Solve the equation
Solution:Multiply both sides of the original equation by 4x to obtain the equation x² - 16 = 0.
Note:This equation need not be equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Factorization yields (x - 4)(x + 4) = 0 and then The Zero Factor Property yields the two equations
x - 4 = 0 OR x + 4 = 0. The solution sets to these two equations are {4} OR {-4}.
The solution set of the second equation x² - 16 = 0 is therefore .
Every solution of the original equation is contained in the set {4, -4}, but some of the elements of the set {4, -4} may not be solutions of the original equation. Therefore we must test each of these possible solutions.
Neither 4 nor - 4 cause any denominator in the original equation to be zero.
Therefore both 4 and -4 are solutions of the original equation.
The solution set for the original equation is {4, -4} .

Observe that

Solve the equation
Solution:Multiply both sides of the original equation by x - 2 to obtain the equation 3 + (x - 2) = 3
Note:This equation is not equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add -1 to both sides ot the second equation to obtain x = 2.
The solution set of the second equation {2}.
Every solution of the original equation is contained in the set {2}, but some of the elements of the set {2} may not be solutions of the original equation. Therefore we must test this possible solutions.
Observe that 2 causes a zero in a denominator of the original equation and is therefore not a solution of the original equation.
Therefore the solution set for the original equation is the empty set .

Observe that

Solve the equation     

Solution:This equation is equivalent to      (Call this Eq 1)

Multiply both sides of the original equation by the LCD (x - 2)(x + 3) to obtain the equation
(x -3)(x + 3) = (2x - 6)(x - 2) + (5x - 15) .       (Call this Eq 2)
Note:Because we have multiplied both sides of Eq 1 by a variable expression, we know that Eq 2 is probably not equivalent to Eq 1. However, we may proceed w9ith confidence because we know the solution set for Eq 2 contains the solution set for Eq 1.
We now continue to solve Eq 2 with familiar techniques all of which yield equations which are equivalent to the previous equation in the process.
(x -3)(x + 3) = (2x - 6)(x - 2) + (5x - 15)
x² - 9 = 2x² -10x + 12 +5x - 15
x² - 5x + 6 = 0
(x - 3)(x - 2) = 0
Using the Zero Fator Property we conclude
x - 3 = 0 OR x - 2 = 0
x = 3 OR x = 2   Note both of these are simplest equations.
Therefore the solution set for Eq 2 is {2, 3}
Every solution of the original equation is contained in the set {2, 3}, but some of the elements of the set {2, 3} might not be solutions of the original equation. Therefore we must test each of these possible solutions.
The possible solution 2 causes a 0 in a denominator of Eq 1 and is therefore not a solution of Eq 1.
On the other hand, the possible solution 3 does not create a 0 in any denominator of Eq 1 and is therefore a solution of Eq 1..
The solution set for the original equation is {3} .

Observe that