$$\frac{2}{3}x(3x+1)-\frac{1}{3}x(6x-2) = 0$$

$$ \begin{aligned} \frac{2}{3}x(3x+1)-\frac{1}{3}x(6x-2) &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]3 \cdot \frac{2}{3}x(3x+1)-3\frac{1}{3}x(6x-2) &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]6x^2+2x-(6x^2-2x) &= 0&& \text{simplify left side} \\[1 em]6x^2+2x-6x^2+2x &= 0&& \\[1 em]6x^2+2x-6x^2+2x &= 0&& \\[1 em]4x &= 0&& \text{ divide both sides by $ 4 $ } \\[1 em]x &= \frac{0}{4}&& \\[1 em]x &= 0&& \\[1 em] \end{aligned} $$

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