Solve by Factoring x/(x-5)+1/(x+5)=50/(x^2-25)

$\frac{x}{x - 5} + \frac{1}{x + 5} = \frac{50}{x^{2} - 25}$

Step 1

Subtract $\frac{50}{x^{2} - 25}$ from both sides of the equation.

$\frac{x}{x - 5} + \frac{1}{x + 5} - \frac{50}{x^{2} - 25} = 0$

Step 2

Simplify $\frac{x}{x - 5} + \frac{1}{x + 5} - \frac{50}{x^{2} - 25}$ .

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

$\frac{x}{x - 5} + \frac{1}{x + 5} - \frac{50}{x^{2} - 5^{2}} = 0$

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 $b = 5$ .

$\frac{x}{x - 5} + \frac{1}{x + 5} - \frac{50}{(x + 5) (x - 5)} = 0$

To write $\frac{x}{x - 5}$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$\frac{x}{x - 5} \cdot \frac{x + 5}{x + 5} + \frac{1}{x + 5} - \frac{50}{(x + 5) (x - 5)} = 0$

To write $\frac{1}{x + 5}$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$\frac{x}{x - 5} \cdot \frac{x + 5}{x + 5} + \frac{1}{x + 5} \cdot \frac{x - 5}{x - 5} - \frac{50}{(x + 5) (x - 5)} = 0$

Write each expression with a common denominator of $(x - 5) (x + 5)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{x - 5}$ by $\frac{x + 5}{x + 5}$ .

$\frac{x (x + 5)}{(x - 5) (x + 5)} + \frac{1}{x + 5} \cdot \frac{x - 5}{x - 5} - \frac{50}{(x + 5) (x - 5)} = 0$

Multiply $\frac{1}{x + 5}$ by $\frac{x - 5}{x - 5}$ .

$\frac{x (x + 5)}{(x - 5) (x + 5)} + \frac{x - 5}{(x + 5) (x - 5)} - \frac{50}{(x + 5) (x - 5)} = 0$

Reorder the factors of $(x - 5) (x + 5)$ .

$\frac{x (x + 5)}{(x + 5) (x - 5)} + \frac{x - 5}{(x + 5) (x - 5)} - \frac{50}{(x + 5) (x - 5)} = 0$

Combine the numerators over the common denominator.

$\frac{x (x + 5) + x - 5}{(x + 5) (x - 5)} - \frac{50}{(x + 5) (x - 5)} = 0$

Combine the numerators over the common denominator.

$\frac{x (x + 5) + x - 5 - 50}{(x + 5) (x - 5)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x \cdot x + x \cdot 5 + x - 5 - 50}{(x + 5) (x - 5)} = 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 5 + x - 5 - 50}{(x + 5) (x - 5)} = 0$

Move $5$ to the left of $x$ .

$\frac{x^{2} + 5 x + x - 5 - 50}{(x + 5) (x - 5)} = 0$

Add $5 x$ and $x$ .

$\frac{x^{2} + 6 x - 5 - 50}{(x + 5) (x - 5)} = 0$

Subtract $50$ from $- 5$ .

$\frac{x^{2} + 6 x - 55}{(x + 5) (x - 5)} = 0$

Factor $x^{2} + 6 x - 55$ 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 $- 55$ and whose sum is $6$ .

$- 5, 11$

Write the factored form using these integers.

$\frac{(x - 5) (x + 11)}{(x + 5) (x - 5)} = 0$

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$\frac{(x - 5) (x + 11)}{(x + 5) (x - 5)} = 0$

Rewrite the expression.

$\frac{x + 11}{x + 5} = 0$

Step 3

Set the numerator equal to zero.

$x + 11 = 0$

Step 4

Subtract $11$ from both sides of the equation.

$x = - 11$

Solve by Factoring x/(x-5)+1/(x+5)=50/(x^2-25)

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