Question

Solve for $ x: $ $ \dfrac{1}{\left( x-1 \right)\left( x-2 \right)}+\dfrac{1}{\left( x-2 \right)\left( x-3 \right)}=\dfrac{2}{3} $ , $ x\ne 1,2,3 $

Answer 1

Hint: Here we have been given an equation and we have to find the value of the unknown variable in it. Firstly we will take the common term from the denominator outside. Then we will use the LCM method to solve the terms left in the bracket. Finally we will cross multiply the values and form an equation which on solving we will get our desired answer.

Complete step-by-step answer:
We have to solve the below value:
 $ \dfrac{1}{\left( x-1 \right)\left( x-2 \right)}+\dfrac{1}{\left( x-2 \right)\left( x-3 \right)}=\dfrac{2}{3} $ , $ x\ne 1,2,3 $
So firstly take $ \left( x-2 \right) $ common from the denominator terms on the left side as follows:
 $ \Rightarrow \dfrac{1}{\left( x-2 \right)}\left( \dfrac{1}{\left( x-1 \right)}+\dfrac{1}{\left( x-3 \right)} \right)=\dfrac{2}{3} $
Now take the LCM of the bracket term as follows:
 $ \Rightarrow \dfrac{1}{\left( x-2 \right)}\left( \dfrac{\left( x-3 \right)+\left( x-1 \right)}{\left( x-1 \right)\left( x-3 \right)} \right)=\dfrac{2}{3} $
 $ \Rightarrow \dfrac{1}{\left( x-2 \right)}\left( \dfrac{2x-4}{\left( x-1 \right)\left( x-3 \right)} \right)=\dfrac{2}{3} $
Taking $ 2 $ common from the numerator term on left side we get,
 $ \Rightarrow \dfrac{2}{\left( x-2 \right)}\left( \dfrac{x-2}{\left( x-1 \right)\left( x-3 \right)} \right)=\dfrac{2}{3} $
Cancel out the common terms on the left side:
 $ \Rightarrow 2\left( \dfrac{1}{\left( x-1 \right)\left( x-3 \right)} \right)=\dfrac{2}{3} $
Now cross multiply as follows:
 $ \Rightarrow 2\times 3=2\times \left( x-1 \right)\left( x-3 \right) $
 $ \Rightarrow 6=2\left( x-1 \right)\left( x-3 \right) $
On dividing both sides by $ 2 $ we get,
 $ \Rightarrow 3=\left( x-1 \right)\left( x-3 \right) $
Now multiply the two brackets on the right hand side,
 $ \Rightarrow 3=x\times x+x\times -3-1\times x-1\times -3 $
On simplifying we get,
 $ \Rightarrow 3={{x}^{2}}-3x-x+3 $
 $ \Rightarrow 3={{x}^{2}}-4x+3 $
Take all the terms on one side,
 $ \Rightarrow {{x}^{2}}-4x+3-3=0 $
 $ \Rightarrow {{x}^{2}}-4x=0 $
Take $ x $ common,
 $ \Rightarrow x\left( x-4 \right)=0 $
So we get,
 $ x=0 $ And $ x-4=0 $
 $ \Rightarrow x=0 $ And $ x=4 $
Hence on solving $ \dfrac{1}{\left( x-1 \right)\left( x-2 \right)}+\dfrac{1}{\left( x-2 \right)\left( x-3 \right)}=\dfrac{2}{3} $ , $ x\ne 1,2,3 $ we get the value of $ x=0,4 $ .
So, the correct answer is “ $ x=0,4 $ ”.

Note: In this question we have taken common from the denominator so that we can cancel out that term in next step because if we try to solve without taking common we will get the equation with variable in power of $ 3 $ and it will get little complicated to solve that equation. Again as the constant was eliminated in the last equation it becomes easy for us to just take the common term and get our answer otherwise we would have used the quadratic formula to get our answer.