Beginning Algebra, 11th Edition

(Marvins-Underground-K-12) #1

SECTION 2.2 The Multiplication Property of Equality^93


In , we must change 3xto 1x, or x. To do this, we multiply each side of


the equation by , the reciprocal of 3, because


Multiply each side by

Associative property

Multiplicative inverse property

x= 5 Multiplicative identity property


1 x= 5


a


1


3


3 bx=^1


3


15


1

3.


1


3


13 x 2 =


1


3


15


3 x= 15


1
3

3 =^3


3 =1.


1
3

3 x= 15


The solution is 5. We can check this result in the original equation.


Just as the addition property of equality permits subtractingthe same number


from each side of an equation, the multiplication property of equality permits


dividingeach side of an equation by the same nonzero number.


Divide each side by 3.

x= 5 Same result as above


3 x


3


=


15


3


3 x= 15


The product of a
number and its
reciprocal is 1.

We can divide each side of an equation by the same nonzero number without


changing the solution. Do not, however, divide each side by a variable, since the


variable might be equal to 0.


NOTE In practice, it is usually easier to multiply on each side if the coefficient of


the variable is a fraction, and divide on each side if the coefficient is an integer. For


example, to solve


it is easier to multiply by than to divide by


On the other hand, to solve


5 x= 20, it is easier to divide by 5 than to multiply by^15.


3

4.


4
3

3

4 x= 12,


NOW TRY
EXERCISE 1
Solve. 8 x= 80


Applying the Multiplication Property of Equality

Solve


5 x
5 =

5

x = 12 5 x= 1 x=x


5 x


5


=


60


5


5 x= 60


5 x= 60.


EXAMPLE 1

Divide each side by 5,
the coefficient of x.

CHECK Substitute 12 for xin the original equation.


Original equation
Let

✓ True


Since a true statement results, the solution set is 5126. NOW TRY


60 = 60


51122  60 x=12.


5 x= 60


Our goal is
to isolate x.

Dividing by 5 is the same
as multiplying by.^15

NOW TRY ANSWER



  1. 5106

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