Question

How do you multiply \[\dfrac{3}{{x - 1}} + \dfrac{1}{{x\left( {x - 1} \right)}} = \dfrac{2}{x}\] ?

Answer 1

Hint: To solve this question, we have to assume that this question is a solve for $x$ type questions. As we can see that the denominators of all the terms are different, we have to take LCM. After that, we will get one linear equation by solving which we will get our final answer.

Complete step-by-step answer:
We have \[\dfrac{3}{{x - 1}} + \dfrac{1}{{x\left( {x - 1} \right)}} = \dfrac{2}{x}\] .
First we will find the LCM of the denominator terms $x - 1$, $x(x - 1)$and $x$. It is clear that the LCM will be $x(x - 1)$.
To solve this question by LCM method, we have to multiply the first term $3$by $x$, second term 1 with $1$ as it already has the LCM in its denominator and the third term \[2\] by $x - 1$.
Therefore, we will get
 \[\dfrac{{3x + 1}}{{x(x - 1)}} = \dfrac{{2(x - 1)}}{{x(x - 1)}}\]
Here, we have similar denominator terms on both sides, therefore, it will get cancelled out.
 \[
   \Rightarrow 3x + 1 = 2(x - 1) \\
   \Rightarrow 3x + 1 = 2x - 2 \;
 \]
Now, we just have to solve this linear equation. For this, we will bring the terms with variable $x$ on the left hand side and other constant terms at the right hand side.
 \[
   \Rightarrow 3x - 2x = - 2 - 1 \\
   \Rightarrow x = - 3 \;
 \]
So, the correct answer is “ x = - 3”.

Note: Here, we have solved the equation by taking LCM both the sides. However, it is also possible to solve the same question by taking LCM only on the left hand side and then simplifying the equation, we can reach our final answer.
For example,
 \[\dfrac{3}{{x - 1}} + \dfrac{1}{{x\left( {x - 1} \right)}} = \dfrac{2}{x}\]
 \[ \Rightarrow \dfrac{{3x + 1}}{{x(x - 1)}} = \dfrac{2}{x}\]
As we can see, here, we have kept the right hand side term as it is. Now, $x$is common in the denominators of both the sides, therefore, it will get cancelled out.
 \[
   \Rightarrow \dfrac{{3x + 1}}{{(x - 1)}} = 2 \\
   \Rightarrow 3x + 1 = 2(x - 1) \\
   \Rightarrow x = - 3 \;
 \]