$$5 \cdot \frac{x}{x+4} = 4 \cdot \frac{x}{x+2}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 6 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5 \cdot \frac{x}{x+4} &= 4 \cdot \frac{x}{x+2}&& \text{multiply ALL terms by } \color{blue}{ (x+4)(x+2) }. \\[1 em](x+4)(x+2)\cdot5 \cdot \frac{x}{x+4} &= (x+4)(x+2)\cdot4 \cdot \frac{x}{x+2}&& \text{cancel out the denominators} \\[1 em]5x^2+10x &= 4x^2+16x&& \text{move all terms to the left hand side } \\[1 em]5x^2+10x-4x^2-16x &= 0&& \text{simplify left side} \\[1 em]x^2-6x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}-6x = 0 } $, first we need to factor our $ x $.
$$ x^{2}-6x = x \left( x-6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x-6 = 0$.
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