Question

Solve for:

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

Answer 1

Step 1: Take the L.C.M of the denominator and simplify the expression in LHS

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

L.C.M of $\left(x-1\right)\left(x-2\right)$ and $\left(x-3\right)\left(x-2\right)$ is $\left(x-1\right)\left(x-2\right)\left(x-3\right)$

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

$\Rightarrow$$\frac{\left(x-3\right)+\left(x-1\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{2}{3}$

$\Rightarrow$$\frac{x-3+x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{2}{3}$

$\Rightarrow$$\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{2}{3}$

$\Rightarrow$$\frac{2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{2}{3}$

$\Rightarrow$ $\frac{2}{\left(x-1\right)\left(x-3\right)}=\frac{2}{3}$

$\Rightarrow$ $\frac{2}{x^2-3x-x+3}=\frac{2}{3}$

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

$\Rightarrow$ $x^2-4x+3=3$

Step 2: Solve the above quadratic equation

$x^2-4x+3=3$

$\Rightarrow$$x^2-4x+3-3=0$

$\Rightarrow$ $x^2-4x=0$

$\Rightarrow$ $x\left(x-4\right)=0$

$$\begin{gathered} \Rightarrow x=0orx-4=0 \\ \Rightarrow x=0orx=4 \end{gathered}$$

Hence, $x=0$ and $x=4$ are the solution of given equation.