11-3

1. Solving Equations with Variables on Both Sides 11-3 Course 3 Warm Up Problem of the Day Lesson Presentation

2. Solving Equations with Variables on Both Sides 11-3 1 1 2x 7 9x 8 16 4 x 2 7 7 Course 3 Warm Up Solve. 1.2x + 9x – 3x + 8 = 16 2.–4 = 6x + 22 – 4x 3. + = 5 4. – = 3 x = 1 x = -13 x = 34 x = 50

3. Solving Equations with Variables on Both Sides 11-3 Course 3 Problem of the Day An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle? 22.5 in.

4. Solving Equations with Variables on Both Sides 11-3 Course 3 Learn to solve equations with variables on both sides of the equal sign.

5. Solving Equations with Variables on Both Sides 11-3 Course 3 Some problems produce equations that have variables on both sides of the equal sign. Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.

6. Solving Equations with Variables on Both Sides 11-3 –3x = –3 6 –3 Course 3 Additional Example 1A: Solving Equations with Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x – 4x– 4x Subtract 4x from both sides. 6 = –3x Divide both sides by –3. –2 = x

7. Solving Equations with Variables on Both Sides 11-3 Helpful Hint Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2. Course 3

8. Solving Equations with Variables on Both Sides 11-3 4b 24 = 4 4 Course 3 Additional Example 1B: Solving Equations with Variables on Both Sides Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b– 5b Subtract 5b from both sides. 4b – 6 = 18 + 6+ 6 Add 6 to both sides. 4b = 24 Divide both sides by 4. b = 6

9. Solving Equations with Variables on Both Sides 11-3 9w + 3 = 9w + 7 – 9w– 9w Subtract 9w from both sides. Course 3 Additional Example 1C: Solving Equations with Variables on Both Sides Solve. 9w + 3 = 9w + 7 3 ≠ 7 No solution. There is no number that can be substituted for the variable w to make the equation true.

10. Solving Equations with Variables on Both Sides 11-3 Helpful Hint if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Course 3

11. Solving Equations with Variables on Both Sides 11-3 –4x = –4 8 –4 Course 3 Check It Out: Example 1A Solve. 5x + 8 = x 5x + 8 = x – 5x– 5x Subtract 5x from both sides. 8 = –4x Divide both sides by –4. –2 = x

12. Solving Equations with Variables on Both Sides 11-3 Course 3 Check It Out: Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b– 2b Subtract 2b from both sides. b – 2 = 12 + 2+ 2 Add 2 to both sides. b = 14

13. Solving Equations with Variables on Both Sides 11-3 3w + 1 = 3w + 8 – 3w– 3w Subtract 3w from both sides. Course 3 Check It Out: Example 1C Solve. 3w + 1 = 3w + 8 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true.

14. Solving Equations with Variables on Both Sides 11-3 Course 3 To solve multi-step equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

15. Solving Equations with Variables on Both Sides 11-3 8z8 = 8 8 Course 3 Additional Example 2A: Solving Multi-Step Equations with Variables on Both Sides Solve. 10z – 15 – 4z = 8 – 2z - 15 10z – 15 – 4z = 8 – 2z – 15 6z– 15 = –2z– 7 Combine like terms. + 2z+ 2z Add 2z to both sides. 8z – 15 = – 7 + 15+15 Add 15 to both sides. 8z = 8 Divide both sides by 8. z = 1

16. Solving Equations with Variables on Both Sides 11-3 7 10 7 10 7 10 7 10 3y 5 3y 5 3y 5 3y 5 y 5 y 5 y 5 y 5 3 4 3 4 3 4 3 4 + – = y – 20( ) = 20( ) + – y – 20() + 20( ) – 20( )= 20(y) – 20( ) Course 3 Additional Example 2B: Solving Multi-Step Equations with Variables on Both Sides + – = y – Multiply by the LCD, 20. 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14 Combine like terms.

17. Solving Equations with Variables on Both Sides 11-3 4y 4 –1 –1 4 = y = 4 Course 3 Additional Example 2B Continued 16y – 15 = 20y – 14 – 16y– 16y Subtract 16y from both sides. –15 = 4y – 14 + 14+ 14 Add 14 to both sides. –1 = 4y Divide both sides by 4.

18. Solving Equations with Variables on Both Sides 11-3 10z50 = 10 10 Course 3 Check It Out: Example 2A Solve. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z– 12 = –2z+ 38 Combine like terms. + 2z+ 2z Add 2z to both sides. 10z – 12 = 38 + 12+12 Add 12 to both sides. 10z = 50 Divide both sides by 10. z = 5

19. Solving Equations with Variables on Both Sides 11-3 6 8 6 8 6 8 6 8 5y 6 5y 6 5y 6 5y 6 y 4 y 4 y 4 y 4 3 4 3 4 3 4 3 4 + + = y – 24( ) = 24( ) + + y – 24() + 24( )+ 24( )= 24(y) – 24( ) Course 3 Check It Out: Example 2B + + = y – Multiply by the LCD, 24. 6y + 20y + 18 = 24y – 18 26y + 18 = 24y – 18 Combine like terms.

20. Solving Equations with Variables on Both Sides 11-3 2y 2 –36 2 = Course 3 Check It Out: Example 2B Continued 26y + 18 = 24y – 18 – 24y– 24y Subtract 24y from both sides. 2y + 18 = – 18 – 18– 18 Subtract 18 from both sides. 2y = –36 Divide both sides by 2. y = –18

21. Solving Equations with Variables on Both Sides 11-3 Course 3 Additional Example 3: Business Application Daisy’s Flowers sell a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florist’s bouquets cost the same price.

22. Solving Equations with Variables on Both Sides 11-3 13.95 1.55r1.55 = 1.55 Course 3 Additional Example 3 Continued Let r represent the price of one rose. 39.95 + 2.95r = 26.00 + 4.50r Subtract 2.95r from both sides. – 2.95r– 2.95r 39.95 = 26.00 + 1.55r Subtract 26.00 from both sides. – 26.00– 26.00 13.95 = 1.55r Divide both sides by 1.55. 9 = r The two services would cost the same when using 9 roses.

23. Solving Equations with Variables on Both Sides 11-3 Course 3 Check It Out: Example 3 Marla’s Gift Baskets sell a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both gift basket companies muffin baskets cost the same price.

24. Solving Equations with Variables on Both Sides 11-3 6.00 0.75b0.75 = .75 Course 3 Check It Out: Example 3 Continued Let b represent the price of one balloon. 22.00 + 2.25b = 16.00 + 3.00b Subtract 2.25b from both sides. – 2.25b– 2.25b 22.00 = 16.00 + 0.75b Subtract 16.00 from both sides. – 16.00– 16.00 6.00 = 0.75b Divide both sides by 0.75. 8 = b The two services would cost the same when using 8 balloons.

25. Solving Equations with Variables on Both Sides 11-3 Course 3 Additional Example 4: Multi-Step Application Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

26. Solving Equations with Variables on Both Sides 11-3 = 3d 3 0.75 3 Course 3 Additional Example 4 Continued First solve for the price of one doughnut. Let d represent the price of one doughnut. 1.25 + 2d = 0.50 + 5d – 2d– 2d Subtract 2d from both sides. 1.25 = 0.50 + 3d Subtract 0.50 from both sides. – 0.50– 0.50 0.75 = 3d Divide both sides by 3. The price of one doughnut is $0.25. 0.25 = d

27. Solving Equations with Variables on Both Sides 11-3 1.75 0.25 = 0.25n 0.25 Course 3 Additional Example 4 Continued Now find the amount of money Jamie spends each morning. Choose one of the original expressions. 1.25 + 2d 1.25 + 2(0.25) = 1.75 Jamie spends $1.75 each morning. Find the number of doughnuts Jamie buys on Tuesday. Let n represent the number of doughnuts. 0.25n = 1.75 Divide both sides by 0.25. n = 7; Jamie bought 7 doughnuts on Tuesday.

28. Solving Equations with Variables on Both Sides 11-3 Course 3 Check It Out: Example 4 Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?

29. Solving Equations with Variables on Both Sides 11-3 2 = 2 2x 2 Course 3 Check It Out: Example 4 Continued First solve for distance around the track. Let x represent the distance around the track. 2x + 4 = 4x + 2 – 2x– 2x Subtract 2x from both sides. 4 = 2x + 2 – 2– 2 Subtract 2 from both sides. 2 = 2x Divide both sides by 2. The track is 1 mile around. 1 = x

30. Solving Equations with Variables on Both Sides 11-3 Course 3 Check It Out: Example 4 Continued Now find the total distance Helene walks each day. Choose one of the original expressions. 2x + 4 2(1) + 4 = 6 Helene walks 6 miles each day. Find the number of laps Helene walks on Saturdays. Let n represent the number of 1-mile laps. 1n = 6 n = 6 Helene walks 6 laps on Saturdays.

31. Solving Equations with Variables on Both Sides 11-3 1 1 2 4 Course 3 Insert Lesson Title Here Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = 15 + 5x 3. 2(3x + 11) = 6x + 4 4.x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = –8 x = 6 no solution x = 36 An orange has 45 calories. An apple has 75 calories.