<s>[INST] $4x≡2\mod5$ can you divide both sides by $2$ to get $2x≡1\mod5\,?$ Since gcd$(2,5)=1$ , could you treat $4x$ as $2(2x)$ and cancel the $2$ on both sides?
i.e. $$2(2x)≡2\mod5\implies 2x≡1\mod5$$
Thanks!
 [/INST] Not really.  
$4x \equiv  2\pmod 6$ (so $x$ could be $2\pmod 6$ or $x$ could be $5\pmod 6$) does not mean $2x \equiv 1\pmod 6$ (which is impossible).
But you can say
$ka \equiv kb \pmod n \implies a \equiv b \pmod {\frac n{\gcd(k, n)}}$
And so 
$4x \equiv 2 \pmod 6$ meams $\frac {4x}2\equiv \frac 2{2}\pmod {\frac {6}{\gcd(2,6)}}$ or $2x =1 \pmod 3$ and that's fine. $x \equiv 2\pmod 3$ so $x\equiv 2\pmod 6$ or $x \equiv 3 \pmod 6$.
ANd to get to your problem:
$4x \equiv 2 \pmod 5$ means
$\frac {4x}2 \equiv \frac 22 \pmod {\frac 5{\gcd(2,5)}}$ so
$2x \equiv 1 \pmod {\frac 51}$ and
$2x \equiv 1  \pmod  5$.
So you CAN but only because $2$ and $5$ are relatively prime.  You couldn't if they weren't.  (but you could if you devide the modulus but the gcd.)
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