Solve for x and y: $\dfrac{15}{x-y}+\dfrac{22}{x+y}=5,\dfrac{40}{x-y}+\dfrac{55}{x+y}=13,$ $x\ne y$ and $x\ne -y$.
A. $x=-8\text{ and }y=-\text{3}$
B. $x=8\text{ and y=8}$
C. $x=3\text{ and y=3}$
D. $x=8\text{ and y=3}$

Hint: To solve this question first we will convert the given equation into a linear equation into two variables. Then we solve the obtained equations by using the elimination method. We eliminate one of the variables from equations and find the value of another variable. Then substitute the value in one of the equations to get the value of the variable.

Complete answer:
We have been given two equations $\dfrac{15}{x-y}+\dfrac{22}{x+y}=5,\dfrac{40}{x-y}+\dfrac{55}{x+y}=13,$
Let us assume $\dfrac{1}{x-y}=u$ and $\dfrac{1}{x+y}=v$.
Let us take the first equation $\dfrac{15}{x-y}+\dfrac{22}{x+y}=5$
Now substituting the values in the given equation we get
$15u+22v=5............(i)$
Let us take the second equation $\dfrac{40}{x-y}+\dfrac{55}{x+y}=13$
Now substituting the values in the equation we get
$40u+55v=13............(ii)$
Now, to solve equations we multiply the equation (i) from 40 and equation (ii) from 15, we get
$\begin{align}
  & \Rightarrow 15\times 40u+22\times 40v=5\times 40 \\
 & \Rightarrow 600u+880v=200.........(iii) \\
\end{align}$
$\begin{align}
  & \Rightarrow 40\times 15u+55\times 15v=13\times 15 \\
 & \Rightarrow 600u+825v=195...........(iv) \\
\end{align}$
Now, subtract equation (iv) from equation (iii), we get
$\Rightarrow \left( 600u+880v \right)-\left( 600u+825v \right)=200-195$
Now, simplifying further we get
$\begin{align}
  & \Rightarrow 600u+880v-600u-825v=200-195 \\
 & \Rightarrow 55v=5 \\
 & \Rightarrow v=\dfrac{5}{55} \\
 & \Rightarrow v=\dfrac{1}{11} \\
\end{align}$
Now, substituting the value of $v$ in equation (i) we get
$\begin{align}
  & \Rightarrow 15u+22\times \dfrac{1}{11}=5 \\
 & \Rightarrow 15u+2=5 \\
 & \Rightarrow 15u=5-2 \\
 & \Rightarrow 15u=3 \\
 & \Rightarrow u=\dfrac{3}{15} \\
 & \Rightarrow u=\dfrac{1}{5} \\
\end{align}$
Now, we have $\dfrac{1}{x-y}=u$ and $\dfrac{1}{x+y}=v$.
Substituting the values of $u\And v$ in the above equations we get
$\begin{align}
  & \Rightarrow \dfrac{1}{x-y}=\dfrac{1}{5} \\
 & \Rightarrow x-y=5.........(v) \\
\end{align}$
And
 $\begin{align}
  & \Rightarrow \dfrac{1}{x+y}=\dfrac{1}{11} \\
 & \Rightarrow x+y=11........(vi) \\
\end{align}$
Now, add both the equation (v) and equation (vi), we get
$\begin{align}
  & \Rightarrow \left( x-y \right)+\left( x+y \right)=5+11 \\
 & \Rightarrow 2x=16 \\
 & \Rightarrow x=\dfrac{16}{2} \\
 & \Rightarrow x=8 \\
\end{align}$
Now, substituting the value of x in equation (vi) we get
$\begin{align}
  & \Rightarrow 8+y=11 \\
 & \Rightarrow y=11-8 \\
 & \Rightarrow y=3 \\
\end{align}$
So, we get the values $x=8\And y=3$

Option D is the correct answer.

Note:
Alternatively one can directly solve the pair of equations but conversion of equation and using the substitution method to solve the equations is the easiest way. Here we use the elimination method to solve the equations. Students can use a substitution method also.