How to solve a congruence system? Here is a tricky congruence system to solve, I have tried to use the Chinese Remainder Theorem without success so far.
$$2x \equiv3\;(mod\;7)\\
x\equiv8\;(mod\;15)$$
Thank you very much
Li
 A: You need to "get rid of" the $2$ from the right hand side of the first congruence in your system:
\begin{aligned}
2x & \equiv 3 \mod 7 & (1) \\
x & \equiv 8 \mod 15 & (2)
\end{aligned}
So here, you have to multiply both sides of equation $(1)$ by the inverse of $2$ modulo $7$, which is $4$. This means the first equation becomes $x \equiv 12 \equiv 5 \mod 7$ So this leaves us with the system:
$$\color{blue}{\begin{aligned}
x & \equiv 5 \mod 7 \\
x & \equiv 8 \mod 15
\end{aligned}}$$
Since $\gcd(7, 15) = 1$, there is a unique solution for $x$ modulo $(7)(15) = 105$. So we can use the Chinese remainder theorem on the system that is in $\color{blue}{\mathrm{blue}
 }$.
Now try using the Chinese remainder theorem. Recall for a system of two congruences:
$$x \equiv a_1 \mod n_1$$
$$x \equiv a_2 \mod n_2,$$
if $\gcd(n_1, n_2) = 1$, then the solution is given by:
$$x \equiv a_1 n_2 [n_2^{-1}]_{n_1} + a_2 n_1[n_1^{-1}]_{n_2},$$
where $[p^{-1}]_{q}$ means "the inverse of $p$ modulo $q$".
You will find this is the solution:
$$x \equiv 5\cdot15\cdot1 + 8\cdot7\cdot13 \equiv \color{green}{803} \mod 105$$
and
$$\color{green}{803} \equiv \color{red}{68} \mod 105,$$ so
$$\boxed{\color{red}{x=68}}.$$
$\color{red}{\mathrm{That}}$ is the least positive residue solution, modulo $105$. Since anything congruent to $68$ modulo $105$ will also be a solution, you could also say $\boxed{x=105n+68}$ for each integer $n$.
A: $$ 2x = 3(mod 7)    $$
$$ x+x= 3(mod 7)    $$
$$ x =  5(mod 7)    $$
$$ x =  8(mod 15)   $$
$$ x = 15k_1 + 8    $$
$$ x = 7k_2  + 5    $$
$$ x = 105k_3 + c   $$
$$ 0 \le c < 7*15   $$
$$ c(mod 7)=5       $$
$$ c = 15k_4+8      $$
$$ 0 \le k_4 < 7    $$
$$(15k_4+8)(mod 7)=3$$
$$ k_4 + 1 = 5      $$
$$ k_4 = 4          $$
$$ c=15*4+4 => c=68 $$
$$ x = 105k_3 + 68  $$
$$ 0 \le k_3 <\infty$$
$$ k_3 \in Z  $$
A: \begin{split}
X&=15L+8\\
2X&=30L+16=30(7m+u)+16 
\end{split}
[ where $u$ can be $0,1,2,3,4,5,6$]
$$
2X=(210m+28u+14)+(2u+2)
$$
So $2u+2$ must leave remainder $3$ when divided by $7$,thus
\begin{split}
u&=4\\
2X&=210m+136.
\end{split}
Therefore $X=105m+68$.
