1. Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 2. – 4 = 6x + 22 – 4x 3. + = 5 4. – = 3 x = 1 x = -13 x = 34 Course 3 11-3 Solving Equations with Variables on Both Sides 2 7 x 77 1 9x9x 16 2x2x 4 1 8 x = 50

2. 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. Course 3 11-3 Solving Equations with Variables on Both Sides

3. Learn to solve equations with variables on both sides of the equal sign. Course 3 11-3 Solving Equations with Variables on Both Sides TB P. 593-597

4. 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. Course 3 11-3 Solving Equations with Variables on Both Sides

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

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

7. Solve. 9b – 6 = 5b + 18 – 5b 4b – 6 = 18 4b4b 4 24 4 = Subtract 5b from both sides. Divide both sides by 4. b = 6 + 6 4b = 24 Add 6 to both sides. Course 3 11-3 Solving Equations with Variables on Both Sides

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

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

10. 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. Course 3 11-3 Solving Equations with Variables on Both Sides

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

12. Multiply by the LCD, 20. 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14Combine like terms. y5y5 3434 3y53y5 7 10 + – = y – y5y5 3434 3y53y5 7 10 + – = y – 20 ( ) = 20 ( ) y5y5 3434 3y53y5 7 10 + – y – 20 ( ) + 20 ( ) – 20 ( ) = 20(y) – 20 ( ) y5y5 3y53y5 3434 7 10 Course 3 11-3 Solving Equations with Variables on Both Sides

13. Add 14 to both sides. –15 = 4y – 14 –1 = 4y + 14 –1 4 4y4y 4 = Divide both sides by 4. –1 4 = y 16y – 15 = 20y – 14 – 16y Subtract 16y from both sides. Course 3 11-3 Solving Equations with Variables on Both Sides

14. 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 florists’ bouquets cost the same price. Course 3 11-3 Solving Equations with Variables on Both Sides

15. 39.95 + 2.95r = 26.00 + 4.50r Let r represent the price of one rose. – 2.95r 39.95 = 26.00 + 1.55r Subtract 2.95r from both sides. – 26.00 Subtract 26.00 from both sides. 13.95 = 1.55r 13.95 1.55 1.55r 1.55 = Divide both sides by 1.55. 9 = r The two services would cost the same when purchasing 9 roses. Course 3 11-3 Solving Equations with Variables on Both Sides

16. 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? Course 3 11-3 Solving Equations with Variables on Both Sides

17. First solve for the price of one doughnut. 1.25 + 2d = 0.50 + 5d Let d represent the price of one doughnut. – 2d 1.25 = 0.50 + 3d Subtract 2d from both sides. – 0.50 Subtract 0.50 from both sides. 0.75 = 3d 0.75 3 3d3d 3 = Divide both sides by 3. 0.25 = d The price of one doughnut is $0.25. Course 3 11-3 Solving Equations with Variables on Both Sides

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

19. 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 = 6 x = –8 Insert Lesson Title Here no solution x = 36 1 4 1 2 An orange has 45 calories. An apple has 75 calories. Course 3 11-3 Solving Equations with Variables on Both Sides