Question

The value of $x$ and $y$ satisfying the equation $\frac{\left(1+i\right)x-2i}{3+i}+\frac{\left(2-3i\right)y+i}{3-i}=i$

• A. $x=-1,y=3$
• B. $x=3,y=-1$
• C. $x=0,y=1$
• D. $x=1,y=0$

Answer 1

The correct option is B

$x=3,y=-1$

Explanation for correct option:

Given, $\frac{\left(1+i\right)x-2i}{3+i}+\frac{\left(2-3i\right)y+i}{3-i}=i$

$\begin{array}{rrcl}\Rightarrow & \frac{x+\left(x-2\right)i}{3+i}+\frac{2y+\left(1-3y\right)i}{3-i}& =& i\\ \Rightarrow & \frac{\left(x+\left(x-2\right)i\right)\left(3-i\right)+\left(2y+\left(1-3y\right)i\right)\left(3+i\right)}{9-i^2}& =& i\\ \Rightarrow & x\left(3-i\right)+i\left(x-2\right)\left(3-i\right)+2y\left(3+i\right)+i\left(1-3y\right)\left(3+i\right)& =& 10i\\ \Rightarrow & 3x-ix+3xi-i^{2}x-6i+2i^2+6y+2iy+3i+i^2-9yi-3y{i}^2& =& 10i\\ \Rightarrow & 3x-ix+3xi+x-6i-2+6y+2iy+3i-1-9yi+3y& =& 10i\\ \Rightarrow & 4x+9y-3+2xi-7yi-13i& =& 0\\ \Rightarrow & 4x+9y-3+\left(2x-7y-13\right)i& =& 0\end{array}$

Comparing the real and imaginary part, we get

$$\begin{gathered} 4x+9y-3=0........\left(i\right) \\ 2x-7y-13=0.......\left(ii\right) \end{gathered}$$

Solving both we get

$$\begin{gathered} x=3 \\ y=-1 \end{gathered}$$

Hence, option B is the correct.