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Algebraic Expressions and Equations: Solving Equations of the Form ax=b and x/a=b

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses solving equations of the form ax=bax=b size 12{"ax"=b} {} and xa=bxa=b size 12{ { {x} over {a} } =b} {}. By the end of the module students should be familiar with the multiplication/division property of equality, be able to solve equations of the form ax=bax=b size 12{ ital "ax"=b} {} and xa=bxa=b size 12{ { {x} over {a} } =b} {} and be able to use combined techniques to solve equations.

Section Overview

  • Multiplication/ Division Property of Equality
  • Combining Techniques in Equations Solving

Multiplication/ Division Property of Equality

Recall 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. From this, we can suggest the multiplication/division property of equality.

Multiplication/Division Property of Equality

Given any equation,

  1. We can obtain an equivalent equation by multiplying both sides of the equa­tion by the same nonzero number, that is, if c0c0 size 12{c <> 0} {}, then a=ba=b size 12{a=b} {} is equivalent to
    a c = b c a c = b c size 12{a cdot c=b cdot c} {}
  2. We can obtain an equivalent equation by dividing both sides of the equation by the same nonzero number, that is, if c0c0 size 12{c <> 0} {}, then a=ba=b size 12{a=b} {} is equivalent to
    a c = b c a c = b c size 12{ { {a} over {c} } = { {b} over {c} } } {}

The multiplication/division property of equality can be used to undo an association with a number that multiplies or divides the variable.

Sample Set A

Use the multiplication / division property of equality to solve each equation.

Example 1

6y=546y=54 size 12{6y="54"} {}
6 is associated with y by multiplication. Undo the association by dividing both sides by 6

6 y 6 = 54 6 6 y 6 = 54 9 6 y = 9 6 y 6 = 54 6 6 y 6 = 54 9 6 y = 9 alignl { stack { size 12{ { {6y} over {6} } = { {"54"} over {6} } } {} # size 12{ { { { {6}}y} over { { {6}}} } = { { { { {5}} { {4}}} cSup { size 8{9} } } over {6} } } {} # {} # y=9 {} } } {}

Check: When y=9y=9 size 12{y=9} {}

6 y = 54 6 y = 54 size 12{6y="54"} {}

becomes
Does 6 times 9 equal 54? Yes.,
a true statement.

The solution to 6 y=546 y=54 size 12{6y="54"} {} is y=9y=9 size 12{y=9} {}.

Example 2

x2=27x2=27 size 12{ { {x} over {-2} } ="27"} {}.
-2 is associated with xx size 12{x} {} by division. Undo the association by multiplying both sides by -2.

2 x 2 = 2 27 2 x 2 = 2 27 alignl { stack { size 12{ left (-2 right ) { {x} over {-2} } = left (-2 right )"27"} {} # {} } } {}

-2 x -2 = 2 27 -2 x -2 = 2 27 alignl { stack { size 12{ left ( - 2 right ) { {x} over { - 2} } = left ( - 2 right )"27"} {} # {} } } {}

x = 54 x = 54 size 12{x= - "54"} {}

Check: When x=54x=54 size 12{x= - "54"} {},

x 2 = 27 x 2 = 27 size 12{ { {x} over { - 2} } ="27"} {}

becomes
Does negative 54 over negative 2 equal 27? Yes.
a true statement.

The solution to x2=27x2=27 size 12{ { {2} over { - 2} } ="27"} {} is x=54x=54 size 12{x= - "54"} {}

Example 3

3a7=63a7=6 size 12{ { {3a} over {7} } =6} {}.
We will examine two methods for solving equations such as this one.

Method 1: Use of dividing out common factors.

3 a7=63 a7=6 size 12{ { {3a} over {7} } =6} {}
7 is associated with aa size 12{a} {} by division. Undo the association by multiplying both sides by 7.

7 3 a 7=767 3 a 7=76 size 12{7 cdot { {3a} over {7} } =7 cdot 6} {}
Divide out the 7’s.

7 3 a 7 = 42 7 3 a 7 = 42 size 12{ { {7}} cdot { {3a} over { { {7}}} } ="42"} {}

3 a=423 a=42 size 12{3a="42"} {}
3 is associated with aa size 12{a} {} by multiplication. Undo the association by dviding both sides by 3.

3 a 3 = 42 3 3 a 3 = 42 3 size 12{ { {3a} over {3} } = { {"42"} over {3} } } {}

3 a 3 = 14 3 a 3 = 14 size 12{ { { { {3}}a} over { { {3}}} } ="14"} {}

a = 14 a = 14 size 12{a="14"} {}

Check: When a=14a=14 size 12{a="14"} {},

3 a 7 = 6 3 a 7 = 6 size 12{ { {3a} over {7} } =6} {}

becomes
Does the quantity 3 times 14, divided by 7 equal 6? Yes.,
a true statement.

The solution to 3 a7=63 a7=6 size 12{ { {3a} over {7} } =6} {} is a=14a=14 size 12{a="14"} {}.

Method 2: Use of reciprocals

Recall that if the product of two numbers is 1, the numbers are reciprocals. Thus 3737 size 12{ { {3} over {7} } } {} and 7373 size 12{ { {7} over {3} } } {} are reciprocals.

3 a7=63 a7=6 size 12{ { {3a} over {7} } =6} {}
Multiply both sides of the equation by 7373 size 12{ { {7} over {3} } } {}, the reciprocal of 3737 size 12{ { {3} over {7} } } {}.

7 3 3 a 7 = 7 3 6 7 3 3 a 7 = 7 3 6 size 12{ { {7} over {3} } cdot { {3a} over {7} } = { {7} over {3} } cdot 6} {}

7 1 3 1 3 a 1 7 1 = 7 3 1 6 2 1 7 1 3 1 3 a 1 7 1 = 7 3 1 6 2 1 size 12{ { { { { {7}}} cSup { size 8{1} } } over { { { {3}}} cSub { size 8{1} } } } cdot { { { { {3}}a} cSup { size 8{1} } } over { { { {7}}} cSub { size 8{1} } } } = { {7} over { { { {3}}} cSub { size 8{1} } } } cdot { { { { {6}}} cSup { size 8{2} } } over {1} } } {}

1 a = 14 a = 14 1 a = 14 a = 14 alignl { stack { size 12{1 cdot a="14"} {} # size 12{a="14"} {} } } {}

Notice that we get the same solution using either method.

Example 4

8 x=248 x=24 size 12{-8x="24"} {}
-8 is associated with x x by multiplication. Undo the association by dividing both sides by -8.

8 x 8 = 24 8 8 x 8 = 24 8 alignl { stack { size 12{ { {-8x} over {-8} } = { {"24"} over {-8} } } {} # {} } } {}

8 x 8 = 24 8 8 x 8 = 24 8 size 12{ { {-8x} over {-8} } = { {"24"} over {-8} } } {}

x = - 3 x = - 3 size 12{x"=-"3} {}

Check: When x=3x=3 size 12{x= - 3} {},

8 x = 24 8 x = 24 size 12{ - 8x="24"} {}

becomes
Does negative 8 times negative 3 equal 24? Yes.,
a true statement.

Example 5

x=7.x=7. size 12{-x=7 "." } {}
Since x x is actually 1x1x size 12{-1 cdot x} {} and 11=111=1 size 12{ left (-1 right ) left (-1 right )=1} {}, we can isolate x x by multiplying both sides of the equation by 11 size 12{-1} {}.

1 x = - 1 7 x =- 7 1 x = - 1 7 x =- 7 alignl { stack { size 12{ left (-1 right ) left (-x right )"=-"1 cdot 7} {} # size 12{x"=-"7} {} } } {}

Check: When x=7x=7 size 12{x=7} {},

x = 7 x = 7 size 12{ - x=7} {}

becomes
graphics5.png

The solution to x=7x=7 size 12{ - x=7} {} is x=7x=7 size 12{x= - 7} {}.

Practice Set A

Use the multiplication/division property of equality to solve each equation. Be sure to check each solution.

Exercise 1

7x=217x=21 size 12{7x="21"} {}

Solution

x=3x=3 size 12{x=3} {}

Exercise 2

5x=655x=65 size 12{-5x="65"} {}

Solution

x=-13x=-13 size 12{x"=-""13"} {}

Exercise 3

x4= -8x4= -8 size 12{ { {x} over {4} } "=-"8} {}

Solution

x=-32x=-32 size 12{x"=-""32"} {}

Exercise 4

3x8=63x8=6 size 12{ { {3x} over {8} } =6} {}

Solution

x=16x=16 size 12{x="16"} {}

Exercise 5

y=3y=3 size 12{-y=3} {}

Solution

y=-3y=-3 size 12{y"=-"3} {}

Exercise 6

k=-2k=-2 size 12{-k"=-"2} {}

Solution

k=2k=2 size 12{k=2} {}

Combining Techniques in Equation Solving

Having examined solving equations using the addition/subtraction and the multi­plication/division principles of equality, we can combine these techniques to solve more complicated equations.

When beginning to solve an equation such as 6x-4=166x-4=16 size 12{6 ital "x-"4= - "16"} {}, it is helpful to know which property of equality to use first, addition/subtraction or multiplication/di­vision. Recalling that in equation solving we are trying to isolate the variable (disas­sociate numbers from it), it is helpful to note the following.

To associate numbers and letters, we use the order of operations.

  1. Multiply/divide
  2. Add/subtract

To undo an association between numbers and letters, we use the order of opera­tions in reverse.

  1. Add/subtract
  2. Multiply/divide

Sample Set B

Solve each equation. (In these example problems, we will not show the checks.)

Example 6

6x4= -166x4= -16 size 12{6x-4"=-""16"} {}
-4 is associated with x x by subtraction. Undo the association by adding 4 to both sides.

6x 4 + 4 = - 16 + 4 6x 4 + 4 = - 16 + 4 size 12{6x-4+4"=-""16"+4} {}

6x= -126x= -12 size 12{6x"=-""12"} {}
6 is associated with x x by multiplication. Undo the association by dividing both sides by 6

6x 6 = 12 6 6x 6 = 12 6 size 12{ { {6x} over {6} } = { {-"12"} over {6} } } {}

x = - 2 x = - 2 size 12{x"=-"2} {}

Example 7

8k+3=-45.8k+3=-45. size 12{-8k+3"=-""45" "." } {}
3 is associated with k k by addition. Undo the association by subtracting 3 from both sides.

8k + 3 3 = -45 3 8k + 3 3 = -45 3 size 12{-8k+3-3"=-""45"-3} {}

8k=- 488k=- 48 size 12{-8k"=-""48"} {}
-8 is associated with k k by multiplication. Undo the association by dividing both sides by -8.

8 k 8 = 48 8 8 k 8 = 48 8 size 12{ { {-8k} over {-8} } = { {-"48"} over {-8} } } {}

k = 6 k = 6 size 12{k=6} {}

Example 8

5m64m=4m8+3m.5m64m=4m8+3m. size 12{5m-6-4m=4m-8+3m "." } {} Begin by solving this equation by combining like terms.

m 6 = 7 m 8 m 6 = 7 m 8 size 12{m-6=7m-8} {} Choose a side on which to isolate m. Since 7 is greater than 1, we'll isolate m on the right side.
Subtract m from both sides.

m 6 m = 7m 8 m m 6 m = 7m 8 m size 12{-m-6-m=7m-8-m} {}

6=6m86=6m8 size 12{-6=6m-8} {}
8 is associated with m by subtraction. Undo the association by adding 8 to both sides.

6 + 8 = 6m 8 + 8 6 + 8 = 6m 8 + 8 size 12{-6+8=6m-8+8} {}

2=6m2=6m size 12{2=6m} {}
6 is associated with m by multiplication. Undo the association by dividing both sides by 6.

26=6m626=6m6 size 12{ { {2} over {6} } = { {6m} over {6} } } {} Reduce.

1 3 = m 1 3 = m size 12{ { {1} over {3} } =m} {}

Notice that if we had chosen to isolate m on the left side of the equation rather than the right side, we would have proceeded as follows:

m6=7m8m6=7m8 size 12{m-6=7m-8} {}
Subtract 7m7m from both sides.

m 6 7m = 7m 8 7m m 6 7m = 7m 8 7m size 12{m-6-7m=7m-8-7m} {}

6m6=-86m6=-8 size 12{-6m-6"=-"8} {}
Add 6 to both sides,

6m 6 + 6 = -8 + 6 6m 6 + 6 = -8 + 6 size 12{-6m-6+6"=-"8+6} {}

6m=- 26m=- 2 size 12{-6m"=-"2} {}
Divide both sides by -6.

6m 6 = 2 6 6m 6 = 2 6 size 12{ { {-6m} over {-6} } = { {-2} over {-6} } } {}

m = 1 3 m = 1 3 size 12{m= { {1} over {3} } } {}

This is the same result as with the previous approach.

Example 9

8 x 7=-28 x 7=-2 size 12{ { {8x} over {7} } "=-"2} {}
7 is associated with x x by division. Undo the association by multiplying both sides by 7.

7 8x 7 = 7 2 7 8x 7 = 7 2 size 12{7 cdot { {8x} over {7} } =7 left (-2 right )} {}

7 8 x 7 = - 14 7 8 x 7 = - 14 size 12{ { {7}} cdot { {8x} over { { {7}}} } "=-""14"} {}

8 x=-148 x=-14 size 12{8x"=-""14"} {}
8 is associated with x x by multiplication. Undo the association by dividing both sides by 8.

8 x 8 = 7 4 8 x 8 = 7 4 size 12{ { { { {8}}x} over { { {8}}} } = { {-7} over {4} } } {}

x = 7 4 x = 7 4 size 12{x= { {-7} over {4} } } {}

Practice Set B

Solve each equation. Be sure to check each solution.

Exercise 7

5m+7=-135m+7=-13 size 12{5m+7"=-""13"} {}

Solution

m=-4m=-4 size 12{m"=-"4} {}

Exercise 8

3a6=93a6=9 size 12{-3a-6=9} {}

Solution

a=-5a=-5 size 12{a"=-"5} {}

Exercise 9

2a+103a=92a+103a=9 size 12{2a+"10"-3a=9} {}

Solution

a=1a=1 size 12{a=1} {}

Exercise 10

11x413x=4x+1411x413x=4x+14 size 12{"11"x-4-"13"x=4x+"14"} {}

Solution

x=-3x=-3 size 12{x"=-"3} {}

Exercise 11

3m+8= -5m+13m+8= -5m+1 size 12{-3m+8"=-"5m+1} {}

Solution

m=-72m=-72 size 12{m"=-" { {7} over {2} } } {}

Exercise 12

5y+8y11=-115y+8y11=-11 size 12{5y+8y-"11""=-""11"} {}

Solution

y=0y=0 size 12{y=0} {}

Exercises

Solve each equation. Be sure to check each result.

Exercise 13

7x=427x=42 size 12{7x="42"} {}

Solution

x=6x=6 size 12{x=6} {}

Exercise 14

8x=818x=81 size 12{8x="81"} {}

Exercise 15

10x=12010x=120 size 12{"10"x="120"} {}

Solution

x=12x=12 size 12{x="12"} {}

Exercise 16

11x=12111x=121 size 12{"11"x="121"} {}

Exercise 17

6a=486a=48 size 12{-6a="48"} {}

Solution

a=-8a=-8 size 12{a"=-"8} {}

Exercise 18

9y=549y=54 size 12{-9y="54"} {}

Exercise 19

3y=-423y=-42 size 12{-3y"=-""42"} {}

Solution

y=14y=14 size 12{y="14"} {}

Exercise 20

5a=-1055a=-105 size 12{-5a"=-""105"} {}

Exercise 21

2m=-622m=-62 size 12{2m"=-""62"} {}

Solution

m=-31m=-31 size 12{m"=-""31"} {}

Exercise 22

3m=-543m=-54 size 12{3m"=-""54"} {}

Exercise 23

x4=7x4=7 size 12{ { {x} over {4} } =7} {}

Solution

x=28x=28 size 12{x="28"} {}

Exercise 24

y3=11y3=11 size 12{ { {y} over {3} } ="11"} {}

Exercise 25

z6=-14z6=-14 size 12{ { {-z} over {6} } "=-""14"} {}

Solution

z=84z=84 size 12{z="84"} {}

Exercise 26

w5=1w5=1 size 12{ { {-w} over {5} } =1} {}

Exercise 27

3m1=-133m1=-13 size 12{3m-1"=-""13"} {}

Solution

m= -4m= -4 size 12{m"=-"4} {}

Exercise 28

4x+7=-174x+7=-17 size 12{4x+7"=-""17"} {}

Exercise 29

2+9x=-72+9x=-7 size 12{2+9x"=-"7} {}

Solution

x= -1x= -1 size 12{x"=-"1} {}

Exercise 30

511x=27511x=27 size 12{5-"11"x="27"} {}

Exercise 31

32=4y+632=4y+6 size 12{"32"=4y+6} {}

Solution

y=132y=132 size 12{y= { {"13"} over {2} } } {}

Exercise 32

5+4= -8m+15+4= -8m+1 size 12{-5+4"=-"8m+1} {}

Exercise 33

3k+6=5k+103k+6=5k+10 size 12{3k+6=5k+"10"} {}

Solution

k=-2k=-2 size 12{k"=-"2} {}

Exercise 34

4a+16=6a+8a+64a+16=6a+8a+6 size 12{4a+"16"=6a+8a+6} {}

Exercise 35

6x+5+2x1=9x3x+156x+5+2x1=9x3x+15 size 12{6x+5+2x-1=9x-3x+"15"} {}

Solution

x=112 or 512x=112 or 512 size 12{x= { {"11"} over {2} } " or 5" { {1} over {2} } } {}

Exercise 36

9y8+3y+7= -7y+8y5y+99y8+3y+7= -7y+8y5y+9 size 12{-9y-8+3y+7"=-"7y+8y-5y+9} {}

Exercise 37

3a=a+53a=a+5 size 12{-3a=a+5} {}

Solution

a= -54a= -54 size 12{a"=-" { {5} over {4} } } {}

Exercise 38

5b=-2b+8b+15b=-2b+8b+1 size 12{5b"=-"2b+8b+1} {}

Exercise 39

3m+28m4= -14m+m43m+28m4= -14m+m4 size 12{-3m+2-8m-4"=-""14"m+m-4} {}

Solution

m=-1m=-1 size 12{m"=-"1} {}

Exercise 40

5a+3=35a+3=3 size 12{5a+3=3} {}

Exercise 41

7x+3x=07x+3x=0 size 12{7x+3x=0} {}

Solution

x=0x=0 size 12{x=0} {}

Exercise 42

7g+411g=-4g+1+g7g+411g=-4g+1+g size 12{7g+4-"11"g"=-"4g+1+g} {}

Exercise 43

5 a7=10 5 a7=10 size 12{ { {5a} over {7} } ="10"} {}

Solution

a=14a=14 size 12{a="14"} {}

Exercise 44

2m9=42m9=4 size 12{ { {2m} over {9} } =4} {}

Exercise 45

3x4=923x4=92 size 12{ { {3x} over {4} } = { {9} over {2} } } {}

Solution

x=6x=6 size 12{x=6} {}

Exercise 46

8k3=328k3=32 size 12{ { {8k} over {3} } ="32"} {}

Exercise 47

3a832=03a832=0 size 12{ { {3a} over {8} } - { {3} over {2} } =0} {}

Solution

a=4a=4 size 12{a=4} {}

Exercise 48

5m6253=05m6253=0 size 12{ { {5m} over {6} } - { {"25"} over {3} } =0} {}

Exercises for Review

Exercise 49

((Reference)) Use the distributive property to compute 40284028 size 12{"40" cdot "28"} {}.

Solution

40302=120080=112040302=120080=1120 size 12{"40" left ("30"-2 right )="1200"-"80"="1120"} {}

Exercise 50

((Reference)) Approximating π π by 3.14, find the approximate circumference of the circle.

A circle with radius 8cm.

Exercise 51

((Reference)) Find the area of the parallelogram.

A parallelogram with base 20cm and height 11cm.

Solution

220 sq cm

Exercise 52

((Reference)) Find the value of 341525341525 size 12{ { {-3 left (4-"15" right )-2} over {-5} } } {}.

Exercise 53

((Reference)) Solve the equation x14+8= -2.x14+8= -2. size 12{x-"14"+8"=-"2 "." } {}

Solution

x=4x=4 size 12{x=4} {}

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