Beginning Algebra, 11th Edition

What happens in Example 4if we start by subtracting from each side?

Original equation from Example 4

Subtract from each side.

;

Subtract 17 from each side.

Combine like terms; additive inverse

This result gives the value of but not of xitself. However, it does say that the

additive inverse of xis which means that xmust be 17.

Same result as in Example 4

We can make the following generalization:

Ifais a number andxa, thenxa.

x= 17

- 17,

- x,

- x=- 17

17 - x- 17 = 0 - 17

8

5 x-

8

5 x=^0

3

5 x-

8

5 x=-

5

17 - x= 0 5 x=- 1 x=-x

8

5 x

3

5

x+ 17 -

8

5

x=

8

5

x-

8

5

x

3

5

x+ 17 =

8

5

x

8

5 x

SECTION 2.1 The Addition Property of Equality 89

Combining Like Terms When Solving

Solve

Combine like terms.

Subtract 3tfrom each side.

Combine like terms.

Add 10 to each side.

t= 17 Combine like terms.

t- 10 + 10 = 7 + 10

t- 10 = 7

4 t- 10 - 3 t= 7 + 3 t- 3 t

4 t- 10 = 7 + 3 t

3 t- 12 +t+ 2 = 5 + 3 t+ 2

3 t- 12 +t+ 2 = 5 + 3 t+2.

EXAMPLE 6

NOW TRY
EXERCISE 5
Solve. 6 x- 8 = 12 + 5 x

Applying the Addition Property of Equality Twice

Solve.

Add 7pto each side.

Combine like terms.

Subtract 8 from each side.

Combine like terms.

CHECK Original equation

Let

Multiply.

✓ True

The check results in a true statement, so the solution set is 5 - 36. NOW TRY

26 = 26

8 + 18  21 + 5

8 - 61 - 32 - 71 - 32 + 5 p=-3.

8 - 6 p=- 7 p+ 5

p=- 3

8 + p- 8 = 5 - 8

8 + p= 5

8 - 6 p+ 7 p=- 7 p+ 5 + 7 p

8 - 6 p=- 7 p+ 5

8 - 6 p=- 7 p+ 5

EXAMPLE 5

Use parentheses

when substituting

to avoid errors.

NOTE There are often several correct ways to solve an equation.In Example 5,we

could begin by adding 6pto each side. Combining like terms and subtracting 5 from

each side gives. (Try this.) If , then , and the variable has

been isolated on the right side of equation. The same solution results.

3 =-p 3 =-p - 3 =p

OBJECTIVE 3 Simplify, and then use the addition property of equality.

NOW TRY ANSWER

5. 5206