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

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

Vedclass · Accepted Answer

$\left(\frac{7}{3}, \frac{-7}{15}\right)$

Vedclass · Answer

(A) Given the equation: $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$ Multiply the first term by $\frac{2-i}{2-i}$ and the second term by $\frac{2+i}{2+i}$: $\frac{[(1+i)x-i](2-i)}{5} + \frac{[(1+2i)y+i](2+i)}{5} = 1$ Expanding the numerators: $\frac{(2-i+2i-i^2)x - 2i + i^2}{5} + \frac{(2+i+4i+2i^2)y + 2i + i^2}{5} = 1$ Since $i^2 = -1$: $\frac{(3+i)x - 2i - 1}{5} + \frac{(5i)y + 2i - 1}{5} = 1$ $(3+i)x + (5i)y - 2 = 5$ $(3x-7) + i(x+5y) = 0$ Comparing real and imaginary parts: $3x-7 = 0 \Rightarrow x = \frac{7}{3}$ $x+5y = 0 \Rightarrow y = -\frac{x}{5} = -\frac{7}{15}$ Thus,$(x, y) = \left(\frac{7}{3}, -\frac{7}{15}\right)$.

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

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