Question

If we have given the following expression:
$\dfrac{2x-2}{3}-\dfrac{2}{x+1}=\dfrac{2{{x}^{2}}-8}{3x+m}$
Find the value of “m”.

Answer 1

Hint: First of all, simplify the L.H.S of the given equation by taking the L.C.M of this expression. The L.C.M of the L.H.S expression is $3\left( x+1 \right)$ and then multiply $\left( x+1 \right)$ with $\left( 2x-2 \right)$ and also multiply 2 with 3. Now, simplify the numerator of the L.H.S of the given expression. After that, compare the L.H.S and R.H.S of the given equation and from that comparison find the value of “m”.

Complete step by step solution:
The equation given in the above problem is as follows:
$\dfrac{2x-2}{3}-\dfrac{2}{x+1}=\dfrac{2{{x}^{2}}-8}{3x+m}$
Let us simplify the L.H.S of the above equation by taking the L.C.M of the denominators of the L.H.S expression.
$\begin{align}
  & \dfrac{2x-2}{3}-\dfrac{2}{x+1} \\
 & =\dfrac{\left( 2x-2 \right)\left( x+1 \right)-2\left( 3 \right)}{3\left( x+1 \right)} \\
 & =\dfrac{2x\left( x \right)+2x-2x-2-6}{3\left( x+1 \right)} \\
\end{align}$
The same terms with opposite sign gets canceled out in the numerator of the above expression and we get,
$=\dfrac{2{{x}^{2}}+2x-2x-8}{3\left( x+1 \right)}$
Now, multiplying 3 inside the bracket in the denominator of the above expression and we get,
$\dfrac{2{{x}^{2}}-8}{3x+3}$
Now, comparing the above expression to the R.H.S of the given equation and we get,
$\dfrac{2{{x}^{2}}-8}{3x+3}=\dfrac{2{{x}^{2}}-8}{3x+m}$
In the above equation, as you can see that the numerators of the two sides are same so they get canceled out and we get,
$\begin{align}
  & \dfrac{2{{x}^{2}}-8}{3x+3}=\dfrac{2{{x}^{2}}-8}{3x+m} \\
 & \Rightarrow \dfrac{1}{3x+3}=\dfrac{1}{3x+m} \\
\end{align}$
On cross multiplying the above equation we get,
$3x+m=3x+3$
As you can see that $3x$ lies on both the sides of the above equation so it will get canceled out and the above expression will look like:
$\begin{align}
  & 3x+m=3x+3 \\
 & \Rightarrow m=3 \\
\end{align}$
Hence, we have found the value of m as 3.

Note: You can check the value of “m” which we have found in the above solution by substituting this value of “m” as 3 in the given equation and we get,
$\begin{align}
  & \dfrac{2x-2}{3}-\dfrac{2}{x+1}=\dfrac{2{{x}^{2}}-8}{3x+m} \\
 & \Rightarrow \dfrac{2x-2}{3}-\dfrac{2}{x+1}=\dfrac{2{{x}^{2}}-8}{3x+3} \\
\end{align}$
Taking L.C.M of the denominators in the L.H.S of the above expression and we get,
$\begin{align}
& \dfrac{\left( 2x-2 \right)\left( x+1 \right)-2\left( 3 \right)}{3\left( x+1 \right)}=\dfrac{2{{x}^{2}}-8}{3x+3} \\
& \Rightarrow \dfrac{2x\left( x \right)+2x-2x-2-6}{3\left( x+1 \right)}=\dfrac{2{{x}^{2}}-8}{3x+3} \\
\end{align}$
$\Rightarrow \dfrac{2{{x}^{2}}-8}{3x+3}=\dfrac{2{{x}^{2}}-8}{3x+3}$
As you can see that L.H.S is coming as equal to R.H.S so the value of “m” which we have found out is correct.