Can someone help me with this algebra 2 problem

When "i" is considered an imaginary number:
(1+i)x-(1-2i)y=1+4i
I understand somehow you need to make 1+i=1 and -(1-2i)=4i. However, the solution book says that (1+i)x turns into (x-y) and -(1-2i) turns into (x-2y)i but I don't understand how they did that...? Thanks!

Basically the i terms combine to find b and the "non-i" terms combine to form a, in a+ bi

(1+i)x-(1-2i)y=1+4i

X+xi -y +2yi= 1+4i

(x-y) +i(x+2y)=1+4i

Then x-y=1 for the "a" part
And x+2y =4, for the "b" part...doublecheck your signs.

Now solve the system ( subtract the equations)
-3y=-3
y=1
x=2

Hoping this helps!

(1+i)x - (1-2i)y = 1+4i
x + ix - y + 2yi = 1 + 4i
Equate real parts: x - y = 1
Equate imaginary parts: x + 2y = 4
Subtract to get 3y = 3.
So y = 1 and x = 2.