<s>[INST] 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).
 [/INST] \begin{align}
x&\equiv 1\phantom {8} \pmod {{\color{blue}{12}}}\tag {1}\\
x&\equiv 4\phantom {8} \pmod {{\color{teal}{21}}}\tag {2}\\
x&\equiv 18 \pmod {{\color{blue}{35}}}\tag{3}
\end{align}
Since $\gcd(12,35)=1 $ you can first apply the Chinese Remainder Theorem to find the solution to the two simultaneous congruences $(1)$ and $(3) $, which  is of the form of 
$$x\equiv x_0\pmod{12\cdot 35} \equiv x_0\pmod { 420}.\tag {4}$$
To compute the smallest positive value $x_{\min} $ of $ x_0 $, using the Chinese Remainder Theorem proceed as follows: 


*

*since $12$ and $35$ are relatively prime, there is an $r$ and an $s$ such that $r\cdot 12 + s\cdot 35 = 1$. Indeed, $3\cdot 12 + (-1)\cdot 35 = 1$, so $r = 3$ and $s = -1$; 

*in general, if $\gcd(m,n)=1 $, then the set of all solutions of the system
\begin{align}
x& \equiv a \pmod {m}\\
x& \equiv b \pmod {\; n} 
\end{align}
is given by the condition $x\equiv a\cdot s\cdot n + b\cdot r\cdot m \pmod {m\cdot n}$. Hence 
$$x \equiv 1\cdot (-1)\cdot 35 + 18\cdot 3\cdot 12   \equiv 613\equiv 613-420\equiv 193\pmod {420};\tag {5}$$


Finally solve the following two simultaneous congruences
\begin{align}
x\equiv &193\pmod{420} \tag {5}\\
x\equiv &4\phantom{93}\pmod{\phantom { 4}{\color {teal}{21}}};\tag {2}
\end{align}
Since $420\equiv 0\pmod{\color{teal}{21}}$ and $193\equiv 4\pmod{\color {teal}{21}}$, the solution of the given system is 
$$x\equiv 193\pmod{420}. \tag {6}$$
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