Question
$$2 \cdot \frac{x}{x}+3-x+1 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 6 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 2 \cdot \frac{x}{x}+3-x+1 &= 0&& \text{multiply ALL terms by } \color{blue}{ x }. \\[1 em]x\cdot2 \cdot \frac{x}{x}+x\cdot3-xx+x\cdot1 &= x\cdot0&& \text{cancel out the denominators} \\[1 em]2x+3x-x^2+x &= 0&& \text{simplify left side} \\[1 em]-x^2+6x &= 0&& \\[1 em] \end{aligned} $$
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$.
This page was created using
Equations Solver