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1-8 Solving Equations by Multiplying or Dividing Lesson Presentation Course 3 Solving Equations by 1-8 Multiplying or Dividing Learn to solve equations using multiplication and division. Course 3 Solving Equations by 1-8 Multiplying or Dividing You can solve a multiplication equation using the Division Property of Equality. DIVISION PROPERTY OF EQUALITY Words You can divide both sides of an equation by the same nonzero number, and the equation will still be true. Course 33 Course Numbers 4 • 3 = 12 4 • 3 = 12 2 2 12 = 6 2 Algebra x=y x=y z z Solving Equations by 1-8 Multiplying or Dividing Example 1A: Solving Equations Using Division Solve 6x = 48. 6x = 48 6x = 48 6 6 1x = 8 x=8 Divide both sides by 6. 1•x=x Check 6x = 48 ? 48 6(8) = Substitute 8 for x. ? 48 48 = Course 3 Solving Equations by 1-8 Multiplying or Dividing Example 1B: Solving Equations Using Division Solve –9y = 45. –9y –9y –9 1y y = 45 = 45 –9 = –5 = –5 Divide both sides by –9. 1•y=y Check –9y = 45 –9(–5) ?= 45 Substitute –5 for y. ? 45 45 = Course 3 Solving Equations by 1-8 Multiplying or Dividing You can solve a division equation using the Multiplication Property of Equality. MULTIPLICATION PROPERTY OF EQUALITY Words You can multiply both sides of an equation by the same number, and the statement will still be true. Course 33 Course Numbers 2•3=6 4• 2•3=4•6 8 • 3 = 24 Algebra x=y z x = zy Solving Equations by 1-8 Multiplying or Dividing Example 2: Solving Equations Using Multiplication b = 5. –4 b = –4 • 5 –4 • –4 b = –20 Solve Check Course 3 b =5 –4 –20 =? 5 –4 ? 5 5= Multiply both sides by –4. Substitute –20 for b. Solving Equations by 1-8 Multiplying or Dividing 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 1 4 total amount needed = amount raised so far • x = 670 1 x = 670 Write the equation. 4 4 • 1 x = 4 • 670 Multiply both sides by 4. 4 x = 2680 Helene needs $2680 total. Course 3 Solving Equations by 1-8 Multiplying or Dividing 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 Multiplication 6x 2 = 10 Subtraction To solve this equation, add to isolate the term with the variable in it. Then divide to solve. Course 3 Solving Equations by 1-8 Multiplying or Dividing Example 4: Solving a Simple Two-Step Equation Solve 3x + 2 = 14. Step 1: 3x + 2 = 14 Subtract 2 to both sides to – 2 – 2 isolate the term with x in it. 3x = 12 Step 2: 3x = 12 Divide both sides by 3. 3 3 x=4 Course 3 Solving Equations by 1-8 Multiplying or Dividing Check It Out: Example 4 Solve 4y + 5 = 29. Step 1: Step 2: Course 3 4y + 5 = 29 Subtract 5 from both sides to – 5 – 5 isolate the term with y in it. 4y = 24 4y = 24 Divide both sides by 4. 4 4 y=6 Solving Equations by 1-8 Multiplying or Dividing Lesson Quiz Solve. 1. 3t = 9 t=3 2. –15 = 3b b = –5 x 3. –4 = –7 x = 28 4. z ÷ 4 = 22 z = 88 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? 5 seconds Course 3