The values of x and y satisfying the equation | vedclass

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(B) Given equation: $\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$Multiply by the common denominator $(3+i)(3-i) = 9 - i^2 = 10$:$((1

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$x = 3, y = -1$

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(B) Given equation: $\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$Multiply by the common denominator $(3+i)(3-i) = 9 - i^2 = 10$:$((1 + i)x - 2i)(3 - i) + ((2 - 3i)y + i)(3 + i) = 10i$$(3x - ix + 3ix - i^2x - 6i + 2i^2) + (6y + 2iy - 9iy - 3i^2y + 3i + i^2) = 10i$$(3x - ix + 3ix + x - 6i - 2) + (6y + 2iy - 9iy + 3y + 3i - 1) = 10i$$(4x + 2ix - 6i - 2) + (9y - 7iy + 3i - 1) = 10i$$(4x + 9y - 3) + i(2x - 7y - 3) = 10i$Equating real and imaginary parts:Real part: $4x + 9y - 3 = 0 \implies 4x + 9y = 3$Imaginary part: $2x - 7y - 3 = 10 \implies 2x - 7y = 13$Solving the system:From $2x - 7y = 13$,$x = \frac{13 + 7y}{2}$.Substitute into $4x + 9y = 3$: $2(13 + 7y) + 9y = 3 \implies 26 + 14y + 9y = 3 \implies 23y = -23 \implies y = -1$.Then $x = \frac{13 + 7(-1)}{2} = \frac{6}{2} = 3$.Thus,$x = 3$ and $y = -1$.

The values of $x$ and $y$ satisfying the equation $\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$ are

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