(1-2i)*x + (1+2i)*y = 1 + i

Let us expacd the brackets:

==> x - 2xi + y + 2yi = 1 + i

Now combine like terms:

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

==> x+y = 1 ......(1)

==> 2y-2x = 1 .......(2)

Let us use the substitution method :

x+y = 1 ==> x = 1-y

==> 2y- 2x = 1

==> 2y - 2(1-y) = 1

==> 2y - 2 + 2y = 1

==> 4y = 3

**==> y= 3/4**

**==> x= 1/4**

We'll note the complex number from the left side as z1 and the complex number from the right side as z2.

For z1 = z2, we'll have to impose the following conditions:

Re(z1) = Re(z2)

Im(z1) = Im(z2)

To determine the real and imaginar parts of the complex number from the left side, we'll have to remove the brackets:

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

We'll combine the real parts and imaginary parts:

Re(z1) = x+y

Im(z1) = -2x + 2y

Re(z2) = 1

Im(z2) = 1

x+y = 1 (1)

-2x + 2y = 1 (2)

We'll multiply by 2 (1):

2x + 2y = 2 (3)

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

2x + 2y - 2x + 2y = 2+1

We'll eliminate like terms:

4y = 3

**y = 3/4**

We'll substitute y in (1):

x+y = 1

x + 3/4 = 1

x = 1 - 3/4

x = (4-3)/4

**x = 1/4**