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

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

Step 1

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

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

Step 2

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

Simplify the denominator.

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 1$ .

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Combine the opposite terms in $\frac{b - 1 + b + 1}{(b + 1) (b - 1)} - \frac{2}{(b + 1) (b - 1)}$ .

Add $- 1$ and $1$ .

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

Add $b$ and $0$ .

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

Combine the numerators over the common denominator.

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

Add $b$ and $b$ .

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

Factor $2$ out of $2 b - 2$ .

Factor $2$ out of $2 b$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (b) + 2 (- 1)$ .

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

Cancel the common factor of $b - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{2}{b + 1} = 0$

Step 3

Set the numerator equal to zero.

$2 = 0$

Step 4

Since $2 \neq 0$ , there are no solutions.

No solution

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

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