$$\frac{1}{x+1}+\frac{1}{10-x} = \frac{11}{30}$$
Answer
$$ \begin{matrix}x_1 = 4 & x_2 = 5 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x+1}+\frac{1}{10-x} &= \frac{11}{30}&& \text{multiply ALL terms by } \color{blue}{ (x+1)\cdot(10-x)\cdot30 }. \\[1 em](x+1)\cdot(10-x)\cdot30\cdot\frac{1}{x+1}+(x+1)\cdot(10-x)\cdot30\cdot\frac{1}{10-x} &= (x+1)\cdot(10-x)\cdot30\cdot\frac{11}{30}&& \text{cancel out the denominators} \\[1 em]-30x+300+30x+30 &= -11x^2+99x+110&& \text{simplify left side} \\[1 em]330 &= -11x^2+99x+110&& \text{move all terms to the left hand side } \\[1 em]330+11x^2-99x-110 &= 0&& \text{simplify left side} \\[1 em]11x^2-99x+220 &= 0&& \\[1 em] \end{aligned} $$
$ 11x^{2}-99x+220 = 0 $ is a quadratic equation.
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