1-8 Solving Equations by Multiplying or Dividing

Course 3

Warm UpWarm Up

Lesson Presentation

Problem of the Day

Warm UpWrite 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

1-8 Solving Equations by Multiplying or Dividing

Course 3

4c__

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.

1-8 Solving Equations by Multiplying or Dividing

Course 3

Course 3

You can solve a multiplication equation using the Division Property of Equality.

You can divide both sides of an equation by the same nonzero number, and the equation will still be true.

4 • 3 = 122 2

x = yx = y

DIVISION PROPERTY OF EQUALITY

Words Numbers Algebra

z

4 • 3 = 12

12 = 62

z

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve 6x = 48.

Additional Example 1A: Solving Equations Using Division

6x = 48

1x = 6

Divide both sides by 6.

Check

6x = 486 6

x = 8

6x = 486(8) = 48?

48 = 48? Substitute 8 for x.

1 • x = x

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve –9y = 45.

Additional Example 1B: Solving Equations Using Division

–9y = 45

1y = –5

Divide both sides by –9.

Check

–9y = 45–9 –9

y = –5

–9y = 45–9(–5) = 45?

45 = 45? Substitute –5 for y.

1 • y = y

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve 9x = 36.

Check It Out: Example 1A

9x = 36

1x = 4

Divide both sides by 9.

Check

9x = 369 9

x = 4

9x = 369(4) = 36?

36 = 36? Substitute 4 for x.

1 • x = x

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve –3y = 36.

Check It Out: Example 1B

–3y = 36

1y = –12

Divide both sides by –3.

Check

–3y = 36–3 –3

y = –12

–3y = 36–3(–12) = 36?

36 = 36? Substitute –12 for y.

1 • y = y

1-8 Solving Equations by Multiplying or Dividing

Course 3

Course 3

You can multiply both sides of an equation by the same number, and the statement will still be true.

2 • 3 = 6x = yx = yz z

MULTIPLICATION PROPERTY OF EQUALITY

Words Numbers Algebra

2 • 3 = 6 4 • 4 •

8 • 3 = 24

You can solve a division equation using the Multiplication Property of Equality.

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve = 5.

Additional Example 2: Solving Equations Using Multiplication

b–4b–4 = 5–4 • –4 • Multiply both sides by –4.

b = –20

Checkb–4= 5

Substitute –20 for b.–20–4 = 5

?

5 = 5?

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve = 5.

Check It Out: Example 2

c–3c–4 = 5–3 • –3 • Multiply both sides by –3.

c = –15

Checkc–3= 5

Substitute –15 for c.–15–3 = 5

?

5 = 5?

1-8 Solving Equations by Multiplying or Dividing

Course 3

Additional Example 3: Money Application

fraction of amount raised so far

amount raised so far =

• =x 670

4 •

x = 2680

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?

Write the equation.

Multiply both sides by 4.

total amount needed

14

x = 67014

4 • x = 67014

Helene needs $2680 total.

1-8 Solving Equations by Multiplying or Dividing

Course 3

Check It Out: Example 3

amount raised so far =

• =x 750

8 •

x = 6000

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?

Write the equation.

Multiply both sides by 8.

total amount needed

18

x = 75018

8 • x = 75018

The library needs to raise a total of $6000.

fraction of total amount raised so far

1-8 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.

6x 2 = 10

Variable term

Subtraction

Multiplication

To solve this equation, add to isolate the term with the variable in it. Then divide to solve.

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve 3x + 2 = 14.

Additional Example 4: Solving a Simple Two-Step Equation

3x + 2 = 14Step 1: Subtract 2 to both sides to isolate the term with x in it. – 2 – 2

3x = 12

Step 2: 3x = 123 3 x = 4

Divide both sides by 3.

1-8 Solving Equations by Multiplying or Dividing

Course 3

Solve 4y + 5 = 29.

Check It Out: Example 4

4y + 5 = 29Step 1: Subtract 5 from both sides to isolate the term with y in it.– 5 – 5

4y = 24

Step 2: 4y = 244 4 y = 6

Divide both sides by 4.

1-8 Solving Equations by Multiplying or Dividing

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

z = 88

5 seconds

b = –5

x = 28x–4

1-8 Solving Equations by Multiplying or Dividing

Course 3