# Solve Each Question before you click

## Presentation on theme: "Solve Each Question before you click"— Presentation transcript:

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

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

Insert Lesson Title Here
Course 2 2-12 Solving Equations by Multiplying or Dividing Insert Lesson Title Here 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

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

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

Lesson Quiz Combine like terms. 1. 3x + 4 + 2x 2. 13k + 6  8k + 9 + k
Simplify. 3. 4(3x + 6)  7x (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? 5x + 4 6k + 15 5x + 24 9x + 30 y = 6 \$3

Solving Two-Step Equations Insert Lesson Title Here
Course 2 11-1 Solving Two-Step Equations Insert Lesson Title Here Lesson Quiz Solve. Check your answers. 1. 6x + 8 = 44 2. 14y – 14 = 28 3. 4. x = 6 y = 3 m 7 + 3 = 12 m = 63 v –8 – 6 = 8 v = –112 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

Insert Lesson Title Here
Lesson Quiz Solve. 1. c c = 63 2. –x – x = 53 3. w – w = 59 4. 4k + 6 c = 7 x = 4 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

Solving Equations with Variables on Both Sides
Course 2 11-3 Solving Equations with Variables on Both Sides 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 3 5 = k = –6 –25

Insert Lesson Title Here
Course 2 11-3 Solving Equations with Variables on Both Sides Insert Lesson Title Here 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. y = –4y – 19 4. – 3n = 81 4k = –12 y = –11 3 4 1 8 x = x – 14 x = 16

Insert Lesson Title Here
Course 2 11-3 Solving Equations with Variables on Both Sides Insert Lesson Title Here 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. y = –4y – 19 4. – 3n = 81 4k = –12 y = –11 3 4 1 8 x = x – 14 x = 16

Additional Example 3: Consumer Math Application
Course 2 11-3 Solving Equations with Variables on Both Sides Additional Example 3: Consumer Math Application Christine can buy a new snowboard for \$ 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?

Course 2 11-3 Solving Equations with Variables on Both Sides Additional Example 3 Continued Let d represent the number of days. 18.25d = d Subtract 8.5d from both sides. 18.25d – 8.5d = d – 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.

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

Homework Work Sheet – go see Mr. Todd