Solve by Factoring x/(x-2)+1/(x+2)=8/(x^2-4)

$\frac{x}{x - 2} + \frac{1}{x + 2} = \frac{8}{x^{2} - 4}$

Step 1

Subtract $\frac{8}{x^{2} - 4}$ from both sides of the equation.

$\frac{x}{x - 2} + \frac{1}{x + 2} - \frac{8}{x^{2} - 4} = 0$

Step 2

Simplify $\frac{x}{x - 2} + \frac{1}{x + 2} - \frac{8}{x^{2} - 4}$ .

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$\frac{x}{x - 2} + \frac{1}{x + 2} - \frac{8}{x^{2} - 2^{2}} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$\frac{x}{x - 2} + \frac{1}{x + 2} - \frac{8}{(x + 2) (x - 2)} = 0$

To write $\frac{x}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

$\frac{x}{x - 2} \cdot \frac{x + 2}{x + 2} + \frac{1}{x + 2} - \frac{8}{(x + 2) (x - 2)} = 0$

To write $\frac{1}{x + 2}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{x}{x - 2} \cdot \frac{x + 2}{x + 2} + \frac{1}{x + 2} \cdot \frac{x - 2}{x - 2} - \frac{8}{(x + 2) (x - 2)} = 0$

Write each expression with a common denominator of $(x - 2) (x + 2)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{x - 2}$ by $\frac{x + 2}{x + 2}$ .

$\frac{x (x + 2)}{(x - 2) (x + 2)} + \frac{1}{x + 2} \cdot \frac{x - 2}{x - 2} - \frac{8}{(x + 2) (x - 2)} = 0$

Multiply $\frac{1}{x + 2}$ by $\frac{x - 2}{x - 2}$ .

$\frac{x (x + 2)}{(x - 2) (x + 2)} + \frac{x - 2}{(x + 2) (x - 2)} - \frac{8}{(x + 2) (x - 2)} = 0$

Reorder the factors of $(x - 2) (x + 2)$ .

$\frac{x (x + 2)}{(x + 2) (x - 2)} + \frac{x - 2}{(x + 2) (x - 2)} - \frac{8}{(x + 2) (x - 2)} = 0$

Combine the numerators over the common denominator.

$\frac{x (x + 2) + x - 2}{(x + 2) (x - 2)} - \frac{8}{(x + 2) (x - 2)} = 0$

Combine the numerators over the common denominator.

$\frac{x (x + 2) + x - 2 - 8}{(x + 2) (x - 2)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x \cdot x + x \cdot 2 + x - 2 - 8}{(x + 2) (x - 2)} = 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 2 + x - 2 - 8}{(x + 2) (x - 2)} = 0$

Move $2$ to the left of $x$ .

$\frac{x^{2} + 2 x + x - 2 - 8}{(x + 2) (x - 2)} = 0$

Add $2 x$ and $x$ .

$\frac{x^{2} + 3 x - 2 - 8}{(x + 2) (x - 2)} = 0$

Subtract $8$ from $- 2$ .

$\frac{x^{2} + 3 x - 10}{(x + 2) (x - 2)} = 0$

Factor $x^{2} + 3 x - 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

$\frac{(x - 2) (x + 5)}{(x + 2) (x - 2)} = 0$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{(x - 2) (x + 5)}{(x + 2) (x - 2)} = 0$

Rewrite the expression.

$\frac{x + 5}{x + 2} = 0$

Step 3

Set the numerator equal to zero.

$x + 5 = 0$

Step 4

Subtract $5$ from both sides of the equation.

$x = - 5$

Solve by Factoring x/(x-2)+1/(x+2)=8/(x^2-4)

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