<s>[INST] 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 $2(2) + 4 \equiv 8 \pmod{5} \equiv 3 \pmod{5}$.

 [/INST] It's the last step, where you're solving for $x$.
$4x + 12 \equiv 0 \Rightarrow 4x\equiv 3 \Rightarrow x\equiv 2$.
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