EXAMPLE 1 Solve an equation with variables on both sides

7 – 8x = 4x – 17

7 – 8x + 8x = 4x – 17 + 8x

7 = 12x – 17

24 = 12x

Write original equation.

Add 8x to each side.

Simplify each side.

Add 17 to each side.

Divide each side by 12.

ANSWER

The solution is 2. Check by substituting 2 for x in the original equation.

Solve 7 – 8x = 4x – 17.

2 = x

EXAMPLE 1 Solve an equation with variables on both sides

– 9 = – 9

Write original equation.

Substitute 2 for x.

Simplify left side.

Simplify right side. Solution checks.

29 = 4(2) – 17?

7 – 8(2) = 4(2) – 17?

CHECK 7 – 8x = 4x – 17

EXAMPLE 2 Solve an equation with grouping symbols

14

(16x + 60).9x – 5 =

9x – 5 = 4x + 15

5x – 5 = 15

5x = 20

x = 4

Write original equation.

Distributive property

Subtract 4x from each side.

Add 5 to each side.

Divide each side by 5.

9x – 5 =

14 (16x + 60).Solve

GUIDED PRACTICE for Examples 1 and 2

24 – 3m = 5m

24 – 3m + 3m = 5m + 3m

24 = 8m

3 = m

Write original equation.

Add 3m to each side.

Simplify each side.

Divide each side by 8.

ANSWER

The solution is 3. Check by substituting 3 for m in the original equation.

1. 24 – 3m = 5m.

15 = 15

Write original equation.

Substitute 3 for m.

Simplify left side.

Simplify right side. Solution checks.

15 = 5(3)?

24 – 3(3) = 5(3)?

24 – 3m = 5m

GUIDED PRACTICE for Examples 1 and 2

CHECK

GUIDED PRACTICE for Examples 1 and 2

20 + c = 4c – 7

20 + c – c = 4c – c – 7

20 = 3c – 7

27 = 3c

Write original equation.

Subtract c from each side.

Simplify each side.

Divide each side by 3.

ANSWER

The solution is 9. Check by substituting 9 for c in the original equation.

2. 20 + c = 4c – 7 .

9 = c

Add 7 to each side.

29 = 29

Write original equation.

Substitute 9 for c.

Simplify left side.

Simplify right side. Solution checks.

29 = 4(9) – 7?

20 + 9 = 4(9) – 7?

20 + c = 4c – 7

GUIDED PRACTICE for Examples 1 and 2

CHECK

GUIDED PRACTICE for Examples 1 and 2

9 – 3k = 17k – 2k

9 – 3k + 3k = 17k – 2k + 3k

9 = 17k + k

Write original equation.

Add 3k to each side.

Simplify each side.

Subtract 17 from each side.

ANSWER

The solution is – 8. Check by substituting – 8 for k in the original equation.

3. 9 – 3k = 17k – 2k .

– 8 = k

33 = 33

Write original equation.

Substitute – 8 for k.

Simplify left side.

Simplify right side. Solution checks.

33 = 17 – (– 8)2?

9 –3(– 8) = 17 – (– 8)2?

9 – 3k = 17 – 2k

GUIDED PRACTICE for Examples 1 and 2

CHECK

GUIDED PRACTICE for Examples 1 and 2

5z – 2 = 2(3z – 4)

5z – 2 = 6z – 8

– z – 2 = – 8

Write original equation.

Distributive property.

Subtract 6z from each side.

Add z to each side.

ANSWER

The solution is 6. Check by substituting 6 for z in the original equation.

4. 5z – 2 = 2(3z – 4) .

z = 6

28 = 28

Write original equation.

Substitute 6 for z.

Simplify left side.

Simplify right side. Solution checks.

28 = 2(3(6) – 4)?

5(6) – 2 = 2(3(6) – 4)?

5z – 2 = 2(3z – 4)

GUIDED PRACTICE for Examples 1 and 2

CHECK

GUIDED PRACTICE for Examples 1 and 2

3 – 4a = 5(a – 3)

3 – 4a = 5a – 15

3 – 9a = – 15

Write original equation.

Distributive property.

Subtract 5a from each side.

Subtract 3 from each side.

ANSWER

The solution is 2. Check by substituting 2 for a in the original equation.

5. 3 – 4a = 5(a – 3) .

– 9a = – 18

a = 2 Divide each side by – 9.

– 5 = – 5

Write original equation.

Substitute 2 for a.

Simplify left side.

Simplify right side. Solution checks.

– 5 = 5(2 – 3)?

3 – 4(2) = 5(2 – 3) ?

3 – 4a = 5(a – 3)

GUIDED PRACTICE for Examples 1 and 2

CHECK

GUIDED PRACTICE for Examples 1 and 2

8y – 6 = 4y + 10

4y – 6 = 10

4y = 16

y = 4

Write original equation.

Distributive property

Subtract 4y from each side.

Add 6 to each side.

Divide each side by 4.

8y – 6 =23 (6y + 15).6.

ANSWERThe solution is 4. Check by substituting 4 for y in the original equation.

8y – 6 =23 (6y + 15).

26 = 26

Write original equation.

Substitute 4 for y.

Simplify left side.

Simplify right side. Solution checks.

GUIDED PRACTICE for Examples 1 and 2

8y – 6 =23 (6y + 15).

8(4) – 6 = (6(4) + 15) ? 2

3

26 = (6(4) + 15)? 2

3

CHECK

CAR SALES

Solve a real-world problem

EXAMPLE 3

A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new cars sold be twice the number of used cars sold?

SOLUTION

Solve a real-world problem

EXAMPLE 3

Let x represent the number of years from now. So, 6x represents the increase in the number of new cars sold over x years and – 4x represents the decrease in the number of used cars sold over x years. Write a verbal model.

6778 + 6x = 2 ( + (– 4 x) )

Solve a real-world problem

EXAMPLE 3

78 + 6x = 2(67 – 4x)

78 + 6x = 134 – 8x

78 + 14x = 134

14x = 56

x = 4

Write equation.

Distributive property

Add 8x to each side.

Subtract 78 from each side.

Divide each side by 14.

ANSWER

The number of new cars sold will be twice the number of used cars sold in 4 years.

Solve a real-world problem

EXAMPLE 3

CHECKYou can use a table to check your answer.

YEAR 0 1 2 3 4

Used car sold 67 63 59 55 51

New car sold 78 84 90 96 102

GUIDED PRACTICE for Example 3

7.

WHAT IF? Suppose the car dealership sold 50 new cars this year instead of 78. In how many years will the number of new cars sold be twice the number of used cars sold?

GUIDED PRACTICE for Example 3

SOLUTION

Let x represent the number of years from now. So, 6x represents the increase in the number of new cars sold over x years and – 4x represents the decrease in the number of used cars sold over x years. Write a verbal model.

6750 + 6x = 2 ( + (–4x) )

GUIDED PRACTICE for Example 3

50 + 6x = 134 – 8x

50 + 14x = 134

14x = 84

x = 6

Write equation.

Distributive property

Add 8x to each side.

Subtract 50 from each side.

Divide each side by 14.

ANSWER

The number of new cars sold will be twice the number of used cars sold in 6 years.

50 + 6x = 2(67 +(– 4x))

SOLUTION

EXAMPLE 4 Identify the number of solutions of an equation

Solve the equation, if possible.

a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5)

a. 3x = 3(x + 4) Original equation

3x = 3x + 12 Distributive property

The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

ANSWER

The statement 0 = 12 is not true, so the equation hasno solution.

Simplify.

3x – 3x = 3x + 12 – 3x Subtract 3x from each side.

0 = 12

EXAMPLE 4 Identify the number of solutions of an equation

EXAMPLE 1

b. 2x + 10 = 2(x + 5) Original equation

2x + 10 = 2x + 10 Distributive property

ANSWER

Notice that the statement 2x + 10 = 2x + 10 is true for all values of x.So, the equation is an identity, and the solution is all real numbers.

EXAMPLE 4 Identify the number of solutions of an equation

GUIDED PRACTICE for Example 4

8. 9z + 12 = 9(z + 3)

SOLUTION

9z + 12 = 9(z + 3) Original equation

9z + 12 = 9z + 27 Distributive property

The equation 9z + 12 = 9z + 27 is not true because the number 9z + 12 cannot be equal to 27 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

GUIDED PRACTICE for Example 4

ANSWER

The statement 12 = 27 is not true, so the equation hasno solution.

Simplify.

9z – 9z + 12 = 9z – 9z + 27 Subtract 9z from each side.

12 = 27

GUIDED PRACTICE for Example 4

9. 7w + 1 = 8w + 1

SOLUTION

– w + 1 = 1

Subtract 1 from each side.– w = 0

Subtract 8w from each side.

ANSWER

w = 0

GUIDED PRACTICE for Example 4

10. 3(2a + 2) = 2(3a + 3)

SOLUTION

3(2a + 2) = 2(3a + 3)

Distributive property6a + 6 = 6a + 6

Original equation

ANSWER

The statement 6a + 6 = 6a + 6 is true for all values of a. So, the equation is an identity, and the solution is all real numbers.