$$\frac{19}{4x}+\frac{2}{3} = \frac{11}{4x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{19}{4x}+\frac{2}{3} &= \frac{11}{4x}&& \text{multiply ALL terms by } \color{blue}{ 4x\cdot3 }. \\[1 em]4x\cdot3\cdot\frac{19}{4x}+4x\cdot3\cdot\frac{2}{3} &= 4x\cdot3\cdot\frac{11}{4x}&& \text{cancel out the denominators} \\[1 em]57x^2+8x &= 33x^2&& \text{move all terms to the left hand side } \\[1 em]57x^2+8x-33x^2 &= 0&& \text{simplify left side} \\[1 em]24x^2+8x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 24x^{2}+8x = 0 } $, first we need to factor our $ x $.
$$ 24x^{2}+8x = x \left( 24x+8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 24x+8 = 0$.
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