we have, (1 – i) x + (1 + i) y = 1 – 3i

⇒ x-ix+y+iy = 1-3i

⇒ (x+y)+i(-x+y) = 1-3i

On equating the real and imaginary coefficients we get,

⇒ x+y = 1 (i) and –x+y = -3 (ii)

From (i) we get

x = 1-y

substituting the value of x in (ii), we get

-(1-y)+y=-3

⇒ -1+y+y = -3

⇒ 2y = -3+1

⇒ y = -1

⇒ x=1-y = 1-(-1)=2

Hence, x=2 and y = -1