Solving Complex Equation: Real & Imaginary Parts of z=x+iy

[DryRun]

Homework Statement
Given that the real and imaginary parts of the complex number $z=x+iy$ satisfy the equation $(2-i)x-(1+3i)y=7$. Find x and y.

The attempt at a solution
I know it's quite simple. Just equate the real and imaginary parts, but i checked and redid it again, but the answer still evades me!
$$(2x-y-7) + i(-x-3y)=0 \\2x-y-7=x \\x-y=7\, (1) \\-x-3y=y \\4y+x=0\, (2) \\x=28/5 \\y=-7/5$$
I replaced in the original equation but i can't get 7 on the L.H.S.
The correct answers: x=3 and y=-1.

 

[LCKurtz]

Homework Statement
Given that the real and imaginary parts of the complex number $z=x+iy$ satisfy the equation $(2-i)x-(1+3i)y=7$. Find x and y.

The attempt at a solution
I know it's quite simple. Just equate the real and imaginary parts, but i checked and redid it again, but the answer still evades me!
[tex](2x-y-7) + i(-x-3y)=0
\\2x-y-7=x
\\x-y=7\, (1)
\\-x-3y=y
\\4y+x=0\, (2)
\\x=28/5
\\y=-7/5
[/tex]
I replaced in the original equation but i can't get 7 on the L.H.S.
The correct answers: x=3 and y=-1.

Why aren't those 0 on the right side?

 

[SammyS]

Given that the real and imaginary parts of the complex number $z=x+iy$ satisfy the equation $(2-i)x-(1+3i)y=7$. Find x and y.

$$(2x-y-7) + i(-x-3y)=0 \\2x-y-7=x \\ \dots \\-x-3y=y \\ \dots$$

When you write:

$2x-y-7=x$

and

$-x-3y=y\ ,$​

you are saying that

$(2x-y-7) + i(-x-3y)=z\ .$​

That's not what you're trying to solve !

 

[DryRun]

I was confused about z=x+iy. I thought i had to compare the real and imaginary parts of z with those of the equation in order to solve it. I now realize that it has absolutely nothing to do with z. All i had to do was solve the equation independently and ignore whatever was given for z.

Solving:$$2x-y=7 \\-x-3y=0$$I get the correct answers.

Thank you, LCKurtz and SammyS.