Question

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

Answer 1

Hint: The main objective here is to eliminate all the fractions. At first, we multiply $x-1$ on both sides of the above equation in order to simplify it and get rid of the $x-1$ in the denominator. Then, we multiply $2x-1$ on both sides of the equation. Finally, we get a quadratic equation $2{{x}^{2}}-5x+3=0$ which we need to solve using the Sridhar Acharya formula.

Complete step-by-step solution:
The given equation that we need to solve in this problem is,
$\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$
We see x in the numerator as well as the denominator and that too on both sides of the above equation. At first, we multiply $x-1$ on both sides of the above equation in order to simplify it. After doing so, the above equation thus becomes,
$\Rightarrow \dfrac{x+1}{x-1}\times \left( x-1 \right)=\left( \dfrac{2}{2x-1}+\dfrac{2}{x-1} \right)\times \left( x-1 \right)$
After a little bit of simplification by multiplying and inserting the term $\left( x-1 \right)$ within the brackets, the above equation thus becomes,
$\Rightarrow x+1=\dfrac{2\left( x-1 \right)}{2x-1}+2$
After simplification of the above equation, we still see a fraction in the right-hand side of the equation. To remove or eliminate this, we need to get rid of the denominator. This can be done by multiplying $2x-1$ on both sides of the equation. After doing so, the above equation thus becomes,
$\Rightarrow \left( x+1 \right)\left( 2x-1 \right)=2\left( x-1 \right)+2\left( 2x-1 \right)$
After this, we need to apply the distributive property to the right-hand side of the above equation. The distributive property goes as follow: an expression of the form $a\left( b+c \right)$ is equal to $ab+ac$ . Doing so in the RHS of the above equation we get,
$\begin{align}
  & \Rightarrow \left( x+1 \right)\left( 2x-1 \right)=2x-2+4x-2 \\
 & \Rightarrow \left( x+1 \right)\left( 2x-1 \right)=6x-4 \\
\end{align}$
As we can see, even the LHS requires the distributive property. Doing so gives,
$\begin{align}
  & \Rightarrow 2{{x}^{2}}+x-1=6x-4 \\
 & \Rightarrow 2{{x}^{2}}-5x+3=0 \\
\end{align}$
This can be solved by the Sridhar Acharya formula which is, for a quadratic expression of the form $a{{x}^{2}}+bx+c$ , $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . So, for the above equation, it gets,
$\begin{align}
  & x=\dfrac{5\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-24}}{4} \\
 & \Rightarrow x=\dfrac{5\pm 1}{4} \\
 & \Rightarrow x=\dfrac{6}{4},\dfrac{4}{4} \\
 & \Rightarrow x=\dfrac{3}{2},1 \\
\end{align}$
Therefore, we can conclude that the solution of the given equation is $x=\dfrac{3}{2},1$ .

Note: The equation is too complicated as it involves the variable x in both the numerator and denominator of both sides of the equation. Thus, this problem requires careful solution of it. The multiplication of the terms using the distributive property keeping the proper signs in mind is very necessary. Finally, the Sridhar Acharya formula must be carried out correctly by putting the correct coefficients with their signs in the formula.