Answer
$\{12\}$.
Work Step by Step
First we will determine the Least Common Denominator (LCD) and use it to clear fractions.
Factor $x^2-10x+16$
Rewrite the middle term $-10x$ as $-8x-2x$.
$\Rightarrow x^2-8x-2x+16$
Group the terms.
$\Rightarrow (x^2-8x)+(-2x+16)$
Factor each group.
$\Rightarrow x(x-8)-2(x-8)$
Factor out $(x-8)$.
$\Rightarrow (x-8)(x-2)$
Back substitute the factor into the given equation.
$\Rightarrow \frac{x}{x-8}+\frac{6}{x-2}=\frac{x^2}{(x-8)(x-2)}$
The LCD of the fractions is $(x-8)(x-2)$.
Multiply the equation by $(x-8)(x-2)$.
$\Rightarrow (x-8)(x-2)\left (\frac{x}{x-8}+\frac{6}{x-2}\right )=(x-8)(x-2)\left (\frac{x^2}{(x-8)(x-2)}\right )$
Use the distributive property.
$\Rightarrow (x-8)(x-2)\left (\frac{x}{x-8}\right )+(x-8)(x-2)\left (\frac{6}{x-2}\right )=(x-8)(x-2)\left (\frac{x^2}{(x-8)(x-2)}\right )$
Cancel common factors.
$\Rightarrow x(x-2)+6(x-8)=x^2$
Use distributive property.
$\Rightarrow x^2-2x+6x-48=x^2$
Add $-x^2+48$ to both sides.
$\Rightarrow x^2-2x+6x-48-x^2+48=x^2-x^2+48$
Simplify.
$\Rightarrow 4x=48$
Divide both sides by $4$.
$\Rightarrow \frac{4x}{4}=\frac{48}{4}$
Simplify.
$\Rightarrow x=12$
The solution set is $\{12\}$.
Note: The equation is defined for all real values of $x$ except the zeros of the denominators, which are $2$ and $8$. Because $x=12$ is not a zero of a denominator, the solution is correct.
