8.3 Solving Equations Using Properties of Equality 66325
3
75Since 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
x2
3
( 9 ) 6
2
3
x 6x 9 1 xx(^3 2) 32 1 32 6 (^182) 9
1 x 9
a 32 23
3
2
2
3
b x3
2
6
233 23
2
2
3
x3
2
6
2
3
x 623 32x 23 xxa number5
4
x 32
3
x 6Self Check 6
Solve:a.b.Now TryProblems 61 and 673
8
b 27
2
x 21WHYTo 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 25x 75 1 xx31
x
31 x33 x
1 x 753 x
375
31
3
1
x
33 25
3
x
33 25
x
325
xxa number