Solve linear congruence for $x$ : $34x ≡ 51( \text{mod}\; 85)$ 
Solve the linear congruence for $x$ : $$34x ≡ 51( \text{mod}\; 85)$$

I found using the Euclidean Algorithm that the GCD is $17$. Because the GCD evenly divides $51$, there equivalence should be solvable. I made a Diophantine equation to solve: $34p - 85q = 17$
From using the Euclidean Algorithm I have:
$$85 = 2 \cdot34+17$$
$$34=2 \cdot 17+0$$
$$17=85-2 \cdot34$$
I do not know where to go from here in order to make this model the Diophantine equation in order to solve for $p$, and I'm not entirely sure what to do with the solution when I get it, because I am solving for $x$.
 A: In this case, it may be easier to reduce by dividing by $17$.  You know that
$$
34x\equiv51\pmod{85}.
$$
This equivalence is the same as
$$ 
34x-51=85k
$$
for some integer $k$.  Since all of $34$, $51$, and $85$ are divisible by $17$, we may divide through by $17$ to get
$$
2x-3=5k.
$$
In other words, the original equivalence has the same solutions as
$$
2x\equiv3\pmod{5}.
$$
Can you finish it from here?
A: We want to solve for $x$ where $$34x-51=85q$$ for some $q \in \mathbb{Z}$.
$$17(2x-3)=17(5q)$$
$$2x \equiv 3 \pmod{5}$$
Multiply both sides by $3$,
$$x \equiv 9 \equiv -1 \equiv 4 \pmod{5}$$
A: $\implies34x=51+85y$ where $y$ is an arbitrary integer
$\iff2x=3+5y$
$\iff2(x+1)=5(y+1)$
$\implies\dfrac{2(x+1)}5= y+1$ which is an integer
$\implies5|2(x+1)\iff5|(x+1)$ as $(2,5)=1$
A: $$\begin{eqnarray*}
34x & \equiv & 51 \mod 85  \stackrel{34 = 85-51}{\Leftrightarrow}\\
-51x & \equiv & 51 \mod 85 \stackrel{:17}{\Leftrightarrow}\\
-3x & \equiv & 3 \mod 5 \stackrel{:(-3)}{\Leftrightarrow}\\
x &\equiv &-1  \equiv 4 \mod 5
\end{eqnarray*}$$
A: Theorem: If $\gcd\{a,m\}=d$ and suppose $d \vert b$, then the linear congruence $$ax \equiv b \;(\text{mod}\;m)$$ has exactly $d$ solutions modulo $m$. These are given by $$t,t+\frac{m}{d},t+\frac{2m}{d},\cdots,t+\frac{m(d-1)}{d}$$ where $t$ is the solution , unique modulo $\frac{m}{d}$, of the linear congruence $$\frac{a}{d}x \equiv \frac{b}{d}\;(\text{mod}\;\frac{m}{d})$$

In your case, $a=34$, $b=51,m=85$ and $\gcd\{34,85\}=17$. so $t$ is the solution of $$2x \equiv 3\;(\text{mod 5})$$ Now since $\gcd\{2,5\}=1$, we have $$t \equiv3\cdot2^{\phi(5)-1}(\text{mod}\;5)$$That is $$t \equiv 24\;(\text{mod}\;5) \equiv 4$$

So the $17$ solutions are $$4, 4+\frac{85}{17},4+\frac{2 \cdot 85}{17}, \cdots,4+\frac{16\cdot 85}{17}$$
Reference : Tom M. Apostol : Introduction to Analytic number theory 
