Complex numbers finding real values

Question

I am trying to solve for x and y using the following equation: $4i + 2 = \frac{x + iy + 5 + 4i}{2x + 2iy - 5}$

I got it down to real and imaginary, but am unsure what to do next. $3x-8y-15=(-8x-3y+24)i$

Answer 1

Hint

Assuming $\ x,y\ $ are required to be real numbers, then equating the real and imaginary parts of both sides of the equation $$ 3x-8y-15=(-8x-3y+24)i $$ will give you two linear equations in $\ x\ $ and $\ y\ $ which you can solve to obtain their valuea.

Answer 2

Suppose a,b are two real numbers,if we are having a=ib as some condition,as a,b are real they can be equal only when both of the sides of equation are zero. So a=b=0.you can proceed with your problem now.