$$10x \equiv 15 \pmod{35}$$
This is equivalent to $$35 \mid 10x-15$$
There exists an integer $k \in \mathbb{Z}$ such that
$$10x-15 = 35k$$
$$2x-3 = 7k$$
Going back to the notation of modular arithmetic, this is the same as:
$$2x \equiv 3 \pmod{7}$$
(We could have skipped the intermediate steps of converting to a normal equation, but I wanted to show why it was possible to cancel the fives in the equivalence)
Now we can test values of $x$ so that the (easier) congruence modulo $7$ is satisfied. It suffices to test $x=0,\,1,\,2,\,3,\,4,\,5,\,6$, because those cover all of the possible remainders, modulo seven.
$$2(0) \equiv 0 \not\equiv 3 \pmod{7}$$
$$2(1) \equiv 2 \not\equiv 3 \pmod{7}$$
$$2(2) \equiv 4 \not\equiv 3 \pmod{7}$$
$$2(3) \equiv 6 \not\equiv 3 \pmod{7}$$
$$2(4) \equiv 8 \equiv 1 \not\equiv 3 \pmod{7}$$
$$2(5) \equiv 10 \equiv 3 \pmod{7}$$
$$2(6) \equiv 12 \equiv 5 \not\equiv 3 \pmod{7}$$
Thus, the solutions are exactly the integers $x$ which leave a remainder of five when divided by seven.
Notice that if you multiply by $5$ on both sides, you can get rid of the $200$ very quickly and save yourself the trouble of having to use the extended Euclidean algorithm. Also, your mistake is not having a negative sign on both sides of your congruence.
$200x \equiv 13 \pmod{1001} \Rightarrow 1000x \equiv 65 \pmod{1001} \Rightarrow -x \equiv 65 \pmod{1001}$.
Now, multiply both sides by $-1$:
$x \equiv -65 \pmod{1000} \equiv 936 \pmod{1001}$
Best Answer
In congruence $\pmod m$, two integers $a,b$ are equivalent if $m$ divides $(a-b)$
Here $x\equiv-8\pmod7\equiv6$ as $7$ divides $6-(-8)=14$
$x\equiv-8\pmod7\implies x=7m-8$ where $m$ is any integer
We can safely write $x\equiv7m_1-8\pmod7$ where $m_1$ is any integer
But it is often customary to keep the Right hand side in $[0,m-1]$ or $(-\frac m2,\frac m2]$ for $\pmod m$