Question
$$3-\frac{x}{x} = x+\frac{1}{2}x$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 4 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3-\frac{x}{x} &= x+\frac{1}{2}x&& \text{multiply ALL terms by } \color{blue}{ x\cdot2 }. \\[1 em]x\cdot2\cdot3-x\cdot2 \cdot \frac{x}{x} &= x\cdot2x+x\cdot2 \cdot \frac{1}{2}x&& \text{cancel out the denominators} \\[1 em]6x-2x &= 2x^2+x^2&& \text{simplify left and right hand side} \\[1 em]4x &= 3x^2&& \text{move all terms to the left hand side } \\[1 em]4x-3x^2 &= 0&& \text{simplify left side} \\[1 em]-3x^2+4x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -3x^{2}+4x = 0 } $, first we need to factor our $ x $.
$$ -3x^{2}+4x = x \left( -3x+4 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -3x+4 = 0$.
This page was created using
Equations Solver