Question
$$\frac{5}{3x}+\frac{3}{2x}-\frac{x-4}{6x} = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 23 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5}{3x}+\frac{3}{2x}-\frac{x-4}{6x} &= 0&& \text{multiply ALL terms by } \color{blue}{ 3x\cdot2\cdot6 }. \\[1 em]3x\cdot2\cdot6\cdot\frac{5}{3x}+3x\cdot2\cdot6\cdot\frac{3}{2x}-3x\cdot2\cdot6\frac{x-4}{6x} &= 3x\cdot2\cdot6\cdot0&& \text{cancel out the denominators} \\[1 em]60x^2+54x^2-(6x^3-24x^2) &= 0&& \text{simplify left side} \\[1 em]114x^2-(6x^3-24x^2) &= 0&& \\[1 em]114x^2-6x^3+24x^2 &= 0&& \\[1 em]-6x^3+138x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -6x^{3}+138x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -6x^{3}+138x^{2} = x^2 \left( -6x+138 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The second root can be found by solving equation $ -6x+138 = 0$.
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