Solve the algebraic expression $a\left( x+y \right)+b\left( x-y \right)={{a}^{2}}-ab+{{b}^{2}}$ and $a\left( x+y \right)-b\left( x-y \right)={{a}^{2}}+ab+{{b}^{2}}$.

Hint: We will solve this question by using elimination method. With the help of elimination method we will get rid of one of the terms with the variable and we will be able to solve it further. Since there are algebraic equations therefore we will use elimination carefully.

Complete step-by-step answer:
We will consider $a\left( x+y \right)+b\left( x-y \right)={{a}^{2}}-ab+{{b}^{2}}...(i)$ and $a\left( x+y \right)-b\left( x-y \right)={{a}^{2}}+ab+{{b}^{2}}...(ii)$. Now we will use the elimination method. With the help of this method we will eliminate the term $b\left( x-y \right)$ by adding the equations (i) and (ii). Therefore we have
$\begin{align}
  & a\left( x+y \right)+b\left( x-y \right)={{a}^{2}}-ab+{{b}^{2}} \\
 & \underline{a\left( x+y \right)-b\left( x-y \right)={{a}^{2}}+ab+{{b}^{2}}} \\
 & 2a\left( x+y \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2{{a}^{2}}\,\,\,\,\,\,\,\,+2{{b}^{2}} \\
\end{align}$
Thus we get
$\begin{align}
  & 2a\left( x+y \right)=2{{a}^{2}}+2{{b}^{2}} \\
 & \Rightarrow 2ax+2ay=2{{a}^{2}}+2{{b}^{2}} \\
 & \Rightarrow ax+ay={{a}^{2}}+{{b}^{2}} \\
\end{align}$
Now we will take the term ‘a’ as a common term from the left side of the equation. Thus we get a new equation $x+y=\dfrac{{{a}^{2}}+{{b}^{2}}}{a}$ which can also be written as $x+y=a+\dfrac{{{b}^{2}}}{a}...(iii)$.
Now we will eliminate the term $a\left( x+y \right)$ by subtracting the equations (i) and (ii). Therefore we have
$\begin{align}
  & a\left( x+y \right)+b\left( x-y \right)={{a}^{2}}-ab+{{b}^{2}} \\
 & \underline{\pm a\left( x+y \right)\mp b\left( x-y \right)=\pm {{a}^{2}}\pm ab\pm {{b}^{2}}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2b\left( x-y \right)=\,\,\,\,\,\,\,\,-2ab \\
\end{align}$
Thus we get
$\begin{align}
  & 2b\left( x-y \right)=-2ab \\
 & \Rightarrow 2bx-2by=-2ab \\
 & \Rightarrow bx-by=-ab \\
\end{align}$
Now we will take the term ‘b’ as a common term from the left side of the equation. Therefore we get $b\left( x-y \right)=-ab$ which after solving can be written as $x-y=\dfrac{-ab}{b}$ or we have $x-y=-a...(iv)$.
Now we again use the elimination method here on the equations (iii) and (iv) and with the help of it we will eliminate the variable y. So, we will add the equations (iii) and (iv). Therefore, we get
$\begin{align}
  & x+y=a+\dfrac{{{b}^{2}}}{a} \\
 & \underline{x-y=-a} \\
 & 2x\,\,\,\,\,\,=\,\,\,\,\,\,\,\,\dfrac{{{b}^{2}}}{a} \\
\end{align}$
Therefore we now have a new equation which is as $2x=\dfrac{{{b}^{2}}}{a}$ or $x=\dfrac{{{b}^{2}}}{2a}$. Now we will use substitution in this step. That is we will substitute the value of x given as $x=\dfrac{{{b}^{2}}}{2a}$ in equation (iv). Therefore, we get
$\begin{align}
  & x-y=-a \\
 & \Rightarrow \dfrac{{{b}^{2}}}{2a}-y=-a \\
 & \Rightarrow y=-a+\dfrac{{{b}^{2}}}{2a} \\
 & \Rightarrow y=\dfrac{-2{{a}^{2}}+{{b}^{2}}}{2a} \\
\end{align}$
Therefore, the value of $y=\dfrac{-2{{a}^{2}}+{{b}^{2}}}{2a}$.
Hence the required values of $x=\dfrac{{{b}^{2}}}{2a}$ and $y=\dfrac{-2{{a}^{2}}+{{b}^{2}}}{2a}$.

Note: We can also solve this question using only substitution methods but that will make the solution complex also it will take time to solve. With substitution we take the value of x by keeping all the terms to the right side of equation (i) except for x. And then substitute the value of x into equation (ii). Then proceed with the usual steps. Moreover we can also solve it with the help of Cramer’s rule in which we can convert the equation into specific determinants for finding the variables x and y. But elimination is the best method to solve this question as this question carries so many terms that in substitution one can get confused and result in the wrong answer.