Question: Solve the system of linear congruences below by finding all $x$ that satisfy it. Hint — try rewriting each congruence in the form $x \equiv a \pmod b$.
$2x \equiv 1 \pmod3$
$3x \equiv 2 \pmod5$
$5x \equiv 4 \pmod7$
So I tried following that hint, and I have:
$x \equiv 2 \pmod3$
$x \equiv 4 \pmod5$
$x \equiv 5 \pmod7$
Firstly, is this correct? Second, where do I go from here? I calculated that $M = 3 \times 5 \times 7 = 105$ and the individual $M_i$'s, but now I'm stuck in a circle, because reducing from there gives me back the original equations. Any tips?