frac{(1+i)x-2i}{3+i} + frac{(2-3i)y}{3-i} = i | vedclass

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(A) Given: $\frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y}{3-i} = i$ Multiply by $(3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10$: $((1+i)x-2i)(3-i) + (2-3i)y(3+i) =

Vedclass · Accepted Answer

$49/23$

Vedclass · Answer

(A) Given: $\frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y}{3-i} = i$ Multiply by $(3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10$: $((1+i)x-2i)(3-i) + (2-3i)y(3+i) = 10i$ $(3x - ix + 3ix - i^2x - 6i + 2i^2) + y(6 + 2i - 9i - 3i^2) = 10i$ $(3x - ix + 3ix + x - 6i - 2) + y(6 - 7i + 3) = 10i$ $(4x + 2ix - 2 - 6i) + y(9 - 7i) = 10i$ $(4x - 2 + 9y) + i(2x - 6 - 7y) = 10i$ Equating real and imaginary parts: Real part: $4x + 9y - 2 = 0 \Rightarrow 4x + 9y = 2$ Imaginary part: $2x - 7y - 6 = 10 \Rightarrow 2x - 7y = 16$ Solving the system: Multiply the second equation by $2$: $4x - 14y = 32$ Subtract from the first: $(4x + 9y) - (4x - 14y) = 2 - 32$ $23y = -30 \Rightarrow y = -30/23$ Substitute $y$ into $2x - 7y = 16$: $2x - 7(-30/23) = 16 \Rightarrow 2x + 210/23 = 368/23$ $2x = 158/23 \Rightarrow x = 79/23$ Therefore,$x+y = 79/23 - 30/23 = 49/23$.

$\frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y}{3-i} = i$. Find the value of $x+y$.

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