 # 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.

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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 1 4 (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 =9x – 5 = 1 4 (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 =8y – 6 = 2 3 (6y + 15). 6.6. ANSWER The solution is 4. Check by substituting 4 for y in the original equation. 8y – 6 =8y – 6 = 2 3 (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 =8y – 6 = 2 3 (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 + 6x6x = 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 CHECK You can use a table to check your answer. YEAR 0 1 2 3 4 Used car sold 6763 59 55 51 New car sold 7884 9096102

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 + 6x6x = 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 has no 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 has no 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 property 6a + 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.

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