The following system of linear congruences in its given form cannot be solved using the Chinese Remainder Theorem. Can you help me transform the system sufficiently such that the Chinese Remainder Theorem can be applied, without actually solving the system?
$x \equiv 1 \pmod {15}$ $2x \equiv 11 \pmod {21}$ $4x \equiv −6 \pmod {35}$
(I can reduce $x \equiv 1 \pmod {15}$ to $x \equiv 1 \pmod 5$ and $x \equiv 1 \pmod 3$.
But I'm not sure how to reduce the other two congruences to end up with mod $3,5$ and $7$?)
Thanks in advance