Answer
$\{7\}$.
Work Step by Step
First we will find the Least Common Denominator (LCD) of all denominators and use it to clear fractions.
Factor $x^2-25$.
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+5)(x-5)$
Substitute the factor into the equation.
$\frac{1}{x-5}-\frac{3}{x+5}=\frac{6}{(x+5)(x-5)}$
Multiply the equation by the LCD $(x+5)(x-5)$.
$(x+5)(x-5)\left ( \frac{1}{x-5}-\frac{3}{x+5}\right )=(x+5)(x-5)\left (\frac{6}{(x+5)(x-5)}\right )$
Use the distributive property.
$(x+5)(x-5)\cdot \frac{1}{x-5}-(x+5)(x-5)\cdot \frac{3}{x+5}=(x+5)(x-5)\cdot \frac{6}{(x+5)(x-5)}$
Simplify.
$x+5-3(x-5)=6$
$x+5-3x+15=6$
$-2x+20=6$
Add $-20$ to both sides.
$-2x+20-20=6-20$
Simplify.
$-2x=-14$
Divide both sides $-2$.
$\frac{-2x}{-2}=\frac{-14}{-2}$
Simplify.
$x=7$.
The solution set is $\{7\}$.
Note: Check if the solution is correct. The given equation is defined for all real values of $x$ except the zeros of the denominators, which are $-5$ and $5$. Since $x=7$ is not a zero of a denominator, the solution is correct.
