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1-8 Solving Equations by Multiplying or Dividing Course 3 Lesson Presentation Lesson Presentation.

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Presentation on theme: "1-8 Solving Equations by Multiplying or Dividing Course 3 Lesson Presentation Lesson Presentation."— Presentation transcript:

1 1-8 Solving Equations by Multiplying or Dividing Course 3 Lesson Presentation Lesson Presentation

2 Learn to solve equations using multiplication and division. 1-8 Solving Equations by Multiplying or Dividing Course 3

3 You can solve a multiplication equation using the Division Property of Equality. You can divide both sides of an equation by the same nonzero number, and the equation will still be true. 4 3 = 12 22 x = y DIVISION PROPERTY OF EQUALITY WordsNumbersAlgebra z 4 3 = 12 12 = 6 2 z 1-8 Solving Equations by Multiplying or Dividing Course 3

4 Solve 6x = 48. Example 1A: Solving Equations Using Division 6x = 48 1x = 8 Divide both sides by 6. Check 6x = 48 66 x = 8 6x = 48 6(8) = 48 ? 48 = 48 ? Substitute 8 for x. 1 x = x 1-8 Solving Equations by Multiplying or Dividing Course 3

5 Solve –9y = 45. Example 1B: Solving Equations Using Division –9y = 45 1y = –5 Divide both sides by –9. Check –9y = 45 –9 y = –5 –9y = 45 –9(–5) = 45 ? 45 = 45 ? Substitute –5 for y. 1 y = y 1-8 Solving Equations by Multiplying or Dividing Course 3

6 You can multiply both sides of an equation by the same number, and the statement will still be true. 2 3 = 6 x = y z z MULTIPLICATION PROPERTY OF EQUALITY WordsNumbersAlgebra 2 3 = 6 4 4 8 3 = 24 You can solve a division equation using the Multiplication Property of Equality. 1-8 Solving Equations by Multiplying or Dividing Course 3

7 Solve = 5. Example 2: Solving Equations Using Multiplication b –4 b = 5 –4 Multiply both sides by –4. b = –20 Check b –4 = 5 Substitute –20 for b. –20 –4 = 5 ? 5 = 5 ? 1-8 Solving Equations by Multiplying or Dividing Course 3

8 Example 3: Money Application fraction of amount raised so far amount raised so far = = x670 4 x = 2680 To go on a school trip, Helene has raised $670, which is one-fourth of the amount she needs. What is the total amount needed? Write the equation. Multiply both sides by 4. total amount needed 1 4 x = 670 1 4 4 1 4 Helene needs $2680 total. 1-8 Solving Equations by Multiplying or Dividing Course 3

9 Sometimes it is necessary to solve equations by using two inverse operations. For instance, the equation 6x 2 = 10 has multiplication and subtraction. 6x 2 = 10 Variable term Subtraction Multiplication To solve this equation, add to isolate the term with the variable in it. Then divide to solve. 1-8 Solving Equations by Multiplying or Dividing Course 3

10 Solve 3x + 2 = 14. Example 4: Solving a Simple Two-Step Equation 3x + 2 = 14Step 1: Subtract 2 to both sides to isolate the term with x in it. – 2 3x = 12 Step 2:3x = 12 3 3 x = 4 Divide both sides by 3. 1-8 Solving Equations by Multiplying or Dividing Course 3

11 Solve 4y + 5 = 29. Check It Out: Example 4 4y + 5 = 29Step 1: Subtract 5 from both sides to isolate the term with y in it. – 5 4y = 24 Step 2:4y = 24 4 4 y = 6 Divide both sides by 4. 1-8 Solving Equations by Multiplying or Dividing Course 3

12 Lesson Quiz Solve. 1. 3t = 9 2. –15 = 3b 3. = –7 4. z ÷ 4 = 22 5. A roller coaster descends a hill at a rate of 80 feet per second. The bottom of the hill is 400 feet from the top. How long will it take the coaster rides to reach the bottom? t = 3 z = 88 5 seconds b = –5 x = 28 x –4 1-8 Solving Equations by Multiplying or Dividing Course 3


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