Basic Mathematics for College Students

(Nandana) #1
8.3 Solving Equations Using Properties of Equality 663

25
 3
75

Since the product of a number and its reciprocal is 1, we can solve equations such
as 23 x 6 , where the coefficient of the variable term is a fraction, as follows.


EXAMPLE 6
Solve: a. b.

StrategyWe will use a property of equality to isolate the variable on one side of
the equation.


WHYTo solve the original equation, we want to find a simpler equivalent
equation of the form , whose solution is obvious.


Solution


a.Since the coefficient of is , we can isolate by multiplying both sides of the


equation by the reciprocal of , which is.

This is the equation to solve.

To undo the multiplication by , multiply both sides
by the reciprocal of.

Use the associative property of multiplication to group and.

On the left side, the product of a number and its reciprocal is 1:

. On the right side,.
The coefficient 1 need not be written since.


Check: This is the original equation.

Substitute 9 for in the original equation.

On the left side,.
Since the statement 66 is true, 9 is the solution of^23 x6.

6  6 23 (9)^183  6


x

2


3


( 9 ) 6


2


3


x 6

x 9 1 xx

(^3 2)  32  1 32  6  (^182)  9
1 x 9
a 32 23


3


2





2


3


b x

3


2


 6


23

3 23


2





2


3


x

3


2


 6


2


3


x 6

23 32

x 23 x

xa number




5


4


x 3

2


3


x 6

Self Check 6
Solve:

a.

b.

Now TryProblems 61 and 67




3


8


b 2

7


2


x 21

WHYTo solve the original equation, we want to find a simpler equivalent
equation of the form , whose solution is obvious.


Solution
To isolate , we use the multiplication property of equality. We can undo the
division by 3 by multiplying both sides by 3.


This is the equation to solve.

Multiply both sides by 3.

Write 3 as.

Do the multiplications.

Simplify by removing the common factor of 3
in the numerator and denominator:.

The coefficient 1 need not be written since.
If we substitute 75 for in , we obtain the true statement. This
verifies that 75 is the solution.


x x 3  25 25  25

x 75 1 xx

3

1
x
31 x

33 x
1 x 75

3 x
3

 75


31


3


1





x
3

 3  25


3 


x
3

 3  25


x
3

 25


x

xa number
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