Basic Mathematics for College Students

670 Chapter 8 An Introduction to Algebra

Self Check 3

Solve:

Now TryProblem 29

7

12 a^6 ^27

Self Check 4

Solve:

Now TryProblem 35

6.6m2.7

EXAMPLE 3

Solve:

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

the equation.

WHYTo solve the original equation, we want to find a simpler equivalent

equation of the form , whose solution is obvious.

Solution

We note that the coefficient of is and proceed as follows.

• To isolate the variable term , we add 2 to both sides to undo the
  subtraction of 2.


• To isolate the variable, , we multiply both sides by to undo the
  multiplication by.

This is the equation to solve.

First, we want to isolate the variable term,.

Use the addition property of equality: Add 2

to both sides to isolate.

Use the multiplication property of equality:

Multiply both sides by which is the

reciprocal of to isolate.

On the left side: and. On the

right side:.

The solution is  16. Check by substituting it into the original equation.

8

5 (10)

8  2  5

1

51 ^16

851852  1 1 mm

m 16

852 m

(^85 1)

8

5

a

5

8

mb

8

5

(10)

Do the additions:  2  2 0 and  12  2 10.

m (^10) Now we want to isolate the variable, m.

5

8

(^85) m

5

8

m 2  2  12  2

85 m

 2  12

5

8

m

58

m 85

58 m

m 58

ma number

5

8

m 2  12

EXAMPLE (^4) Solve:
StrategyFirst, we will use a property of equality to isolate the variable term on
one side of the equation. Then we will use a second property of equality to isolate
the variable itself.
WHYTo solve the original equation, we want to find a simpler equivalent
equation of the form , whose solution is obvious.
Solution
To isolate the variable term on the right side, we eliminate by adding 0.8
to both sides.
Add 0.8 to both sides to isolate.
Do the additions.
Since the term has an understood coefficient of , the equation can be
written as. To isolate , we can either multiply both sides or divide both
sides by. If we choose to divide both sides by , we proceed as follows.
Now we want to isolate the variable y.
On the left side, do the division. The quotient of a positive and a negative
number is negative. On the right side, simplify:.
The solution is 0.6. Check this by substituting it into the original equation.
0.6y
 1
1
y
 11 y^

0.6

 1



 1 y

 1

0.6 1 y

 1  1

0.6 1 y y

y  1

0.6y

0.20.80.8y0.8 y

This is the equation to solve. First, we

 0.20.8 y want to isolate the variable term, y.

y 0.8

a numbery

0.20.8y

0.8

0.2

0.6