$$\frac{2}{3x}-\frac{7}{6} = \frac{1}{6x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 7 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{2}{3x}-\frac{7}{6} &= \frac{1}{6x}&& \text{multiply ALL terms by } \color{blue}{ 3x\cdot6 }. \\[1 em]3x\cdot6\cdot\frac{2}{3x}-3x\cdot6\cdot\frac{7}{6} &= 3x\cdot6\cdot\frac{1}{6x}&& \text{cancel out the denominators} \\[1 em]12x^2-21x &= 3x^2&& \text{move all terms to the left hand side } \\[1 em]12x^2-21x-3x^2 &= 0&& \text{simplify left side} \\[1 em]9x^2-21x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 9x^{2}-21x = 0 } $, first we need to factor our $ x $.
$$ 9x^{2}-21x = x \left( 9x-21 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 9x-21 = 0$.
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