Solving system of linear congruences (3 pairs) So we have the following:
$$2x \equiv 3\pmod {5} \\ 3x \equiv 4\pmod {7} \\ 5x \equiv 7\pmod {11}$$
which reduces to:
$$x \equiv 4\pmod {5} \\ x \equiv 6\pmod {7} \\ x \equiv 8\pmod {11}$$
Now the confusion begins here. At this point, I choose the first two pairs of congruences and equate them, giving:
$$ 5k+4= 7l +6 \\ \\$$
But I'm not sure what to do past this point. I know in essence I need to solve this and pair this new equation with the last one and re-do the steps. It's just past this point I don't know how to solve.
 A: Your first step in simplification of the congruences is correct. Now use the Chinese remainder theorem to solve the system. 
A: For solving the system of linear congruence, we need to use the Chinese Remainder Theorem
$$
\left\{\begin{array}{ll}
2 x \equiv 3 & (\bmod 5) \\
3 x \equiv 4 & (\bmod 7) \\
5 x \equiv 7 & (\bmod 11)
\end{array}\right.
\Leftrightarrow \left\{\begin{array}{ll}
x \equiv 4 & (\bmod 5) \\
x \equiv 6 & (\bmod 7) \\
x \equiv 8 & (\bmod 11)
\end{array}\right. $$
First of all, I am going to solve the following system of linear congruence for $A, B$ and $C$
$$
\left\{\begin{array}{ll}
7 \times 11 A \equiv 4 & (\bmod 5) \\
5 \times 11B \equiv 6 & (\bmod 7) \\
5 \times 7 C \equiv 8 & (\bmod 11)
\end{array}\right. \Leftrightarrow \left\{\begin{array}{ll}
77 A \equiv 4 & (\bmod 5) \\
55 B \equiv 6 & (\bmod 7) \\
35 C \equiv 8 & (\bmod 11)
\end{array}\right. $$
$$\Leftrightarrow\left\{\begin{array}{cc}
2 A \equiv 4 & (\bmod 5) \\
-B \equiv 6 & (\bmod 7) \\
2 C=8 & (\bmod 11)
\end{array}\right. \Leftrightarrow \left\{\begin{array}{ll}
A \equiv 2 & (\bmod 5) \\
B \equiv 1 & (\bmod 7) \\
C \equiv 4 & (\bmod 11)
\end{array}\right. $$
Using the Chinese Remainder Theorem, the integer solutions for the system are
$$385k+77\times 2+55\times 1+35\times 444$$
$\text{i.e. }\boxed{385k+349,}\tag*{} $
where $k\in \mathbb Z$.
:|D Wish you enjoy my solution! Your suggestions, comments and alternate solutions are warmly welcome!
