$$\frac{13}{2x}-\frac{4}{15} = \frac{31}{6x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{13}{2x}-\frac{4}{15} &= \frac{31}{6x}&& \text{multiply ALL terms by } \color{blue}{ 2x\cdot15\cdot6 }. \\[1 em]2x\cdot15\cdot6\cdot\frac{13}{2x}-2x\cdot15\cdot6\cdot\frac{4}{15} &= 2x\cdot15\cdot6\cdot\frac{31}{6x}&& \text{cancel out the denominators} \\[1 em]1170x^2-48x &= 930x^2&& \text{move all terms to the left hand side } \\[1 em]1170x^2-48x-930x^2 &= 0&& \text{simplify left side} \\[1 em]240x^2-48x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 240x^{2}-48x = 0 } $, first we need to factor our $ x $.
$$ 240x^{2}-48x = x \left( 240x-48 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 240x-48 = 0$.
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