find the complex numbers z and w in form of x +yi which satisfy the equation system (1+i)z-iw+i= iz+(1-i)w-3i =6?

(1+i)z - iw = 6 - i ..... (1)
iz + (1-i)w = 6 + 3i ..... (2)
(1) - (2)
z - w = - 4i
z = w - 4i , substitute into (2)
iw + 4 + (1-i)w = 6 + 3i
w = 2 + 3i
z = w - 4i = 2 + 3i - 4i = 2 - i
Ans: z = 2 - i , w = 2 + 3i

The original system=>
(1+i)z-iw=6-i--------(1)
iz+(1-i)w=6+3i-----(2)
=>
....| (6-i)..... ....- i |
....| (6+3i)...(1-i) |
z=---------------------=>
....| (1+i)......-i |
....| i........(1-i) |

z=2-i

w=2+3i from (2)

(1) : (1 + i).z - iw + i = 6

(1) : (1 + i).z = 6 + iw - i

(1) : z = (6 + iw - i)/(1 + i)



(2) : iz + (1 - i).w - 3i = 6

(2) : iz = 6 - (1 - i).w + 3i

(2) : z = (6 - w + iw + 3i)/i


z = z → (1) = (2)

(6 + iw - i)/(1 + i) = (6 - w + iw + 3i)/i

i.(6 + iw - i) = (1 + i).(6 - w + iw + 3i)

6i + i²w - i² = 6 - w + iw + 3i + 6i - iw + i²w + 3i²

- i² = 6 - w + 3i + 3i² → where: i² = - 1

1 = 6 - w + 3i - 3

→ w = 2 + 3i


Recall (2): z = (6 - w + iw + 3i)/i

z = [6 - (2 + 3i) + i.(2 + 3i) + 3i)/i

z = (6 - 2 - 3i + 2i + 3i² + 3i)/i

z = (4 + 2i + 3i²)/i → where: i² = - 1

z = (4 + 2i - 3)/i

z = (1 + 2i)/i

z = [(1 + 2i).i] / (i * i)

z = (i + 2i²) / i² → where: i² = - 1

z = (i - 2) / - 1

→ z = 2 - i