$$\frac{1}{x}+\frac{1}{x-8} = \frac{1}{3}$$
Answer
$$ \begin{matrix}x_1 = 2 & x_2 = 12 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x}+\frac{1}{x-8} &= \frac{1}{3}&& \text{multiply ALL terms by } \color{blue}{ x(x-8)\cdot3 }. \\[1 em]x(x-8)\cdot3\cdot\frac{1}{x}+x(x-8)\cdot3\cdot\frac{1}{x-8} &= x(x-8)\cdot3\cdot\frac{1}{3}&& \text{cancel out the denominators} \\[1 em]3x-24+3x &= x^2-8x&& \text{simplify left side} \\[1 em]6x-24 &= x^2-8x&& \text{move all terms to the left hand side } \\[1 em]6x-24-x^2+8x &= 0&& \text{simplify left side} \\[1 em]-x^2+14x-24 &= 0&& \\[1 em] \end{aligned} $$
$ -x^{2}+14x-24 = 0 $ is a quadratic equation.
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