Basic Mathematics for College Students

Success Tip In Example 4a, the solution, , is an improper fraction. If you

were inclined to write it as the mixed number 1 , that is not necessary. It is

common practice in algebra to leave such solutions in improper fraction form.

Just make sure that they are simplified (the numerator and denomintor have

no common factors other than 1).

Use the multiplication property of equality.

The first scale shown on the right

represents the equation. The scale is

in balance because the weights on the left

side and right side are equal. To find , we

must triple (multiply by 3) the weight on

the left side. To keep the scale in balance,

we must also triple the weight on the right

side. After doing this, we see that is

balanced by 75. Therefore, must be 75.

The procedure that we used to solve

illustrates the following property of

equality.

Multiplication Property of Equality

Multiplying both sides of an equation by the same nonzero number does not

change its solution.

For any numbers , , and , where is not 0,

if , then

When we use this property, the resulting equation is equivalent to the original

one. We will now show how it is used to solve x 3  25 algebraically.

ab cacb

ab c c

x–

3 = 25

x = 75

x–

3

x–

3

x–

3

• x
  3

Triple Triple

25

x 25 25 25

3 ^25

x

x

x

x 3  25

4

(^58)
13
8
662 Chapter 8 An Introduction to Algebra
Self Check 5
Solve:
Now TryProblem 53
b
24

 3

b.To isolate , we use the subtraction property of equality. We can undo the

addition of 54.9 by subtracting 54.9 from both sides.

This is the equation to solve.

Subtract 54.9 from both sides.

On the left side,.

Check: This is the original equation.

Substitute for.

True

The solution is 9.7.

45.245.2

54.9(9.7)45.2 9.7 x

54.9x45.2

x9.7 54.954.9 0

54.9x54.945.254.9

54.9x45.2

x

5

4

4

14

.9

 9.7

45.2

5

4

4

14

.9

 45.2

9.7

EXAMPLE 5

Solve:

StrategyWe will use a property of equality to isolate the variable on one side of

the equation.

x

3

 25