Additional Example 1: Using Inverse Operations to Group Terms with Variables Group the terms with variables on one side of the equal sign, and simplify. A. 60 – 4y = 8y 60 – 4y = 8y 60 – 4y + 4y = 8y + 4y Add 4y to both sides. 60 = 12y Simplify. B. –5b + 72 = –2b –5b + 72 = –2b –5b + 5b + 72 = –2b + 5b Add 5b to both sides. 72 = 3b Simplify.

Check It Out: Example 1 Group the terms with variables on one side of the equal sign, and simplify. A. 40 – 2y = 6y 40 – 2y = 6y 40 – 2y + 2y = 6y + 2y Add 2y to both sides. 40 = 8y Simplify. B. –8b + 24 = –5b –8b + 24 = –5b –8b + 8b + 24 = –5b + 8b Add 8b to both sides. 24 = 3b Simplify.

Additional Example 2A: Solving Equations with Variables on Both Sides Solve. 7c = 2c + 55 7c = 2c + 55 7c – 2c = 2c – 2c + 55 Subtract 2c from both sides. 5c = 55 Simplify. 5c = 55 5 Divide both sides by 5. c = 11

Additional Example 2B: Solving Equations with Variables on Both Sides Solve. 49 – 3m = 4m + 14 49 – 3m = 4m + 14 49 – 3m + 3m = 4m + 3m + 14 Add 3m to both sides. 49 = 7m + 14 Simplify. 49 – 14 = 7m + 14 – 14 Subtract 14 from both sides. 35 = 7m 35 = 7m 7 Divide both sides by 7. 5 = m

Additional Example 2C: Solving Equations with Variables on Both Sides Solve. 2 5 x = 1 x – 12 2 5 1 5 x = x – 12 2 5 x 15 – = 1 – 12 1 5 Subtract x from both sides. 1 5 x –12 = Simplify. 1 5 (5)‏ x = (5)(–12)‏ Multiply both sides by 5. x = –60

Check It Out: Example 2A Solve. 8f = 3f + 65 8f = 3f + 65 8f – 3f = 3f – 3f + 65 Subtract 3f from both sides. 5f = 65 Simplify. 5f = 65 5 Divide both sides by 5. f = 13

Check It Out: Example 2B Solve. 54 – 3q = 6q + 9 54 – 3q = 6q + 9 54 – 3q + 3q = 6q + 3q + 9 Add 3q to both sides. 54 = 9q + 9 Simplify. 54 – 9 = 9q + 9 – 9 Subtract 9 from both sides. 45 = 9q 45 = 9q 9 Divide both sides by 9. 5 = q

Check It Out: Example 2C Solve. 2 3 w = 1 w – 9 2 3 1 3 w = w – 9 1 3 2 3 w 13 – = 1 – 9 Subtract w from both sides. 1 3 w –9 = Simplify. 1 3 (3) w = (3)(–9) Multiply both sides by 3. w = –27

Additional Example 3: Consumer Math Application Christine can buy a new snowboard for $136.50. She will still need to rent boots for $8.50 a day. She can rent a snowboard and boots for $18.25 a day. How many days would Christine need to rent both the snowboard and the boots to pay as much as she would if she buys the snowboard and rents only the boots for the season?

Additional Example 3 Continued Let d represent the number of days. 18.25d = 136.5 + 8.5d Subtract 8.5d from both sides. 18.25d – 8.5d = 136.5 + 8.5d – 8.5d 9.75d = 136.5 Simplify. 9.75d = 136.5 Divide both sides by 9.75. 9.75 9.75 d = 14 Christine would need to rent both the snowboard and the boots for 14 days to pay as much as she would have if she had bought the snowboard and rented only the boots.

Check It Out: Example 3 A local telephone company charges $40 per month for services plus a fee of $0.10 a minute for long distance calls. Another company charges $75.00 a month for unlimited service. How many minutes does it take for a person who subscribes to the first plan to pay as much as a person who subscribes to the unlimited plan?

Check It Out: Example 3 Continued Let m represent the number of minutes. 75 = 40 + 0.10m Subtract 40 from both sides. 75 – 40 = 40 – 40 + 0.10m 35 = 0.10m Simplify. 35 0.10m 0.10 = Divide both sides by 0.10. 350 = m A person who subscribes to the first plan would have to use 350 minutes to pay as much as a person who subscribes to the unlimited plan.