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in Quadratic Equations by (50.9k points)
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Solve for x: (1 – i) x + (1 + i) y = 1 – 3i.

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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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