Question

If ,$\begin{array}{ll}\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}& =i\end{array}$then $(x,y)$ equal

• A. $(3,1)$
• B. $(3,-1)$
• C. $(-3,1)$
• D. $(-3,-1)$

Answer 1

The correct option is B

$(3,-1)$

Step 1: Simplify the given equation

$\begin{array}{ll}\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}& =i\end{array}$

$\Rightarrow$ $\begin{array}{ll}\frac{\left[\right(1+i)x-2i](3-i)}{(3+i)(3-i)}+\frac{\left[\right(2-3i)y+i](3+i)}{(3-i)(3+i)}& =i\end{array}$

$\Rightarrow$ $\begin{array}{ll}\frac{(x+xi-2i)(3-i)+[2y+(1-3y\left)i\right](3+i)}{9+1}& =i\end{array}$

$\Rightarrow$ $\begin{array}{ll}(x+xi-2i)(3-i)+[2y+(1-3y\left)i\right](3+i)& =10i\end{array}$

$\Rightarrow$$\begin{array}{ll}(3x+3xi-6i-xi+x-2)+(6y+3i-9yi+2yi-1+3y)& =10i\end{array}$

$\Rightarrow$ $\begin{array}{ll}(4x+9y-3)+i(2x-7y-13)& =0\end{array}$

Comparing real and imaginary part

$\begin{array}{rl}4x+9y-3& =0\end{array}$ …..$\left(1\right)$

$\begin{array}{rl}2x-7y-13& =0\end{array}$ ..…$\left(2\right)$

Multiply equation($2$) by $2$

$\begin{array}{rl}4x-14y-26& =0\end{array}$ …..$\left(3\right)$

Step 2: Subtract equation(3) from equation(1)

$\begin{array}{rl}23y+23& =0\end{array}$

$\Rightarrow$$\begin{array}{rl}23(y+1)& =0\end{array}$

$\Rightarrow$ $\begin{array}{rl}y+1& =0\end{array}$

$\Rightarrow$ $\begin{array}{rl}y& =-1\end{array}$

Put $y=-1$ in equation ($2$)

$\begin{array}{rl}2x-7(-1)-13& =0\end{array}$

$\Rightarrow$ $\begin{array}{rl}2x+7-13& =0\end{array}$

$\Rightarrow$ $\begin{array}{rl}2x-6& =0\end{array}$

$\Rightarrow$ $\begin{array}{rl}2x& =6\end{array}$

$\Rightarrow$ $\begin{array}{rl}x& =3\end{array}$

$\therefore$The value of $(x,y)$ is $\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}$.

Hence, option ‘B’ is correct option.