chinese remainder theorem

[afif mansib ch]

obtain the two incongruent solutions modulo 210 of the system
\[2x\equiv 3(mod5)
\]
\[4x\equiv 2(mod6)
\]
\[3x\equiv 2(mod7)
\]

[Moon]

First of all, we can divide the second congruence by $2$ to get $2x \equiv 1 \pmod{3}$. Then using chinese remainder theorem you end up with a family of solution mod 105. From there you get two incongruent family of solutions.

[afif mansib ch]

thnx moon via.feeling so good. 2 c u'r sols again.