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Solve Each Question before you click. Check you answers, look at your notes if you need.

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Lesson Quiz: Part 1 Solve the equation. Check your answer. 1. 12 = 4x 2. 18z = 90 3. 12 = 4. 840 = 12y 5. z = 5 x = 3 Insert Lesson Title Here x = 48 y = 70 x4x4 h 22 = 9 h = 198 Course 2 2-12 Solving Equations by Multiplying or Dividing

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Lesson Quiz: Part 2 6. The cost of each ticket at the carnival was $0.25. Li bought $7.50 worth of tickets. How many tickets did she buy? 30 Insert Lesson Title Here Course 2 2-12 Solving Equations by Multiplying or Dividing

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Warm Up Solve. 1. n + 9 = 17 2. 6x = 42 3. 71 – z = 55 4. n = 8 x = 7 z = 16 = 9 y = 72 y8y8

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1) 5x + 4y – x + 3x2) 7(4y -2) + 2x - y 3) 35 + 3x + 4g – x + 3g – g + 7 4) 3(3x + 2) – x Combine Like Terms 7x + 4y 28x – 14 + 2x -y 30x – 14 - y 42 + 2x + 6g 9x + 6 - x 8x + 6

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Lesson Quiz Combine like terms. 1. 3x + 4 + 2x 2. 13k + 6 8k + 9 + k Simplify. 3. 4(3x + 6) 7x 4. 6(x + 5) + 3x Solve. 5. 6y + y = 42 6. The accounting department ordered 15 boxes of pens. The marketing department ordered 9 boxes of pens. If the total cost of the combined order was $72, what is the price of each box of pens? 6k + 15 y = 6 5x + 4 5x + 249x + 30 $3

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Lesson Quiz Solve. Check your answers. 1. 6x + 8 = 44 2. 14y – 14 = 28 3. 4. y = 3 x = 6 Insert Lesson Title Here m = 63 v = –112 m7m7 + 3 = 12 – 6 = 8 5. Last Sunday, the Humane Society had a 3-hour adoption clinic. During the week the clinic is open for 2 hours on days when volunteers are available. If the Humane Society was open for a total of 9 hours last week, how many weekdays was the clinic open? 3 days v –8 Course 2 11-1 Solving Two-Step Equations

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Lesson Quiz Solve. 1. c + 21 + 5c = 63 2. –x – 11 + 17x = 53 3. w – 16 + 4w = 59 4. 4k + 6 x = 4 c = 7 Insert Lesson Title Here w = 15 k = 11 = 10 5 5. Kelly swam 4 times as many laps as Kathy. Adding 5 to the number of laps Kelly swam gives you the number of laps Julie swam. If Julie swam 9 laps, how many laps did Kathy swim? 1 lap

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Warm Up Solve. 1. 6n + 8 – 4n = 20 2. –4w + 16 – 4w = –32 3. 25t – 17 – 13t = 67 4. 4k + 9 n = 6 w = 6 t = 7 Course 2 11-3 Solving Equations with Variables on Both Sides = 3535 k = –6 –25

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Lesson Quiz Group the terms with variables on one side of the equal sign, and simplify. 1. 14n = 11n + 81 2. –14k + 12 = –18k Solve. 3. 58 + 3y = –4y – 19 4. – 4k = –12 3n = 81 Insert Lesson Title Here y = –11 x = 16 3434 x = 1818 x – 14 Course 2 11-3 Solving Equations with Variables on Both Sides

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Lesson Quiz Group the terms with variables on one side of the equal sign, and simplify. 1. 14n = 11n + 81 2. –14k + 12 = –18k Solve. 3. 58 + 3y = –4y – 19 4. – 4k = –12 3n = 81 Insert Lesson Title Here y = –11 x = 16 3434 x = 1818 x – 14 Course 2 11-3 Solving Equations with Variables on Both Sides

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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

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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

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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

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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 =

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Homework Work Sheet – go see Mr. Todd

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