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

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

• $\left(\frac{7}{5}, \frac{- 7}{15}\right)$

• $\left(\frac{7}{5}, \frac{7}{15}\right)$

A.

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

$$\begin{gathered} \frac{\left(1 + \mathrm{i}\right)\mathrm{x} - \mathrm{i}}{2 + \mathrm{i}} + \frac{\left(1 + 2\mathrm{i}\right)\mathrm{y} + \mathrm{i}}{2 - \mathrm{i}} = 1 \\ \Rightarrow \frac{\left[\left(1 + \mathrm{i}\right)\mathrm{x} - \mathrm{i}\right]\left(2 - \mathrm{i}\right)}{4 - {\mathrm{i}}^2} + \frac{\left[\left(1 + 2\mathrm{i}\right)\mathrm{y} + \mathrm{i}\right]\left(2 + \mathrm{i}\right)}{4 - {\mathrm{i}}^2} = 1 \\ \Rightarrow \frac{2\left(1 + \mathrm{i}\right)\mathrm{x} - 2\mathrm{i} - \mathrm{i}\left(1 + \mathrm{i}\right)\mathrm{x} + {\mathrm{i}}^2}{4 + 1} + \frac{2\left(1 + 2\mathrm{i}\right)\mathrm{y} + 2\mathrm{i} + \mathrm{i}\left(1 + 2\mathrm{i}\right)\mathrm{x} + {\mathrm{i}}^2}{4 + 1} = 1 \\ \Rightarrow \frac{\left(2 + 2\mathrm{i} - \mathrm{i} - {\mathrm{i}}^2\right)\mathrm{x} - 2\mathrm{i} + {\mathrm{i}}^2}{5} + \frac{\left(2 + 4\mathrm{i} + \mathrm{i} + 2{\mathrm{i}}^2\right)\mathrm{y} + 2\mathrm{i} + {\mathrm{i}}^2}{5} = 1 \\ \Rightarrow \left(3 + \mathrm{i}\right)\mathrm{x} - 2\mathrm{i} - 1 +\hspace{0.17em}\left(5\mathrm{i}\right)\mathrm{y} +\hspace{0.17em}2\mathrm{i} - 1 = 5 \\ \Rightarrow \left(3 + \mathrm{i}\right)\mathrm{x} + 5\mathrm{iy} = 7 \\ \Rightarrow 3\mathrm{x} + \mathrm{ix} + 5\mathrm{iy} - 7 = 0 \\ \Rightarrow \left(3\mathrm{x} - 7\right) + \left(\mathrm{x} +\hspace{0.17em}5\mathrm{y}\right)\mathrm{i} = 0 + 0\mathrm{i} \\ \mathrm{On} \mathrm{compairing}, \mathrm{we} \mathrm{get} \\ 3\mathrm{x} - 7 = 0 \\ \Rightarrow \mathrm{x} = \frac{7}{3}, \mathrm{x} + 5\mathrm{y} = 0 \\ \Rightarrow \mathrm{y} = \frac{- 7}{15} \\ \mathrm{Hence}, \left(\mathrm{x}, \mathrm{y}\right) = \left(\frac{7}{3}, \frac{- 7}{15}\right) \end{gathered}$$