Question

a^2x+b^2y=c^2,b^2x+a^2y=d^2

Answer 1

multiply the first eq by a^2 and second eq. by b^2
then the two equations would become a^4x+a^2b^2y= a^2c^2 and b^4x+a^2b^2y= b^2d^2 
by solving both the equations you will get x= a^2c^2- b^2d^2 / a^4 - b^4 and y= a^2d^2- b^2c^2 / a^4 - b^4

hope it helps..

Answer 2

Given:

$a^{2} x+b^{2}y=c^{2}; \ b^{2} x+a^{2}y=d^{2}$

Answer:

Given two equations are

$a^{2} x+b^{2}y=c^{2} \rightarrow(1)$

$b^{2} x+a^{2}y=d^{2} \rightarrow(2)$

$Multiply \ the \ equation \ 1\times a^2 \ and \ 2\times b^2$

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

$a^{4} x+a^{2} b^{2} y-a^{2} c^{2}=0 \rightarrow(3)$

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

$b^{4} \mathrm{x}+a^{2} b^{2} y-b^{2} d^{2}=0 \rightarrow(4)$

Subtract the equation 4 from 3

$b^{4} x+a^{2} b^{2} y-b^{2} d^{2}-a^{4} x-a^{2} b^{2} y+a^{2} c^{2}=0$

$x\left(b^{4}-a^{4}\right)=b^{2} d^{2}-a^{2} c^{2}$

$x=\frac{b^{2} d^{2}-a^{2} c^{2}}{b^{4}-a^{4}}$

$x=\frac{a^{2} c^{2}-b^{2} d^{2}}{a^{4}-b^{4}}$

$Similarly, Multiply \ the \ equation \ 1\times a^2 \ and \ 2\times b^2$

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

$a^{2} b^{2} x+b^{4} y-b^{2} c^{2}=0$

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

$a^{2} b^{2} x+a^{24} y-a^{2} d^{2}=0$

$a^{2} b^{2} x+b^{4} y-b^{2} c^{2}-a^{2} b^{2} x-a^{4} y+a^{2} d^{2}=0$

$y\left(b^{4}-a^{4}\right)=b^{2} c^{2}-a^{2} d^{2}$

$y=\frac{b^{2} c^{2}-a^{2} d^{2}}{b^{4}-a^{4}}$