Question
$$\frac{2}{3}+\frac{2}{3(3x-1)} = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{2}{3}+\frac{2}{3(3x-1)} &= 0&& \text{multiply ALL terms by } \color{blue}{ 3(3x-1) }. \\[1 em]3(3x-1)\cdot\frac{2}{3}+3(3x-1)\cdot\frac{2}{3(3x-1)} &= 3(3x-1)\cdot0&& \text{cancel out the denominators} \\[1 em]6x-2+18x^2-12x+2 &= 0&& \text{simplify left side} \\[1 em]18x^2-6x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 18x^{2}-6x = 0 } $, first we need to factor our $ x $.
$$ 18x^{2}-6x = x \left( 18x-6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 18x-6 = 0$.
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