$$\frac{2}{3}-\frac{1}{5} = \frac{7}{3x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{2}{3}-\frac{1}{5} &= \frac{7}{3x}&& \text{multiply ALL terms by } \color{blue}{ 3\cdot5x }. \\[1 em]3\cdot5x\cdot\frac{2}{3}-3\cdot5x\cdot\frac{1}{5} &= 3\cdot5x\cdot\frac{7}{3x}&& \text{cancel out the denominators} \\[1 em]10x-3x &= 35x^2&& \text{simplify left side} \\[1 em]7x &= 35x^2&& \text{move all terms to the left hand side } \\[1 em]7x-35x^2 &= 0&& \text{simplify left side} \\[1 em]-35x^2+7x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -35x^{2}+7x = 0 } $, first we need to factor our $ x $.
$$ -35x^{2}+7x = x \left( -35x+7 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -35x+7 = 0$.
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