Download 1.7 Solving Equations by Multiplication and Division Example 1

Math 40
Prealgebra
Section 1.7 – Solving Equations by Multiplication and Division
1.7 Solving Equations by Multiplication and Division
Multiplication Property of Equality
Multiplying the same number to both sides of an equation does not change its solution.
For a, b, and c, if a  b, then a c  b c .
As long as you perform the same operation on both sides of the equal sign, you maintain equality.
Division Property of Equality
Dividing the same number on both sides of an equation does not change its solution.
For a, b, and c, if a  b, then
a b
 .
c c
As long as you perform the same operation on both sides of the equal sign, you maintain equality.
The inverse operation of Multiplication is Division. Likewise, the inverse operation of Division
is Multiplication.
a x
x
a
and
a
x
x
a
Example 1: Solve 3x  24 for x.
Solution: To isolate the variable, x, we must get rid of the multiplied 3. To get rid of it we do the opposite,
which is to divide by 3. We divide by 3 on both sides of the equal sign to make sure we maintain
equality.
3 x 24

3
3 On the left, the multiplied 3 and divided 3 will cancel since they are opposites. On the right we simplify by dividing.
x8
Therefore, x  8 .
Check:
3 x  24
3  8   24
Since we got a true statement back, we know we got the right answer.
24  24
1
2015 Worrel
Math 40
Prealgebra
Section 1.7 – Solving Equations by Multiplication and Division
Example 2: Solve
x
 12 for x.
7
Solution: To isolate the variable, x, we must get rid of the divided 7. To get rid of it we do the opposite, which
is to multiply by 7. We multiply by 7 to both sides of the equal sign to make sure we maintain
equality.
x
7
 12 7
On the left, the multiplied 7 and divided 7 will cancel since they are opposites. On the right we simplify by multiplying.
7
x  84
Therefore, x  84 .
Check:
x
 12
7
 84   12
7
12  12
Since we got a true statement back, we know we got the right answer.
You Try It 1: Solve 5x  120 for x.
You Try It 2: Solve
x
 19 for x.
2
Translating Words to Math
Key words that translate to or indicate MULTIPLICATION:
Multiply
Times
Product
Of
Triple
Twice
Key words that translate to or indicate DIVISION:
Divide
Goes into
Quotient
Ratio
Double
Shared equally
2
2015 Worrel
Math 40
Prealgebra
Section 1.7 – Solving Equations by Multiplication and Division
REVIEW from Section 1.6
Five Step Word Problem Method
1)
2)
3)
4)
5)
Identify a variable.
Write an equation.
Solve the equation.
State your answer.
Check your answer.
Example 3: Fifteen times a certain number is 45. Find the unknown number.
Solution: Using the five step word problem method, we have
1) Identify a variable. x = the unknown number
2) Write an equation. 15 x  45
Fifteen times a certain number is 45.
15
x

45
3) Solve the equation.
15 x 45

15 15
x3
4) State your answer. The number is 3.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 3: Seven times a certain number is one hundred five. Find the unknown number.
3
2015 Worrel
Math 40
Prealgebra
Section 1.7 – Solving Equations by Multiplication and Division
Example 4: The area of a rectangle is 120 square feet. If the length of the rectangle is 12 feet, find the width of
the rectangle.
Solution: Using the five step word problem method, we have
1) Identify a variable. w = the width of the rectangle
2) Write an equation. 120  12w
Area of a rectangle = length
120

width
w
12
3) Solve the equation.
120 12 w

12
12
10  w
4) State your answer. The width of the rectangle is 10 feet.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 4: The area of a rectangle is 3,500 square meters. If the width is 50 meters, find the length.
Example 5: A class of 23 students averaged 76 points on an exam. How many total points were accumulated
by the class as a whole?
Solution: Using the five step word problem method, we have
1) Identify a variable. T = the total number of points accumulated by the entire class
T
2) Write an equation. 76  T  23 or 76 
23
The average is found by taking the total number of points and dividing by the number of students
76

T

23
3) Solve the equation.
T
23
23
1748  T
4) State your answer. The total number of points received by the entire class was 1,748 points.
5) Check your answer. Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
23 76 
You Try It 5: A class of 30 students averaged 75 points on an exam. How many total points were accumulated
by the class as a whole?
4
2015 Worrel