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