Question

How do you solve $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$?

Answer 1

Hint: As can be seen in the given equation, $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$ that it contains fractions with the variables present in the denominators also. We know that the denominator must not be equal to zero. Applying this condition on the denominator terms $\left( x-1 \right)$ and $\left( 2x-1 \right)$, we will obtain the domain of the solution of the given equation. Then we have to simplify the given equation by multiplying the equation by the LCM of the denominator terms $\left( x-1 \right)$ and $\left( 2x-1 \right)$ which is equal to $\left( x-1 \right)\left( 2x-1 \right)$. On multiplying $\left( x-1 \right)\left( 2x-1 \right)$ with the given equation, we will get a quadratic equation. On solving the quadratic equation obtained, we will get two solutions. Finally we need to check the obtained solutions according to the domain of solutions.

Complete step by step answer:
The equation in the above question is written as
$\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$
We can see that there are variable terms present in the denominators. Since the denominator of a fraction cannot be equal to zero, we have the following conditions
$\begin{align}
  & \Rightarrow \left( x-1 \right)\ne 0 \\
 & \Rightarrow x\ne 1 \\
\end{align}$
And
$\begin{align}
  & \Rightarrow \left( 2x-1 \right)\ne 0 \\
 & \Rightarrow x\ne \dfrac{1}{2} \\
\end{align}$
From the above two conditions, we obtain the domain of solutions for the above equation to as
$D=R-\left\{ \dfrac{1}{2},1 \right\}.......(i)$
Now, we again consider the given equation
$\Rightarrow \dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$
For getting rid of fractions, we multiply both sides of the above equation by $\left( x-1 \right)\left( 2x-1 \right)$ to get
\[\begin{align}
  & \Rightarrow \left( x-1 \right)\left( 2x-1 \right)\left( \dfrac{x+1}{x-1} \right)=\dfrac{2\left( x-1 \right)\left( 2x-1 \right)}{2x-1}+\dfrac{2\left( x-1 \right)\left( 2x-1 \right)}{x-1} \\
 & \Rightarrow \left( 2x-1 \right)\left( x+1 \right)=2\left( x-1 \right)+2\left( 2x-1 \right) \\
 & \Rightarrow 2x\left( x+1 \right)-1\left( x+1 \right)=2x-2+4x-2 \\
 & \Rightarrow 2{{x}^{2}}+2x-x-1=2x+4x-2-2 \\
 & \Rightarrow 2{{x}^{2}}+x-1=6x-4 \\
\end{align}\]
Subtracting $6x$ from both the sides, we get
$\begin{align}
  & \Rightarrow 2{{x}^{2}}+x-1-6x=6x-4-6x \\
 & \Rightarrow 2{{x}^{2}}-5x-1=-4 \\
\end{align}$
Adding $4$ both sides, we have
$\begin{align}
  & \Rightarrow 2{{x}^{2}}-5x-1+4=-4+4 \\
 & \Rightarrow 2{{x}^{2}}-5x+3=0 \\
\end{align}$
Now, by the middle split technique we split the middle term as $-5x=-2x-3x$ we get
$\Rightarrow 2{{x}^{2}}-2x-3x+3=0$
Taking $2x$ common from the first two terms and $-3$ common from the last two terms, we get
$\Rightarrow 2x\left( x-1 \right)-3\left( x-1 \right)=0$
Now, taking $\left( x-1 \right)$ common, we have
\[\begin{align}
  & \Rightarrow \left( x-1 \right)\left( 2x-3 \right)=0 \\
 & \Rightarrow x=1,x=\dfrac{3}{2} \\
\end{align}\]
From (i) the domain of the solutions is $R-\left\{ \dfrac{1}{2},1 \right\}$. So the value $x=1$ does not belong to the domain and therefore it is rejected.

Hence, the solution of the given equation is $x=\dfrac{3}{2}$.

Note: Do not forget to determine the domain of solutions of the given equation before solving it. Also, since the solution of the given question involves many calculations, so the chances of calculation mistakes are huge. Therefore, it is advised to back substitute the final solution obtained into the given equation for the confirmation.