Is there a way to find the solution to $x \equiv 3 $ (mod $4)$ and $x \equiv 10$ (mod $35$) without trial and error in the scope of elementary number theory? Is there a trivial solution you can immediately see from this? (How would you guess that case?) One solution is 115, but I got this from guessing.
EDIT: To clarify, let me give an example of a problem I did. I solved
$x \equiv 3 $ (mod $4)$, $x \equiv 5 $ (mod $21)$, and $x \equiv 7 $ (mod $25)$ by solving the first two congruences first: $x \equiv 3 $ (mod $4)$, $x \equiv 5 $ (mod $21)$. I guessed that $x = 47$ would work in both by using the first congruence with trial and error. Would there have been an easier way to do this? From this, would there be a trivial answer to the original system provided?