$$\frac{7}{3x}+\frac{1}{3x} = 1$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 3 }{ 8 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{7}{3x}+\frac{1}{3x} &= 1&& \text{multiply ALL terms by } \color{blue}{ 3x }. \\[1 em]3x\cdot\frac{7}{3x}+3x\cdot\frac{1}{3x} &= 3x\cdot1&& \text{cancel out the denominators} \\[1 em]7x^2+x^2 &= 3x&& \text{simplify left side} \\[1 em]8x^2 &= 3x&& \text{move all terms to the left hand side } \\[1 em]8x^2-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 8x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ 8x^{2}-3x = x \left( 8x-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 8x-3 = 0$.
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