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

2. 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 __

3. 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

4. 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.

5. 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

6. 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 ? 

7. 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 ? 

8. 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 ? 

9. 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 ? 

10. 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

11. 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 =
–4 •
–4 •
Multiply both sides by –4.
b = –20
Check b –4
= 5
–20 –4
= 5 ?
Substitute –20 for b.
5 = 5 ? 

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

13. = =  • 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 = 1 4
Multiply both sides by 4.
4 •
4 •
x = 2680
Helene needs $2680 total.

14. = = • 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 = 1 8
8 •
8 •
Multiply both sides by 8.
x = 6000
The library needs to raise a total of $6000.

15. 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.

16. 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

17. 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

18. 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