# Solve for x: (1 –

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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