Solve by elimination method:$ax+by=1$ and $bx+ay=\frac{(a+b)^2}{a^2+b^2}-1$.

Given : 

$ax+by=1$ 

$bx+ay=\frac{(a+b)^2}{a^2+b^2}-1$  

To do:

We have to solve the given equations by elimination method.

Solution : 

$ax+by=1$ .....................................(i)

$bx+ay=\frac{(a+b)^2}{a^2+b^2}-1$

Take L C M on Right hand side,

$\displaystyle bx\ +\ ay\ \ =\ \frac{( a+b)^{2} \ -\ \left( a^{2} \ +\ b^{2} \ \right)}{a^{2} \ +\ b^{2} \ }$

$\displaystyle bx\ +\ ay\ \ =\ \frac{a^{2} +b^{2} \ +\ 2ab\ -\ a^{2} \ -\ b^{2} \ }{a^{2} \ +\ b^{2} \ }$ 

$$\displaystyle ( a+b)^{2} \ =a^{2} +b^{2} \ +\ 2ab\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^{2} \ -\ a^{2} \ =\ 0\ ;\ b^{2} \ -\ b^{2} \ =\ 0\ $$

$\displaystyle bx\ +\ ay\ \ =\ \frac{\ 2ab\ \ }{a^{2} \ +\ b^{2} \ }$......................(ii)

Multiply (i)  by a,

$\displaystyle a^{2} x\ +\ aby\ =\ a$............................(iii)

Multiply (ii)  by b,

$\displaystyle b^{2} x\ +\ aby\ \ =\ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ }$..........(iv)

$\displaystyle ( iii) \ -\ ( iv) \ \ \Rightarrow $

$\displaystyle a^{2} x\ -\ b^{2} x\ +\ aby\ \ -\ aby\ =a\ -\ \ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ }$

$\displaystyle \begin{array}{{>{\displaystyle}l}} a^{2} x\ -\ b^{2} x\ \ =a\ -\ \ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ aby\ -\ aby\ =\ 0\ \ \end{array}$

$\displaystyle \begin{array}{{>{\displaystyle}l}} x\left( a^{2} \ -\ b^{2}\right) \ =a\ -\ \ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

Take L C M on Right Hand Side,

$ x\left( a^{2} \ -\ b^{2}\right) \ =\ \ \frac{a\ \left( a^{2} \ +\ b^{2} \ \right) \ -\ \ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ }$

Take a as common on Right Hand Side

$x\left( a^{2} \ -\ b^{2}\right) \ =\ \ \frac{a\ \left[ a^{2} \ +\ b^{2} \ \ -\ 2b^{2}\right] \ \ }{a^{2} \ +\ b^{2} \ }$

$\displaystyle x\left( a^{2} \ -\ b^{2}\right) \ =\ \ \frac{a\ \left[ a^{2} -b^{2}\right] \ \ }{a^{2} \ +\ b^{2} \ } \ \ \ \ \ \ $

$x=\ \ \frac{a\ \left[ a^{2} -b^{2}\right] \ \ }{\left( a^{2} \ +\ b^{2}\right)\left( a^{2} \ -\ b^{2}\right) \ \ }$

$x\ =\ \frac{a}{a^{2} \ +\ b^{2}}$

Substitute 'x' value in (i),

$a x\ +\ by\ =\ 1$

$a\left( \ \frac{a}{a^{2} \ +\ b^{2}} \ \right) \ +\ by\ =\ 1$

$\frac{a^{2}}{a^{2} \ +\ b^{2}} \ +\ by\ =\ 1$

$by\ \ =\ \ 1\ -\ \frac{a^{2}}{a^{2} \ +\ b^{2}}$

Take LCM on Right Hand Side

$by\ \ =\ \ \ \frac{a^{2} \ +\ b^{2} -a^{2}}{a^{2} \ +\ b^{2}}$

$by\ \ =\ \ \ \frac{\ b^{2}}{a^{2} \ +\ b^{2}}$

$y\ \ =\ \ \ \frac{\ b^{2}}{b\left( a^{2} \ +\ b^{2}\right)}$

$y\ =\ \frac{\ b}{a^{2} \ +\ b^{2}}$

$x\ =\ \frac{a}{a^{2} \ +\ b^{2}}$.

$y\ =\ \frac{\ b}{a^{2} \ +\ b^{2}}$.