Question

Solve :
`a/x - b/y = 0`

`(ab^2)/x + (a^2b)/y = a^2 + b^2`

Answer 1

Given equation are `a/x - b/y = 0 and (ab^2)/x + (a^2b)/y = a^2 + b^2`

Taking `1/x = u and 1/y = v`, the above system of equations become
au - bv + 0 = 0
ab²u + a²bv - ( a² + b² ) = 0
By cross-multiplication, we have
`u/[ -b xx [-( a^2 + b^2 )] - a^2b xx 0 ] = [-v]/[ a xx [-( a^2 + b^2 )] - ab^2 xx 0 ] = 1/[ a xx a^2b - ab^2 xx ( - b )]`

⇒ `u/[b( a^2 + b^2 )] = (-v)/[ - a( a^2 + b^2 )] = 1/[a^3b + ab^3 ]`

⇒ `u/[b( a^2 + b^2 )] = v/[ a( a^2 + b^2 )] = 1/[ab(a^2 + b^2 )]`

⇒ `u = [b( a^2 + b^2 )]/[ab( a^2 + b^2 )]  and  v = [a( a^2 + b^2 )]/[ ab( a^2 + b^2 )]`

⇒ u = `1/a`  and v = `1/b`

⇒ `1/x = 1/a  and 1/y = 1/b` 

⇒ x =a and y = b