Answer :

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


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