### Lesson 1-8 - Glynn County Schools

```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
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