As far as I know, Cauchy gave the [first] general solution:
Theorem (Cauchy). If $r,s,t$ are relatively prime integers, the complete solution to $$rX+sY+tZ=0$$ in integers $x,y,z$ is given by
$$ (x,y,z) = (s\delta-t\beta,\ t\alpha-r\delta,\ r\beta-s\alpha),$$
where $\alpha, \beta,\delta$ are arbitrary integers.
Your problem is then, as you point out, handled by Cauchy's after dividing out the gcd of $17$.
Suppose $\,(x',y')\,$ is the minimal solution to the $\,ax+by=1\,$ generated by the extended Euclidean algorithm. Then as you pointed out, $\,(nx',ny')\,$ will be a solution to $\,ax+by=n\,$ but not necessarily minimal. However, note that $\,(nx'-kb,ny'+ka)\,$ will also be a solution for any $\,k \in \Bbb Z.\,$
To find the minimal solution to $\,ax+by=n\,$, we simply need to find $\,k\,$ such that $\,nx'-kb\,$ and $\,ny'+ka\,$ are as close as possible [see note below]. To do that, we solve the equation $\,nx'-kb=ny'+ka\,$ for $\,k\,$; if the equation yields an integral $\,k\,$, then we will have our minimal solution, otherwise either $\,\lfloor k \rfloor\,$ or $\,\lceil k \rceil\,$ (whichever is closer to $\,k\,$) will generate the minimal solution.
In your particular example, the extended Euclidean algorithm should generate the solution $\,(1,-2)\,$ as the minimal solution to $\,7x+3y=1,\,$ so our starting solution to $\,7x+3y=10,\,$ would be $\,(10,-20).\,$ Then solving the equation $\,10-3k=-20+7k\,$ gives $\,k=3\,$ which yields the minimal solution $\,(1,1).\,$
[Note: this assumes that $\,ab \gt 0;\,$ if not then one can just solve $\,ax-by=n\,$ and then use the solution $\,(nx'-kb,-[ny'+ka]).\,$]
Best Answer
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.