Solving to find the general equation with a "mod" equation They probably aren't called "mod" equation but i couldn't think how else to word them, so I have this equation 
$8x + 10y ≡ 8 \pmod 7$
And have been tasked with finding the general solution, I know if it did have a y variable the "mod" part becomes the y variable so in this case would it become
$8x + 10y + 7z = 8$
? but can you use a linear Diophantine equation on this? if so how?
what kind of thing would I expect the general equation to look like?
Any help will be much appreciated 
 A: Some rearranging and reduction:
\begin{align}
8x + 10y &\equiv 8 \pmod 7 \iff \\
x + 3y + 7x + 7y &\equiv 1 + 7 \pmod 7 \iff \\
x + 3y &\equiv 1 \pmod 7
\end{align}
The following will apply the algorithms for linear congruence equations and for linear Diophantine equations (because it was asked) to get solutions.
I use the algorithms here because they are guaranteed to give the number of solutions and all solutions.
Solution as linear congruence equation
We can interpret this as linear congruence $a X \equiv b \pmod m$ in the integer unknown $X$:
\begin{align}
x + 3y &\equiv 1 \pmod 7 \iff \\
7x + 3y &\equiv 1 + 6x \pmod 7 \iff \\
3y & \equiv 1 + 6x \pmod 7 \quad (*)
\end{align}
For this there exists an solution algorithm (link) to find all solutions from $\mathbb{Z}_m$.
The linear congruence $(*)$ has $d = \gcd(a,m) = \gcd(3, 7) = 1$ and 
$d \vert b$ and $d$ solutions (per fixed $x$) in $\mathbb{Z}_m$.
One uses the extended Euclidean algorithm to find $s, t$ with $1 = 3s + 7t$, e.g. $s = -2$, $t = 1$. 
We get the only solution from $\mathbb{Z}_7$: $y = (-2) (1+6x) \bmod 7$ or: 
$$
y = (5 + 2x) \bmod 7
$$
Regarding $\mathbb{Z}$, the general solution is
$$
(x,y) = (x, ((5 + 2x) \bmod 7) + 7k) \quad (k \in \mathbb{Z}) \quad (\#)
$$
Some example solutions

The colour encodes different $k$ parameters.
$$
(x, y) \in \{ \ldots (-2, 1), (-1, 3), (0, 5), (1, 0), (2, 2), (3, 4), (4, 6), (5, 1), (6, 3), (7, 5), (8, 0), \ldots \}
$$ 
Test
\begin{align}
x + 3 y &\equiv 1 \pmod 7 \iff \\
x + 3(5 + 2x) &\equiv 1 \pmod 7 \iff \\
7 x + 15 &\equiv 1 \pmod 7  \iff \\
7 x + 14 &\equiv 0 \pmod 7 \iff \\
0 &\equiv 0 \pmod 7 
\end{align}
Solution as Diophantine equation
We interpret the congruence as linear Diophantine equation
$a X + b Y = c$ in the integer unknowns $X$ and $Y$:
$$
x + 3y \equiv 1 \pmod 7 \iff \\
x + 3y = 1 + 7 k \quad (k \in \mathbb{Z})
$$
For this there exists an algorithm as well (link).
We have $d = \gcd(a, b) = \gcd(1,3) = 1$. So $d \vert c = 1+7k$.
We set $m = 1 + 7k$. Then we need $s, t$ for $s + 3t = 1$, e.g. $s=1$, $t = 0$. This gives the particular solution $x_0 = ms = 1 + 7k$, $y_0 = mt = 0$, thus
$$
(x_0, y_0) = (1+7k, 0)
$$
All solutions (for fixed $k \in \mathbb{Z}$) are
$$
(x_n, y_n) 
= (x_0 + bn, y_0 - an) 
= (1+7k+3n,-n) 
\quad (n \in \mathbb{Z})
$$
E.g. for  $k=0, n=1$ we have the solution $(4,-1)$. But we want to have a similar solution like $(\#)$, in terms of $x$:
Now $7k + 3n = x - 1$ is another Diophantine equation, because $d=\gcd(7,3) = 1$ and $d\vert x - 1$ we have infinite solutions. We get $s = 1$, $t=-2$ and $k_0 = x-1$, $n_0=-2(x-1)=2-2x$, $k_i = x-1+3i$, $n_i = 2-2x - 7i$.   
So we have the solutions
\begin{align}
(x,y) 
&= (1+7k+3n,-n) \\ 
&= (1+7(x-1+3i)+3(2-2x-7i), -(2-2x-7i)) \\ 
&= (1+7x-7+21i+6-6x-21i, -2+2x+7i) \\
&= (x, 2x-2+7i) \quad (i \in \mathbb{Z}) \quad (\#\#)
\end{align}
So this looks more similar to $(\#)$. And $(5+2x)\bmod 7 = (2x + 5) \bmod 7 = (2x-2) \bmod 7$. Thus because of $2x-2 = 7q + ((2x-2) \bmod 7)$ for some $q$ we got the same solution sets, once we add multiples of $7$.

The solution looks the same, the colour encodes different $i$ parameters. So we see the parametrization is different, but set seems the same, looking at the $[0,6]\times[0,6]$ with its $7$ solutions.
