiplying b o t h sides by m . m ( 4 a x m ) = m 3 b 4 a x = 3 b m 4 a is associated with x by multiplication . Undo the association by multiplying b o t h sides by 4 a . 4 a x 4 a = 3 b m 4 a x = 3 b m 4 a

C h e c k : 4 a ( 3 b m 4 a ) m = 3 b Is this correct? 4 a ( 3 b m 4 a ) m = 3 b Is this correct? 3 b m m = 3 b Is this correct? 3 b = 3 b Yes, this is correct .

Practice set a

Solve 6 a = 42 for a .

a = 7

Solve 12 m = 16 for m .

m = 4 3

Solve y 8 = 2 for y .

y = 16

Solve 6.42 x = 1.09 for x .

x = 0.17 (rounded to two decimal places)

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

Solve 5 k 12 = 2 for k .

k = 24 5

Solve a b 2 c = 4 d for b .

b = - 8 c d a

Solve 3 x y 4 = 9 x h for y .

y = 12 h

Solve 2 k 2 m n 5 p q = 6 n for m .

m = 15 p q k 2

Exercises

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

3 x = 42

x = 14

5 y = 75

6 x = 48

x = 8

8 x = 56

4 x = 56

x = 14

3 x = 93

5 a = 80

a = 16

9 m = 108

6 p = 108

p = 18

12 q = 180

4 a = 16

a = 4

20 x = 100

6 x = 42

x = 7

8 m = 40

3 k = 126

k = 42

9 y = 126

x 6 = 1

x = 6

a 5 = 6

k 7 = 6

k = 42

x 3 = 72

x 8 = 96

x = 768

y 3 = 4

m 7 = 8

m = 56

k 18 = 47

f 62 = 103

f = 6386

3.06 m = 12.546

5.012 k = 0.30072

k = 0.06

x 2.19 = 5

y 4.11 = 2.3

y = 9.453

4 y 7 = 2

3 m 10 = 1

m = 10 3

5 k 6 = 8

8 h 7 = 3

h = 21 8

16 z 21 = 4

Solve p q = 7 r for p .

p = 7 r q

Solve m 2 n = 2 s for n .

Solve 2.8 a b = 5.6 d for b .

b = 2 d a

Solve m n p 2 k = 4 k for p .

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

b = 15 c 8

Solve 3 p c b 2 m = 2 b for p c .

Solve 8 r s t 3 p = 2 p r s for t .

t = 3 p 2 4

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

Solve 3 Δ 2 = Δ for .

= 2 3

Exercises for review

( [link] ) Simplify ( 2 x 0 y 0 z 3 z 2 ) 5 .

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

binomial; 3rd degree; 10 , 7

( [link] ) Simplify 3 a 2 2 a + 4 a ( a + 2 ) .

( [link] ) Specify the domain of the equation y = 3 7 + x .

all real numbers except 7

( [link] ) Solve the conditional equation x + 6 = 2 .

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Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
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