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`

⇒ b²x + a²y = (ab)a² + b²

⇒ b²x + a²y = a³b + ab³    ...(i)

Also, x + y = 2ab   ...(ii)

On multiplying (ii) by a², we get:

a²x + a²y = 2a³b   ...(iii)

On subtracting (iii) from (i), we get:

(b² – a²)x = a³b + ab³ – 2a³b

⇒ (b² – a²)x = –a³b + ab³

⇒ (b² – a²)x = ab(b² – a²)

∴ `x = (ab(b^2 - a^2))/((b^2 - a^2)) = ab`

On substituting x = ab in (i), we get:

b²(ab) + a²y = a³b + ab³

⇒ a²y = a³b

⇒ `(a^3b)/(a^2) = ab`

Hence, the solution is x = ab and y = ab.