System of Linear Congruences Find all $x$ such that 
\begin{align}
x&\equiv 1 \pmod {12}\\
x&\equiv 4 \pmod {21}\\
x&\equiv 18 \pmod {35}
\end{align}
Im not quite sure if this system of linear congruence is solvable. Since 
$\gcd(12,21) =3$, $\gcd (12,35)=1$ and $\gcd(21,35) = 7$, and the CRT states that "If(m1, m2) = 1, then the system has its complete solution a single resident class (mod m1.....mr).

When the moduli are not coprime, you can proceed like this:  $x\equiv 1 \pmod{12}$ so $x=1+12y$ for some $y$.  
Then $x\equiv 1+12y \equiv 4 \pmod{21}$, which has solution $y\equiv 2 \pmod{7}$, so $y=2+7z$ for some $z$ and we have $x = 1+12(2+7z)=25+84z.$
Then $x \equiv 25+84z \equiv 18 \pmod{35}$ which has solution $z=2 \pmod{5}$.
So $x = 25+84(2+5w) = 193 + 420 w$, which is to say $x\equiv 193 \pmod{420}.$
