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Solving Equations by Multiplying or Dividing. 1-8. Warm Up. Problem of the Day. Lesson Presentation. Course 3. Solving Equations by Multiplying or Dividing. 1-8. 4. __. c. Course 3. Warm Up Write an algebraic expression for each word phrase. 1. A number x decreased by 9

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  1. Solving Equations by Multiplying or Dividing 1-8 Warm Up Problem of the Day Lesson Presentation Course 3

  2. Solving Equations by Multiplying or Dividing 1-8 4 __ c Course 3 Warm Up Write an algebraic expression for each word phrase. 1. A number x decreased by 9 2. 5 times the sum of p and 6 3. 2 plus the product of 8 and n 4. the quotient of 4 and a number c x – 9 5(p + 6) 2 +8n

  3. Solving Equations by Multiplying or Dividing 1-8 Course 3 Problem of the Day How many pieces do you have if you cut a log into six pieces and then cut each piece into 4 pieces? 24

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

  5. Solving Equations by Multiplying or Dividing 1-8 Words Numbers Algebra DIVISION PROPERTY OF EQUALITY 12 = 6 2 Course 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 x = y 4 • 3 = 12 x = y 2 2 z z Course 3

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

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

  8. Solving Equations by Multiplying or Dividing 1-8 ? 9(4) = 36 ? 36 = 36 Course 3 Check It Out: Example 1A Solve 9x = 36. 9x = 36 9x = 36 Divide both sides by 9. 9 9 1 • x = x 1x = 4 x = 4 Check 9x = 36 Substitute 4 for x. 

  9. Solving Equations by Multiplying or Dividing 1-8 ? –3(–12) = 36 ? 36 = 36 Course 3 Check It Out: Example 1B Solve –3y = 36. –3y = 36 –3y = 36 Divide both sides by –3. –3 –3 1 • y = y 1y = –12 y = –12 Check –3y = 36 Substitute –12 for y. 

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

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

  12. Solving Equations by Multiplying or Dividing 1-8 c c c –3 –4 –3 = 5 = 5 –15 ? = 5 –3 ? 5 = 5 Course 3 Check It Out: Example 2 Solve = 5. –3 • –3 • Multiply both sides by –3. c = –15 Check Substitute –15 for c. 

  13. Solving Equations by Multiplying or Dividing 1-8 1 1 1 4 4 4 x = 670 x = 670 Course 3 Additional Example 3: Money Application 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? fraction of amount raised so far amount raised so far total amount needed  = • = x 670 Write the equation. Multiply both sides by 4. 4• 4• x = 2680 Helene needs $2680 total.

  14. Solving Equations by Multiplying or Dividing 1-8 fraction of total amount raised so far 1 1 1 8 8 8 x = 750 x = 750 Course 3 Check It Out: Example 3 The school library needs money to complete a new collection. So far, the library has raised $750, which is only one-eighth of what they need. What is the total amount needed? amount raised so far total amount needed =  • = x 750 Write the equation. 8• Multiply both sides by 8. 8• x = 6000 The library needs to raise a total of $6000.

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

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

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

  18. Solving Equations by Multiplying or Dividing 1-8 x –4 Course 3 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 b = –5 x = 28 z = 88 5 seconds

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