Solve for x, y: -
\[\begin{align}
  & \left( a+2b \right)x+\left( 2a-b \right)y=2 \\
 & \left( a-2b \right)x+\left( 2a+b \right)y=3 \\
\end{align}\]

Hint: Assume the two equations as equation (1) and equation (2). Take the sum of two equations and form a relation between x and y. Now, take the difference of the two equations and form another relation in x and y. Solve the newly obtained equations by the elimination method and find the value of x and y.

Complete step-by-step solution
We have been provided with two equations: -
\[\left( a+2b \right)x+\left( 2a-b \right)y=2\] - (1)
\[\left( a-2b \right)x+\left( 2a+b \right)y=3\] - (2)
Now, adding equations (1) and (2), we get,
\[\begin{align}
  & \Rightarrow \left( a+2b \right)x+\left( a-2b \right)x+\left( 2a-b \right)y+\left( 2a+b \right)y=2+3 \\
 & \Rightarrow \left( a+2b+a-2b \right)x+\left( 2a-b+2a+b \right)y=5 \\
\end{align}\]
Cancelling the like terms, we get,
\[\begin{align}
  & \Rightarrow 2ax+4ay=5 \\
 & \Rightarrow 2a\left( x+2y \right)=5 \\
\end{align}\]
Dividing both sides by 2a, we get,
\[\Rightarrow x+2y=\dfrac{5}{2a}\] - (3)
Now, subtracting equation (2) from equation (1), we get,
\[\begin{align}
  & \Rightarrow \left( a+2b \right)x-\left( a-2b \right)x+\left( 2a-b \right)y-\left( 2a+b \right)y=2-3 \\
 & \Rightarrow \left( a+2b-a+2b \right)x+\left( 2a-b-2a-b \right)y=-1 \\
\end{align}\]
Cancelling the like terms, we get,
\[\begin{align}
  & \Rightarrow 4bx-2by=-1 \\
 & \Rightarrow 2b\left( 2x-y \right)=-1 \\
\end{align}\]
Dividing both sides by 2b, we get,
\[\Rightarrow 2x-y=\dfrac{-1}{2b}\] - (4)
Multiplying equation (4) with 2 and adding it with equation (3), we get,
\[\begin{align}
  & \Rightarrow 2\left( 2x-y \right)+x+2y=2\times \left( \dfrac{-1}{2b} \right)+\dfrac{5}{2a} \\
 & \Rightarrow 4x-2y+x+2y=\dfrac{-1}{b}+\dfrac{5}{2a} \\
 & \Rightarrow 5x=\dfrac{-2a+5b}{2ab} \\
 & \Rightarrow x=\dfrac{5b-2a}{10ab} \\
\end{align}\]
Substituting the value of x, obtained above, in equation (4), we get,
\[\begin{align}
  & \Rightarrow 2\times \left( \dfrac{5b-2a}{10ab} \right)-y=\dfrac{-1}{2b} \\
 & \Rightarrow \dfrac{5b-2a}{5ab}-y=\dfrac{-1}{2b} \\
 & \Rightarrow y=\dfrac{5b-2a}{5ab}+\dfrac{1}{2b} \\
 & \Rightarrow y=\dfrac{2\times \left( 5b-2a \right)+1\times 5a}{10ab} \\
 & \Rightarrow y=\dfrac{10b-4a+5a}{10ab} \\
 & \Rightarrow y=\dfrac{10b+a}{10ab} \\
\end{align}\]
Hence, the values of x and y obtained by solving the equations are: -
\[x=\dfrac{5b-2a}{10ab}\] and \[y=\dfrac{10b+a}{10ab}\], which is our required solution.

Note: One may note that here we have applied the method of elimination of variables to solve the above question. One can also use the method is substitution or the method of cross – multiplication to solve the question. But remember that in these two processes we have to encounter some difficult calculations. That is why the method of elimination is preferred. One can see that we have added equations (1) and (2) in the first process and subtracted in the second process. The main reason for this was to remove ‘a’ and ‘b’ from the L.H.S of the given equations.