Question

How do you simplify the given equation: $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$?

Answer 1

Hint: We start solving the problem by making use of the fact that $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}$ in the given equation. We then make the necessary calculations to get the quadratic equation in x. We then make use of the quadratic formula that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to proceed through the problem. We then make the necessary calculations to find the roots of the obtained quadratic equation and then neglect the values at which the given function is not defined to get the required answer.

Complete step-by-step answer:
According to the problem, we are asked to simplify the given equation: $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$.
We have given the equation $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$ ---(1).
We know that $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}$. Let us use this result in equation (1).
\[\Rightarrow \dfrac{x+1}{x-1}=\dfrac{2\left( x-1 \right)+2\left( 2x-1 \right)}{\left( 2x-1 \right)\left( x-1 \right)}\].
\[\Rightarrow \dfrac{x+1}{x-1}=\dfrac{2x-2+4x-2}{\left( 2x-1 \right)\left( x-1 \right)}\].
\[\Rightarrow \dfrac{x+1}{x-1}=\dfrac{6x-4}{\left( 2x-1 \right)\left( x-1 \right)}\].
\[\Rightarrow x+1=\dfrac{6x-4}{2x-1}\] ---(2).
Now, let us cross multiply the numerator and denominators of both sides of equation (2).
\[\Rightarrow \left( x+1 \right)\left( 2x-1 \right)=6x-4\].
\[\Rightarrow 2{{x}^{2}}-x+2x-1-6x+4=0\].
\[\Rightarrow 2{{x}^{2}}-5x+3=0\].
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, which is also known as quadratic formula.
Now, let us compare the given quadratic equation $2{{x}^{2}}-5x+3=0$ with $a{{x}^{2}}+bx+c=0$. So, we get $a=2$, $b=-5$, $c=3$. Now, let us substitute these values in the quadratic formula to get the roots of the given quadratic equation $2{{x}^{2}}-5x+3=0$.
So, the roots are $x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)}$.
$\Rightarrow x=\dfrac{5\pm \sqrt{25-24}}{4}$.
$\Rightarrow x=\dfrac{5\pm \sqrt{1}}{4}$.
$\Rightarrow x=\dfrac{5\pm 1}{4}$.
$\Rightarrow x=\dfrac{5+1}{4}$, $\dfrac{5-1}{4}$.
$\Rightarrow x=\dfrac{6}{4}$, $\dfrac{4}{4}$.
$\Rightarrow x=\dfrac{3}{2}$, 1.
But we can see that the given equation doesn’t exist for $x=1$. So, the solution of the given equation $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$ as $x=\dfrac{3}{2}$.
$\therefore $ The solution of the given equation $\dfrac{x+1}{x-1}=\dfrac{2}{2x-1}+\dfrac{2}{x-1}$ is $x=\dfrac{3}{2}$.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. We should not report $x=1$ as the solution of the given problem, which is the common mistake done by the students. We can also verify the obtained answer by substituting it in the given equation and checking the values in both L.H.S (Left Hand Side) and R.H.S (Right Hand Side). Similarly, we can expect the problems to find the solution of the given equation: $\dfrac{2x-1}{x-2}=\dfrac{3}{x-4}+\dfrac{3}{x-2}$.