Question

The values of $x$ and $y$ satisfying the equation $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$, are

Answer 1

Consider the given expressions.

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

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

$\frac{(x+(x-2)i)(3-i)+(2y+(1-3y)i)(3+i)}{9-i^2}=i$

$x(3-i)+i(x-2)(3-i)+2y(3+i)+i(1-3y)(3+i)=(9+1)i$

$3x-ix+i(3x-ix-6+2i)+6y+2iy+i(3+i-9y-3yi)=10i$

$3x-ix+3xi-i^{2}x-6i+2i^2+6y+2iy+3i+i^2-9yi-3y{i}^2=10i$

$3x-ix+3xi+x-6i-2+6y+2iy+3i-1-9yi+3y=10i$

$4x+9y-3+2xi-7yi-13i=0$

$4x+9y-3+(2x-7y-13)i=0$

On comparing real part and imaginary part, we get

$4x+9y-3=0$ …… (1)

$2x-7y-13=0$ …… (2)

On solving both equations, we get

$x=3$

$y=-1$

Hence, the value of $x,y$ is $3,-1$.