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<br /> \\2x-y-7=x<br /> \\x-y=7\, (1)<br /> \\-x-3y=y<br /> \\4y+x=0\, (2)<br /> \\x=28/5<br /> \\y=-7/5<br />
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!
(2x-y-7) + i(-x-3y)=0<br /> \\2x-y-7=x<br /> \\x-y=7\, (1)<br /> \\-x-3y=y<br /> \\4y+x=0\, (2)<br /> \\x=28/5<br /> \\y=-7/5<br />
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<br /> \\2x-y-7=x<br /> \\ \dots<br /> \\-x-3y=y<br /> \\ \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 <br /> \\-x-3y=0I get the correct answers.

Thank you, LCKurtz and SammyS.