$$4 \cdot \frac{x}{x}+\frac{1}{3x} = 9$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 15 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4 \cdot \frac{x}{x}+\frac{1}{3x} &= 9&& \text{multiply ALL terms by } \color{blue}{ x\cdot3 }. \\[1 em]x\cdot3\cdot4 \cdot \frac{x}{x}+x\cdot3\cdot\frac{1}{3x} &= x\cdot3\cdot9&& \text{cancel out the denominators} \\[1 em]12x+x^2 &= 27x&& \text{simplify left side} \\[1 em]x^2+12x &= 27x&& \text{move all terms to the left hand side } \\[1 em]x^2+12x-27x &= 0&& \text{simplify left side} \\[1 em]x^2-15x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}-15x = 0 } $, first we need to factor our $ x $.
$$ x^{2}-15x = x \left( x-15 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x-15 = 0$.
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