$$\frac{x+1}{5x} = \frac{x-3}{6x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -21 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x+1}{5x} &= \frac{x-3}{6x}&& \text{multiply ALL terms by } \color{blue}{ 5x\cdot6 }. \\[1 em]5x\cdot6 \cdot \frac{x+1}{5x} &= 5x\cdot6 \cdot \frac{x-3}{6x}&& \text{cancel out the denominators} \\[1 em]6x^3+6x^2 &= 5x^3-15x^2&& \text{move all terms to the left hand side } \\[1 em]6x^3+6x^2-5x^3+15x^2 &= 0&& \text{simplify left side} \\[1 em]x^3+21x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{3}+21x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ x^{3}+21x^{2} = x^2 \left( x+21 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The second root can be found by solving equation $ x+21 = 0$.
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