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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
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