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

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(A) Given: $\frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2 i$Multiply the first term by $\frac{2-i}{2-i}$ and the second term by $\frac{1

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$4$

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(A) Given: $\frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2 i$Multiply the first term by $\frac{2-i}{2-i}$ and the second term by $\frac{1-2i}{1-2i}$:$\frac{(2-i)^2 x+(1+i)(2-i)}{5}+\frac{(1-2 i)^2 y+(1-i)(1-2 i)}{5}=1-2 i$$\Rightarrow \frac{(3-4i)x + (3+i)}{5} + \frac{(-3-4i)y + (-1-3i)}{5} = 1-2i$$\Rightarrow (3x-3y-1) + i(-4x-4y-2) = 5-10i$Equating real and imaginary parts:$3x-3y-1 = 5 \Rightarrow 3x-3y = 6 \Rightarrow x-y = 2 \quad (i)$$-4x-4y-2 = -10 \Rightarrow -4x-4y = -8 \Rightarrow x+y = 2 \quad (ii)$Adding $(i)$ and $(ii)$,$2x = 4 \Rightarrow x = 2$.Substituting $x=2$ in $(ii)$,$2+y = 2 \Rightarrow y = 0$.Therefore,$2x+4y = 2(2) + 4(0) = 4$.

If $\frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2 i$,then $2 x+4 y=$

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