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

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

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

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

Example 1B: Solving Equations Using Division 1-8 Solving Equations by Multiplying or Dividing Course 3 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 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

Example 2: Solving Equations Using Multiplication 1-8 Solving Equations by Multiplying or Dividing Course 3 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 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 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.

Example 4: Solving a Simple Two-Step Equation 1-8 Solving Equations by Multiplying or Dividing Course 3 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