Question

Solve the following equation by factorisation:

$\frac{x+1}{x-1}+\frac{x-2}{x+2}=3(x\ne 1,-2)$

Answer 1

Step 1: Simplifying the given equation:

We have, $\frac{x+1}{x-1}+\frac{x-2}{x+2}=3(x\ne 1,-2)$

Rearrange the equation in the form of $A{x}^2+Bx+C=0$.

$$\begin{gathered} \Rightarrow \frac{x+1}{x-1}+\frac{x-2}{x+2}-3=0 \\ \Rightarrow \frac{(x+1)(x+2)+(x-2)(x-1)-3(x-1)(x+2)}{(x-1)(x+2)}=0 \\ \Rightarrow \frac{x^2+2x+x+2+x^2-x-2x+2-3(x^2-x+2x-2)}{(x-1)(x+2)}=0 \\ \Rightarrow \frac{2x^2+4-3x^2-3x+6}{(x-1)(x+2)}=0 \end{gathered}$$

After simplifying,

$$\begin{gathered} \Rightarrow \frac{-x^2-3x+10}{(x-1)(x+2)}=0 \\ \Rightarrow -x^2-3x+10=0 \end{gathered}$$

Step 2: Factorizing the simplified equation:

$$\begin{gathered} \Rightarrow -x^2-3x+10=0 \\ \Rightarrow -x^2-5x+2x+10=0 \\ \Rightarrow -x(x+5)+2(x+5)=0 \\ \Rightarrow (-x+2)(x+5)=0 \end{gathered}$$

Using zero-product property find the value of $x$.

$$\begin{gathered} \Rightarrow -x+2=0 \\ \Rightarrow x=2 \end{gathered}$$

Or,

$$\begin{gathered} \Rightarrow x+5=0 \\ \Rightarrow x=-5 \end{gathered}$$

Therefore, the values are $x=2$ or, $x=-5$.