Question

Find the real values of x and y, if
(i) $(x+iy)(2-3i)=4+i$
(ii) $(3x-2iy)(2+i{)}^2=10(1+i)$
(iii) $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}$
(iv) $(1+i)(x+iy)=2-5i$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\left(x+iy\right)\left(2-3i\right)=4+i \\ 2x-3ix+2iy-3i^{2}y=4+i \\ 2x+3y+i\left(-3x+2y\right)=4+i \\ \mathrm{Comparing}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ 2x+3y=4....\left(1\right) \\ -3x+2y=1....\left(2\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(1\right)\mathrm{by}3\mathrm{and}\mathrm{equation}\left(2\right)\mathrm{by}2: \\ 6x+9y=12...\left(3\right) \\ -6x+4y=2...\left(4\right) \\ \mathrm{Adding}\mathrm{equations}\left(3\right)\mathrm{and}\left(4\right): \\ 13y=14 \\ y=\frac{14}{13} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{equation}\left(1\right): \\ 2x+3\times \frac{14}{13}=4 \\ \Rightarrow 2x=4-\frac{42}{13} \\ \Rightarrow 2x=\frac{10}{13} \\ \Rightarrow x=\frac{5}{13} \\ \therefore x=\frac{5}{13}\mathrm{and}y=\frac{14}{13} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\left(3x-2iy\right){\left(2+i\right)}^2=10\left(1+i\right) \\ \Rightarrow \left(3x-2iy\right)\left(4+i^2+4i\right)=10\left(1+i\right) \\ \Rightarrow \left(3x-2iy\right)\left(3+4i\right)=10\left(1+i\right) \\ \Rightarrow 9x+12xi-6iy-8i^{2}y=10+10i \\ \Rightarrow 9x+8y+i\left(12x-6y\right)=10+10i \\ \mathrm{Comparing}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ 9x+8y=10....\left(1\right) \\ 12x-6y=10 \\ \mathrm{or},6x-3y=5...\left(2\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(1\right)\mathrm{by}3\mathrm{and}\mathrm{equation}\left(2\right)\mathrm{by}8, \\ 27x+24y=30....\left(3\right) \\ 48x-24y=40....\left(4\right) \\ \mathrm{Adding}\mathrm{equations}\left(3\right)\mathrm{and}\left(4\right): \\ 75x=70 \\ \therefore x=\frac{14}{15} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{equation}\left(1\right): \\ 9\times \frac{14}{15}+8y=10 \\ \Rightarrow \frac{126}{15}+8y=10 \\ \Rightarrow 8y=10-\frac{126}{15} \\ \Rightarrow 8y=\frac{24}{15} \\ \Rightarrow y=\frac{1}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\frac{\left(1+i\right)x-2i}{3+i}+\frac{\left(2-3i\right)y+i}{3-i}=i \\ \Rightarrow \frac{\left(1+i\right)\left(3-i\right)x-2i\left(3-i\right)+\left(2-3i\right)\left(3+i\right)y+i\left(3+i\right)}{\left(3+i\right)\left(3-i\right)}=i \\ \Rightarrow \frac{3x-ix+3ix-i^{2}x-6i+2i^2+6y+2iy-9iy-3i^{2}y+3i+i^2}{9-i^2}=i \\ \Rightarrow \frac{4x+2ix-3i+9y-7iy-3}{10}=i \\ \Rightarrow \left(4x+9y-3\right)+i\left(2x-3-7y\right)=10i \\ \mathrm{Comparing}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ 4x+9y-3=0 \\ \Rightarrow 4x+9y=3....\left(1\right) \\ 2x-3-7y=10 \\ \Rightarrow 2x-7y=13...\left(2\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(2\right)\mathrm{by}2: \\ 4x-14y=26...\left(3\right) \\ \mathrm{Subtracting}\mathrm{equation}\left(3\right)\mathrm{from}\left(1\right): \\ 4x+9y=3 \\ 4x-14y=26 \\ \underline{-+-} \\ 23y=-23 \\ \therefore y=-1 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{equation}\left(1\right): \\ 4x-9=3 \\ \Rightarrow 4x=12 \\ \Rightarrow x=3 \\ \therefore x=3\mathrm{and}y=-1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\left(1+i\right)\left(x+iy\right)=2-5i \\ \Rightarrow x+iy+ix+i^{2}y=2-5i \\ \Rightarrow x+iy+ix-y=2-5i \\ \Rightarrow \left(x-y\right)+i\left(y+x\right)=2-5i \\ \mathrm{Comparing}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ x-y=2...\left(1\right) \\ x+y=-5...\left(2\right) \\ \mathrm{Adding}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right), \\ 2x=-3 \\ \Rightarrow x=\frac{-3}{2} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{equation}\left(1\right), \\ \frac{-3}{2}-y=2 \\ \Rightarrow y=\frac{-3}{2}-2 \\ \Rightarrow y=\frac{-7}{2} \\ \therefore x=\frac{-3}{2}\mathrm{and}y=\frac{-7}{2} \end{gathered}$$