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

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

3. 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

4. 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 ? 

5. 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 ? 

6. 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 = z
x = y
8 • 3 = 24
Course 3

7. 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 =
–4 •
–4 •
Multiply both sides by –4.
b = –20
Check b –4
= 5
–20 –4
= 5 ?
Substitute –20 for b.
5 = 5 ? 

8. = =  • 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 = 1 4
Multiply both sides by 4.
4 •
4 •
x = 2680
Helene needs $2680 total.

9. 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.

10. 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

11. 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

12. 1-8 Solving Equations by Multiplying or Dividing Lesson Quiz Solve.
Course 3
Lesson Quiz
Solve.
1. 3t = 9
2. –15 = 3b
= –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