If (1 - i)x + (1 + i)y = 1 - 3i,then (x, y) = | vedclass

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(A) Given the equation: $(1 - i)x + (1 + i)y = 1 - 3i$Expanding the terms,we get: $(x - ix) + (y + iy) = 1 - 3i$Grouping real and imaginary parts: $(x

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$(2, -1)$

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(A) Given the equation: $(1 - i)x + (1 + i)y = 1 - 3i$Expanding the terms,we get: $(x - ix) + (y + iy) = 1 - 3i$Grouping real and imaginary parts: $(x + y) + i(y - x) = 1 - 3i$Equating the real and imaginary parts on both sides:$x + y = 1$  (Equation $1$)$y - x = -3$ (Equation $2$)Adding Equation $1$ and Equation $2$:$(x + y) + (y - x) = 1 + (-3)$$2y = -2$$y = -1$Substituting $y = -1$ into Equation $1$:$x + (-1) = 1$$x = 2$Therefore,$(x, y) = (2, -1)$.

If $(1 - i)x + (1 + i)y = 1 - 3i$,then $(x, y) = $

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