Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Let     Z_{1 =} (x – iy) (3 + 5i) = 3c + 5xi - 3yi - 5yi² = (3x + 5y) + i ( 5x - 3y)
   
Let     Z₂    =  -6 -24i
5
It is given  that Z_{1  is} conjugate of Z₂
 
         Z₁     =

   (3x + 5y) + i (5x - 3y) = -6 + 24 i

Equating real and imaginary parts, we get
                     3x + 5y  = - 6                                                             ....(i)
and                 5x - 3y   = 24                                                              ...(ii)

Multiplying (i)  by 5 and (ii) by 3, we get

             15x + 25y = -30                                                                  ...(iii)
              15x - 9y   = 72                                                                   ...(iv)

Subtracting (iv) from (iii), we get

                34y  = -102                    or 

Putting y = -3 in (i), we get
          3x + 5 (-3) = -6
or              3x = -6 +15 = 9   or      x    = 3

Hence,      x = 3,   y = -3