$$\frac{1}{x^2+4x}+\frac{5}{x} = \frac{4}{x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -5 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x^2+4x}+\frac{5}{x} &= \frac{4}{x}&& \text{multiply ALL terms by } \color{blue}{ (x^2+4x)x }. \\[1 em](x^2+4x)x\cdot\frac{1}{x^2+4x}+(x^2+4x)x\cdot\frac{5}{x} &= (x^2+4x)x\cdot\frac{4}{x}&& \text{cancel out the denominators} \\[1 em]x+5x^2+20x &= 4x^2+16x&& \text{simplify left side} \\[1 em]5x^2+21x &= 4x^2+16x&& \text{move all terms to the left hand side } \\[1 em]5x^2+21x-4x^2-16x &= 0&& \text{simplify left side} \\[1 em]x^2+5x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}+5x = 0 } $, first we need to factor our $ x $.
$$ x^{2}+5x = x \left( x+5 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x+5 = 0$.
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