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