How to deal with an additive constant in a linear congruence equation? I am trying to solve the following equation: $10x+3 \equiv 2 \pmod{17}$.
The problem I am having is that I don't know what to do with the number $3$. 
This is what I have done so far: 
$10x+3 = 2+17y$
$10x-17y = 2$ (I don't know what I should do with the $3$ here.) 
I use this to find $\gcd(10,17) = 1$.
Then I calculate $x$ and $y$ which is :
$10*(-5) + 17*3 = 1$
$10*(-10) + 17*6 = 2$
$x= -10$ and $y = 6$
So what I need help with is where I should add the $3$ or if I should add it at all.

Regarding your specific concern on $3$ it is really no different than solving 
$10 x +3 = 2$ (in the rationals). 
What would you do? You subtract $3$ to get $10 x  = -1$. 
You can also do this modulo $17$. (You can then use that $-1$ is $16$ modulo $17$ but this might not even be needed.)
So you have to solve $10x \equiv -1 \pmod{17}$ or equivalently $10x \equiv 16 \pmod{17}$. It seems you know how to do this.
