$$\frac{3x-1}{x+3}+\frac{3x+2}{x+1} = 3$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = -\dfrac{ 4 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{3x-1}{x+3}+\frac{3x+2}{x+1} &= 3&& \text{multiply ALL terms by } \color{blue}{ (x+3)(x+1) }. \\[1 em](x+3)(x+1)\frac{3x-1}{x+3}+(x+3)(x+1)\frac{3x+2}{x+1} &= (x+3)(x+1)\cdot3&& \text{cancel out the denominators} \\[1 em]3x^2+2x-1+3x^2+11x+6 &= 3x^2+12x+9&& \text{simplify left side} \\[1 em]6x^2+13x+5 &= 3x^2+12x+9&& \text{move all terms to the left hand side } \\[1 em]6x^2+13x+5-3x^2-12x-9 &= 0&& \text{simplify left side} \\[1 em]3x^2+x-4 &= 0&& \\[1 em] \end{aligned} $$
$ 3x^{2}+x-4 = 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