Solve the system in complex set: ix-2y=-i (1+i)x-2iy=3+i

We'll solve the system using elimination
method.

We'll note the
equations:

ix - 2y = -i
(1)

(1+i)x - 2iy = 3+i
(2)

We'll multiply (1) by -i and we'll
get:

-i(ix - 2y) = -i*-i

We'll
remove the brackets:

-i^2*x + 2iy =
i^2

We'll substitute i^2 =
-1

-x + 2iy = -1 (3)

We'll add
(3) to (2):

-x + 2iy + (1+i)x - 2iy = -1 + 3 +
i

We'll remove the brackets and eliminate like
terms:

-x + x + ix = 2 + i

ix
= 2 + i

We'll divide by i:

x =
(2+i)/i

We'll multiply by i the result in order to obtain a
real number for denominator:

x =
(2+i)*i/i^2

x = -(2i +
i^2)

x = 1 -
2i

We'll substitute x in
(1):

i(1 - 2i) - 2y = -i

We'll
remove the brackets:

i - 2i^2 - 2y =
-i

i + 2 - 2y = -i

We'll
subtract i+2 both sides:

-2y =
-i-i-2

-2y = -2i - 2

We'll
divide by -2:

y = i +
1

The solution of the system
is: {(1 - 2i ; i +
1)}.