Question

Solve the given pair of linear equations:
$\begin{align}
  & \left( a-b \right)x+\left( a+b \right)y={{a}^{2}}-2ab-{{b}^{2}} \\
 & \left( a+b \right)\left( x+y \right)={{a}^{2}}+{{b}^{2}} \\
\end{align}$

Answer 1

Hint: In the given pair of linear equations, subtract the second equation from the first equation then you will find that y is eliminated and you are left with the variable x then simplify and get the value of x and substitute the value of x in the second equation. This substitution gives you the value of y.

Complete step-by-step answer:
The given pair of linear equations is:
$\begin{align}
  & \left( a-b \right)x+\left( a+b \right)y={{a}^{2}}-2ab-{{b}^{2}} \\
 & \left( a+b \right)\left( x+y \right)={{a}^{2}}+{{b}^{2}} \\
\end{align}$
Simplifying the left hand side of the second equation we get,
$\left( a+b \right)x+\left( a+b \right)y={{a}^{2}}+{{b}^{2}}$ ………….. Eq. (1)
$\left( a-b \right)x+\left( a+b \right)y={{a}^{2}}-2ab-{{b}^{2}}$ ………….. Eq. (2)
Subtracting eq. (2) from eq. (1) we get,
$\begin{align}
  & \left( a+b \right)x+\left( a+b \right)y={{a}^{2}}+{{b}^{2}} \\
 & \dfrac{-\left( \left( a-b \right)x+\left( a+b \right)y={{a}^{2}}-2ab-{{b}^{2}} \right)}{2bx=2{{b}^{2}}+2ab} \\
\end{align}$
Simplifying the result of the above subtraction we get,
$2bx=2{{b}^{2}}+2ab$
As you can see 2 is common on both the left and right hand side of the above equation so it will be cancelled out.
$bx={{b}^{2}}+ab$
Dividing “b” on both the sides of the above equation we get,
$x=b+a$
Substituting the above value of “x” in eq. (1) we get,
$\begin{align}
  & \left( a+b \right)x+\left( a+b \right)y={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow \left( a+b \right)\left( a+b \right)+\left( a+b \right)y={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}+2ab+\left( a+b \right)y={{a}^{2}}+{{b}^{2}} \\
\end{align}$
In the above equation ${{a}^{2}}+{{b}^{2}}$ will be cancelled out from both the left hand and right hand side of the equation.
$\begin{align}
  & 2ab+\left( a+b \right)y=0 \\
 & \Rightarrow y=-\dfrac{2ab}{a+b} \\
\end{align}$
From the above solution, the values of x and y we have got are: $\begin{align}
  & x=a+b; \\
 & y=-\dfrac{2ab}{a+b} \\
\end{align}$

Note: You can verify the values of x and y that you have got above by substituting these values of x and y in any one of the given equations.
The value of x and y that we have got above is:
$\begin{align}
  & x=a+b; \\
 & y=-\dfrac{2ab}{a+b} \\
\end{align}$
We are going to substitute these above values in eq. (1).
$\left( a+b \right)x+\left( a+b \right)y={{a}^{2}}+{{b}^{2}}$
Solving the L.H.S of the above equation we get,
$\begin{align}
  & \left( a+b \right)x+\left( a+b \right)y={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow \left( a+b \right)\left( a+b \right)+\left( a+b \right)\left( -\dfrac{2ab}{a+b} \right)={{a}^{2}}+{{b}^{2}} \\
\end{align}$
$\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}+2ab-2ab={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\
\end{align}$
From the above equation, we can see that the L.H.S = R.H.S which proves that the values of x and y is satisfying the linear pair of equations.