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

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

Step 1

Subtract $\frac{2 x}{x^{2} - 1}$ from both sides of the equation.

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

Step 2

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

Simplify the denominator.

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

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

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

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

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

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

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

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

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

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

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

Reorder the factors of $(x - 1) (x + 1)$ .

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

Combine the numerators over the common denominator.

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

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

Combine the opposite terms in $2 x - 2 - x - 1 - 2 x$ .

Subtract $2 x$ from $2 x$ .

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

Add $- 2 - x - 1$ and $0$ .

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

Subtract $1$ from $- 2$ .

$\frac{- x - 3}{(x + 1) (x - 1)} = 0$

Factor $- 1$ out of $- x$ .

$\frac{- (x) - 3}{(x + 1) (x - 1)} = 0$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{- (x) - 1 (3)}{(x + 1) (x - 1)} = 0$

Factor $- 1$ out of $- (x) - 1 (3)$ .

$\frac{- (x + 3)}{(x + 1) (x - 1)} = 0$

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

$\frac{- 1 (x + 3)}{(x + 1) (x - 1)} = 0$

Move the negative in front of the fraction.

$- \frac{x + 3}{(x + 1) (x - 1)} = 0$

Step 3

Set the numerator equal to zero.

$x + 3 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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