Question

In the following exercises, solve each rational equation.

$\frac{2}{x^2+2x-3}+\frac{3}{x^2+4x+3}=\frac{6}{x^2-1}$

Short answer

The solution is$x=-19$.

Step-by-step solution

Step 1. Given 

The given equation is: $\frac{2}{x^2+2x-3}+\frac{3}{x^2+4x+3}=\frac{6}{x^2-1}$

Step 2. To find  

We have to solve the equation.

Step 3. Calculation 

Factor the denominators.

$$\begin{gathered} \frac{2}{x^2+2x-3}+\frac{3}{x^2+4x+3}=\frac{6}{x^2-1} \\ \frac{2}{\left(x+3\right)\left(x-1\right)}+\frac{3}{\left(x+3\right)\left(x+1\right)}=\frac{6}{\left(x+1\right)\left(x-1\right)} \end{gathered}$$

The denominator is never 0. So,$x\ne -3,1,-1$..

We multiply both sides by LCD and then solve the equation.

localid="1662465746915" $$\begin{gathered} \frac{2}{\left(x+3\right)\left(x-1\right)}+\frac{3}{\left(x+3\right)\left(x+1\right)}=\frac{6}{\left(x+1\right)\left(x-1\right)} \\ \frac{2}{\left(x+3\right)\left(x-1\right)}\left(x+3\right)\left(x-1\right)\left(x+1\right)+\frac{3}{\left(x+3\right)\left(x+1\right)}\left(x+3\right)\left(x-1\right)\left(x+1\right) \\ =\frac{6}{\left(x+1\right)\left(x-1\right)}\left(x+3\right)\left(x-1\right)\left(x+1\right) \\ 2(x+1)+3(x-1)=6(x+3) \\ 2x+2+3x-3=6x+18 \\ 5x-1=6x+18 \\ -x=19 \\ x=-19 \end{gathered}$$