Answer
$\{-7,3\}$.
Work Step by Step
Determine the Least Common Denominator (LCD) of the denominators of all fractions in the given equation in order to clear fractions.
Factor $x^2-16$
$\Rightarrow x^2-4^2 $
Use the special formula $A^2-B^2=(A+B)(A-B)$
$\Rightarrow =(x+4)(x-4)$
Back substitute into the given equation.
$\Rightarrow \frac{x}{x+4}=\frac{11}{(x+4)(x-4)}+2$
The LCD of the denominators is $(x+4)(x-4)$.
Multiply the equation by $(x+4)(x-4)$.
$\Rightarrow (x+4)(x-4)\left (\frac{x}{x+4}\right )=(x+4)(x-4) \left (\frac{11}{(x+4)(x-4)}+2\right )$
Use the distributive property.
$\Rightarrow (x+4)(x-4)\left (\frac{x}{x+4}\right )=(x+4)(x-4) \left (\frac{11}{(x+4)(x-4)}\right )+(x+4)(x-4) \left (2\right )$
Cancel the common factors.
$\Rightarrow x(x-4)=11+2(x+4)(x-4)$
Use the distributive property and the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow x^2-4x=11+2x^2-32$
Simplify.
$\Rightarrow x^2-4x=2x^2-21$
Add $-x^2+4x$ to both sides.
$\Rightarrow x^2-4x-x^2+4x=2x^2-21-x^2+4x$
Simplify.
$\Rightarrow 0=x^2+4x-21$
Rewrite the middle terms $4x$ as $7x-3x$.
$\Rightarrow 0=x^2+7x-3x-21$
Group terms.
$\Rightarrow 0=(x^2+7x)+(-3x-21)$
Factor each group.
$\Rightarrow 0=x(x+7)-3(x+7)$
Factor out $(x+7)$.
$\Rightarrow 0=(x+7)(x-3)$
By using zero product rule set each factor equal to zero.
$\Rightarrow x+7=0$ or $x-3=0$
Isolate $x$.
$\Rightarrow x=-7$ or $x=3$
The solution set is $\{-7,3\}$.
Note: The given equation is defined for all real values of $x$ except the zeros of the denominators, which are $-4$ and $4$. As neither of the solutions is a zero of the denominators, the solution we found is correct.
