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

1-8 Solving Equations by Multiplying or Dividing Warm Up 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 4 c __

1-8 Solving Equations by Multiplying or Dividing Problem of the Day 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

1-8 Solving Equations by Multiplying or Dividing Course 3 Learn to solve equations using multiplication and division; and learn to solve equations involving more than one step by isolating the variable.

1-8 Solving Equations by Multiplying or Dividing Course 3 You can solve a multiplication equation using the Division Property of Equality. DIVISION PROPERTY OF EQUALITY Words Numbers Algebra 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 12 = 6 2 Course 3

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

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

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

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

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

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

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

= =  • 1-8 Solving Equations by Multiplying or Dividing 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  total amount needed = amount raised so far 1 4 • = x 670 x = 670 1 4 Write the equation. x = 670 1 4 Multiply both sides by 4. 4 • 4 • x = 2680 Helene needs $2680 total.

= = • 1-8 Solving Equations by Multiplying or Dividing 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? fraction of total amount raised so far total amount needed = amount raised so far  1 8 • = x 750 x = 750 1 8 Write the equation. x = 750 1 8 8 • 8 • Multiply both sides by 8. x = 6000 The library needs to raise a total of $6000.

1-8 6x  2 = 10 Solving Equations by Multiplying or Dividing 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.

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

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

1-8 Solving Equations by Multiplying or Dividing Lesson Quiz Solve. 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 –4 x = 28 z = 88 5 seconds