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

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### 2 Answers

We have to solve:

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

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

i*(1) - (2)

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

=> -x - x - ix = -2 - i

=> ix = 2 + i

=> -x = 2i - 1

=> x = 1 - 2i

substitute in (1)

ix - 2y = -i

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

=> i + 2 - 2y = -i

=> y = 1 + i

**The solution of the system is x = 1 - 2i and y = 1 + 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)}.