$$\frac{19}{2x}+\frac{5}{7} = \frac{9}{2}x$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 7 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{19}{2x}+\frac{5}{7} &= \frac{9}{2}x&& \text{multiply ALL terms by } \color{blue}{ 2x\cdot7 }. \\[1 em]2x\cdot7\cdot\frac{19}{2x}+2x\cdot7\cdot\frac{5}{7} &= 2x\cdot7 \cdot \frac{9}{2}x&& \text{cancel out the denominators} \\[1 em]133x^2+10x &= 63x^2&& \text{move all terms to the left hand side } \\[1 em]133x^2+10x-63x^2 &= 0&& \text{simplify left side} \\[1 em]70x^2+10x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 70x^{2}+10x = 0 } $, first we need to factor our $ x $.
$$ 70x^{2}+10x = x \left( 70x+10 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 70x+10 = 0$.
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