Using CRT or not solve the following:
$$\begin{cases} x \equiv 19 \mod 49 \\ x \equiv 10 \mod 14 \end{cases} $$
And now, I don't know how to deal with it.
Please help me.
Answer
The first equation implies $x \equiv 5 \pmod 7$, while the second one implies $x \equiv 3 \pmod 7$. Therefore there is no solution to this set of congruences.
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