Question
$$\frac{3}{x+1}-\frac{1}{2} = \frac{1}{3x+3}$$
Answer
$$ \begin{matrix}x_1 = -1 & x_2 = \dfrac{ 13 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{3}{x+1}-\frac{1}{2} &= \frac{1}{3x+3}&& \text{multiply ALL terms by } \color{blue}{ (x+1)\cdot2(3x+3) }. \\[1 em](x+1)\cdot2(3x+3)\cdot\frac{3}{x+1}-(x+1)\cdot2(3x+3)\cdot\frac{1}{2} &= (x+1)\cdot2(3x+3)\cdot\frac{1}{3x+3}&& \text{cancel out the denominators} \\[1 em]18x+18-(3x^2+6x+3) &= 2x+2&& \text{simplify left side} \\[1 em]18x+18-3x^2-6x-3 &= 2x+2&& \\[1 em]-3x^2+12x+15 &= 2x+2&& \text{move all terms to the left hand side } \\[1 em]-3x^2+12x+15-2x-2 &= 0&& \text{simplify left side} \\[1 em]-3x^2+10x+13 &= 0&& \\[1 em] \end{aligned} $$
$ -3x^{2}+10x+13 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
This page was created using
Equations Solver