1. Warm Up Solve. 1. 6n + 8 – 4n = 20 2. –4w + 16 – 4w = –32 3. 25t – 17 – 13t = 67 4. 4k + 9 Course 2 11-3 Solving Equations with Variables on Both Sides = 3535 –25

2. Learn to solve equations that have variables on both sides. Course 2 11-3 Solving Equations with Variables on Both Sides

3. Marilyn can rent a video game console for $2.20 per day or buy one for $74.80. The cost of renting a game is $1.99 per day. How many days would Marilyn have to rent both the game and the console to pay as much as she would if she had bought the console and rented the game instead? Problems such as this require you to solve equations that have the same variable on both sides of the equal sign. To solve this kind of problem, you need to get the terms with variables on one side of the equal sign. Course 2 11-3 Solving Equations with Variables on Both Sides

4. Group the terms with variables on one side of the equal sign, and simplify. Additional Example 1: Using Inverse Operations to Group Terms with Variables A. 60 – 4y = 8y 60 – 4y + 4y = 8y + 4y 60 = 12y B. –5b + 72 = –2b –5b + 72 = –2b –5b + 5b + 72 = –2b + 5b 72 = 3b Add 4y to both sides. Simplify. 60 – 4y = 8y Add 5b to both sides. Simplify. Course 2 11-3 Solving Equations with Variables on Both Sides

5. Group the terms with variables on one side of the equal sign, and simplify. A. 40 – 2y = 6y 40 – 2y + 2y = 6y + 2y 40 = 8y B. –8b + 24 = –5b –8b + 24 = –5b –8b + 8b + 24 = –5b + 8b 24 = 3b Add 2y to both sides. Simplify. 40 – 2y = 6y Add 8b to both sides. Simplify. Try This: Example 1 Course 2 11-3 Solving Equations with Variables on Both Sides

6. Solve. Additional Example 2A: Solving Equations with Variables on Both Sides A. 7c = 2c + 55 7c = 2c + 55 7c – 2c = 2c – 2c + 55 5c = 55 5 5 c = 11 Subtract 2c from both sides. Simplify. Divide both sides by 5. Course 2 11-3 Solving Equations with Variables on Both Sides

7. Additional Example 2B: Solving Equations with Variables on Both Sides Solve. B. 49 – 3m = 4m + 14 49 – 3m = 4m + 14 49 – 3m + 3m = 4m + 3m + 14 49 = 7m + 14 49 – 14 = 7m + 14 – 14 35 = 7m 7 7 5 = m Add 3m to both sides. Simplify. Subtract 14 from both sides. Divide both sides by 7. Course 2 11-3 Solving Equations with Variables on Both Sides

8. Additional Example 2C: Solving Equations with Variables on Both Sides C. 2525 x = 1515 x– 12 2525 x =x = 1515 x 2525 x 15 15 –x = 1515 x 1515 x – 1515 x = 1515 x(5)(–12) = (5) x = –60 Subtract 1515 x from both sides. Simplify. Multiply both sides by 5. Course 2 11-3 Solving Equations with Variables on Both Sides

9. Solve. A. 8f = 3f + 65 8f = 3f + 65 8f – 3f = 3f – 3f + 65 5f = 65 5 5 f = 13 Subtract 3f from both sides. Simplify. Divide both sides by 5. Try This: Example 2A Course 2 11-3 Solving Equations with Variables on Both Sides

10. Solve. B. 54 – 3q = 6q + 9 54– 3q = 6q + 9 54 – 3q + 3q = 6q + 3q + 9 54 = 9q + 9 54 – 9 = 9q + 9 – 9 45 = 9q 9 9 5 = q Add 3q to both sides. Simplify. Subtract 9 from both sides. Divide both sides by 9. Try This: Example 2B Course 2 11-3 Solving Equations with Variables on Both Sides

11. C. 2323 w = 1313 w– 9– 9 2323 w =w = 1313 w– 9 2323 w 13 13 –w = 1313 w 1313 w – 1313 w = 1313 w(3)(–9) = (3) w = –27 Subtract 1313 w from both sides. Simplify. Multiply both sides by 3. Try This: Example 2C Solve. Course 2 11-3 Solving Equations with Variables on Both Sides

12. 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: Consumer Math Application Course 2 11-3 Solving Equations with Variables on Both Sides

13. Additional Example 3 Continued 18.25d = 136.5 + 8.5d 18.25d – 8.5d = 136.5 + 8.5d – 8.5d 9.75d = 136.5 9.75 d = 14 Let d represent the number of days. Subtract 8.5d from both sides. Simplify. Divide both sides by 9.75. 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. Course 2 11-3 Solving Equations with Variables on Both Sides

14. Try This: 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? Insert Lesson Title Here Course 2 11-3 Solving Equations with Variables on Both Sides

15. Try This: Example 3 Continued Insert Lesson Title Here Let m represent the number of minutes. 75 = 40 + 0.10m 75 – 40 = 40 – 40 + 0.10m 350 = m Subtract 40 from both sides. Simplify. If you are going to use more than 350 minutes, it will be cheaper to subscribe to the unlimited plan. Divide both sides by 0.10. 35 = 0.10m Course 2 11-3 Solving Equations with Variables on Both Sides 35 0.10m 0.10 =

16. Assignment Page 570 – 571 –# 8 – 31, 35 - 43