Question

Solve for x and y:

`(bx)/a - (ay)/b + a + b = 0, bx – ay + 2ab = 0`

Answer 1

The given equations are:
`(bx)/a - (ay)/b + a + b = 0`
By taking LCM, we get:
`b^2x – a^2y = -a^2b – b^2a`                          …..(i)
and bx – ay + 2ab = 0
bx – ay = -2ab                                                  …….(ii)
On multiplying (ii) by a, we get:
`abx – a^2y = -2a^2b`                                      ……(iii)
On subtracting (i) from (iii), we get:
`abx – b^2x = 2a^2b + a^2b + b^2a = -a^2b + b^2a`

`⇒x(ab – b^2) = -ab(a – b)`
⇒x(a – b)b = -ab(a – b)
`∴x = (−ab(a−b))/((a−b)b) = -a`
On substituting x = -a in (i), we get:
`b^2(-a) – a^2y = -a^2b – b^2a`
`⇒ -b^2a – a^2y = -a^2b – b^2a`
`⇒ -a^2y = -a^2b`
⇒ y = b
Hence, the solution is x = -a and y = b.