complex number?

Question

Solve (1-i)x +3i/2+i + (3+2i)y+i/2-i = -i where x,y belongs to R whoever will give me the right answer with ans i will rate him

Answer 1

(1-i)x+3i/2+i+(3+2i)y+i/2-i=-i
(x+3y) + (-x+3/2+1+2y+1/2)i=0
(x+3y) + (-x+2y+3)i = 0
x+3y = 0 ----(1)
-x+2y+3 = 0---(2)
(1)+(2)
5y+3=0
y= -3/5
x= 9/5

Answer 2

( 1 - i ) x + ( 3i / 2 )+ i + ( 3 + 2i ) y + ( i / 2 ) - i = - i

x - ix + ( 3i / 2 ) + i +3y +2iy + ( i / 2 ) - i = 0 - i

Clubbing all the real and the imaginary terms we get

( x + 3y ) + i ( - x + 3/2 +1 + 2y + 1/2 - 1 ) = 0 - i

( x + 3y ) + i ( - x + 2y + 2 ) = 0 - i

Comparing the real and imaginary terms we get

( x + 3y ) = 0 and

( - x + 2y +2 ) = - 1 => ( -x + 2y ) = -3

solving the terms simultaneously

x = 9 / 5 and y = ( - 3 / 5 )

Answer 3

(1-i)x+3i/2+i+(3+2i)y+i/2-i=-i
x-xi+3i/2+i+3y+2yi+i/2-i=-i
(x+3y)+i(-x+3/2+1+2y+1/2)= 0+0i
now equting real and imaginary term
x+3y=0 & -x+2y=-3
now on solving both
5y= -3
y= -3/5
x= -3* -3/5=9/5
Hence: x=9/5 & y= --3/5:
ans...............

Answer 4

The way to do this is to create simultaneous equations for the real and imaginary parts:
x -xi +3i/2 + 3y +2yi+i/2 - i = - i
(x+3y) + (-x+3/2+2y+0.5)i = 0
(x+3y) + (-x+2+2y)i = 0

So we know that both the real part and the imaginary part is equal to zero, or:
x+3y = 0
-x+2+2y = 0
and we can add the two together to give:
2+5y=0
y=2/5
and substitute in the first equation:
x+6/5=0
x=-6/5

Answer 5

By opening the brackets using distributive properties and rearranging real part and imaginary parts we get
(x+3y)-(x-2y-2) i = -i
and comparing real and imaginary parts, we get
x+3y=0 & x-2y-2=1 solving these equations
x= 9/5 and y = - 3/5

Answer 6

The answert is wat robcraine has given i.e.x=-6/5 and y=2/5