$$10x \equiv 15 \pmod{35}$$
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.
$$-11 \equiv -11+16 \equiv 5 \equiv 1 \pmod 4$$
So,the congruence becomes:
$$x^2+y^2 \equiv 3 \pmod 4$$
$$x \equiv 0 \pmod 4 \Rightarrow x^2 \equiv 0 \pmod 4$$
$$x \equiv 1 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$x \equiv 2 \pmod 4 \Rightarrow x^2 \equiv 0\pmod 4$$
$$x \equiv 3 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$$$
$$y \equiv 0 \pmod 4 \Rightarrow y^2 \equiv 0 \pmod 4$$
$$y \equiv 1 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
$$y \equiv 2 \pmod 4 \Rightarrow y^2 \equiv 0\pmod 4$$
$$y \equiv 3 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
We can see that it cannot be $x^2+y^2 \equiv 3 \pmod 4$
EDIT:
$$-11 \equiv 1 \pmod 3$$
So,the congruence becomes:
$$x^2+y^2 \equiv 0 \pmod 3$$
$$x \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 0 \pmod 3$$
$$x \equiv 1 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$
$$x \equiv 2 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$
$$$$
$$y \equiv 0 \pmod 3 \Rightarrow y^2 \equiv 0 \pmod 3$$
$$y \equiv 1 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$
$$y \equiv 2 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$
We can see that $x^2+y^2 \equiv 0 \pmod 3$,only if $x \equiv 0 \pmod 3 \text{ AND } y \equiv 0 \pmod 3$
Best Answer
The system
should be prepared more before solving.
The reason for this is that two of the modulus share a common factor: 8 and 6 are both multiples of 2.
This could lead to a clash where the two different equations demand contradictory properties mod 2, in this case it's actually fine though:the mod 2 part of both these equations agree so we should cast the 2 part out of one of the equations (this loses no information)