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

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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)}.**

To solve:

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

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

Solution:

First we will eliminate y .

Eq(1)*i - eq(2):

(ix-2y)i - ((1+i)x -2iy) = -i*i -(3+i)

i^2*x -(1+i)x = 1-3-i

-2x-ix =-2-i

-(2+i)x = -(2+i)

x= 1.

Putx=1 in (2):

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

-2iy = 3+i - 1-i

-2iy = 2

y = 2/(-2i) = -1/i = i^2/i = i.

Therefore x = 1 and y = i

Tally:EQ(1): LHS : ix-2y = i*1-2i = i-2i = -i = RHS

EQ(2): LHS = (1+i)x-2iy = (1+i)*1 -2i*i = 1+i-2(-1) = 1+2+ i -3 +i = RHS.

We need to solve for x and y given:

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

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

Now (2) can be simplified as

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

=> x +xi - 2iy = 3+ i

from (1)

ix - 2y = -i

=> -2y = -i - ix

=> y = i/2 + ix/2

substitute y = i/2 + ix/2 in (2)

=> x +xi - 2iy = 3+ i

=> x + xi - 2i ( i/2 + ix/2) = 3 +i

=> x+ xi - i^2 - i^2x = 3+i

=> x + xi +1 + x = 3 + i

=> 2x + -2 + xi - i = 0

=> x (2+i) = 2+i

=> x =1

y = i/2 + ix/2

=> y = i/2 + i/2 = i

**Therefore x = 1 and y = i**