elementary number theory - Solving a system of equations using modular arithmetic modulo 5

Give the solution to the following system of equations using modular arithmetic modulo 5:

$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$

I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.

$-6 \pmod{5} \equiv 4 \pmod 5$

Then I added the two equations:

$4x + 3y \equiv 0 \pmod{5}$
$-4x - 2y \equiv 4 \pmod{5}$

This simplifies to $y \equiv 4 \pmod{5}$.

I then plug this into the first equation: $4x + 3(4) = 0 \pmod{5}$

Wrong work:

Thus, $x = 3$.
But when I plug the values into the first equation, I get $2(3) + 4 \not\equiv 3 \pmod{5}$.
What am I doing wrong?

EDIT:

Revised work:

$x = -3 \pmod{5} = 2 \pmod{5}$.
Now when I plug the values into the first equation, I get $2(2) + 4 \equiv 8 \pmod{5} \equiv 3 \pmod{5}$.

Answer

Sign error on substitution, it should be $x\equiv -3\pmod{5}$.

You had $4x+(3)(4)\equiv 0$, that is, $4(x+3)\equiv 0$. From this we get $x+3\equiv 0$, so $x\equiv -3\pmod{4}$.

Negative numbers are sometimes troublesome, so we may wish to rewrite as $x\equiv 2\pmod{5}$.