solve for x and y which are both real numbers?

Question

4x+3y(i-1)-12(1-i)=9-2x(1+i)+(1-2y)i

Answer 1

Rearrange these complex numbers so that the "a" and "b" parts of each are expressions of x and y:
4x + 3y (i-1) - 12 (1-i) = 9 - 2x (1+i) + (1 - 2y) i
4x + 3yi - 3y - 12 + 12i = 9 - 2x - 2xi + i - 2yi
4x - 3y - 12 + 3yi + 12i = 9 - 2x + i - 2xi - 2yi
(4x - 3y - 12) + (3y + 12) i = (9 - 2x) + (1 - 2x - 2y) i

You now have 2 equal complex numbers. The real parts are equal, therefore:
4x-3y-12=9-2x

and the coefficients on i must also be equal, therefore:
3y+12=1-2x-2y

Rearranging and solving those 2 equations we get:

6x-3y=21 and 2x+5y=-11

5 (6x-3y) = 5(21)
3 (2x+5y) = 3(-11)

30x-15y=105
6x+15y=-33
36x=72
x=2

6x-3y=21
6(2)-3y=21
12-3y=21
-3y=9
y=-3

Answer 2

You reorder the eq and compare the real coeff and imagine part

4x - 3y - 12 + (3y +12)i = 9 - 2x + (-2x+1-2y)i

solve 4x - 3y - 12 = 9 - 2x and 3y+12 = -2x-2y+1

6x - 3y = 21 and 2x + 5y = -11
x = 2 and y = -3

Answer 3

4x+3y(i-1)-12(1-i)=9-2x(1+i)+(1-2y)i

Seperate the 'rea' and 'complex' part as followsL

(4x - 3y -12) + (3y + 12)i = ( 9-2x ) + ( 1 -2y --2x )i

"equate them" [ complex coefficients with complex coefficients and real ] :

For the "real" we get,
(4x - 3y -12) = ( 9-2x )
6x = 3y + 21
2x = y + 7
y = 2x - 7 ....(1)

For the "complex part", we get
( 1 -2y --2x ) = (3y + 12)
5y = -11 -2x ...(2)

Put (1) in (2),
5( 2x - 7 ) = -11 -2x
10x - 35 = -11 -2x
12x = 35 - 11
12x = 24
x=2

and so y = 2x -7 = 2(2) - 7 = -3

Hence,

(x,y) =(2, -3)

Hope this helped.

:)