Question

Solve for x and y:
`(bx)/a + (ay)/b = a^2 + b^2, x + y = 2ab`

Answer 1

The given equations are:
`(bx)/a + (ay)/b = a^2 + b^2`
By taking LCM, we get:
`(b^2x+a^2y)/(ab) = a^2 + b^2`
`⇒b^2x + a^2y = (ab)a^2 + b^2`
`⇒b^2x + a^2y = a^3b +ab^3`              …..(i)
Also, x + y = 2ab                                    …….(ii)
On multiplying (ii) by a2, we get:
`a^2x + a^2y = 2a^3b`                          ……(iii)
On subtracting (iii) from (i), we get:
`(b^2 – a^2)x = a^3b +ab^3 – 2a^3b`
`⇒(b^2 – a^2)x = -a^3b +ab^3`
`⇒(b^2 – a^2)x = ab(b^2 – a^2)`
`⇒(b^2 – a^2)x = ab(b^2 – a^2)`
`∴x =( ab(b^2− a^2))/((b^2− a^2)) = ab`
On substituting x = ab in (i), we get:
`b^2(ab) + a^2y = a^3b + ab^3`
`⇒ a^2y = a^3b`
`⇒ (a^3b)/(a^2) = ab`
Hence, the solution is x = ab and y = ab.