How to solve this equation if we can't use Chinese remainder theorem. Let consider:
$$\begin{cases}6x \equiv 2 \mod 8\\ 5x \equiv 5\mod 6 \end{cases}$$
We can't use Chinese remainder theorem because $\gcd(8,6) = 2 > 1$
Help me.
 A: $6x \equiv 2 \mod 8$ means that $6x=2+8k$, with $k \in \mathbb{Z}$
It means also that $3x = 1 +4k$, or $3x \equiv 1 \mod 4$ ...
A: The system is equivalent to
$$6x \equiv 2 \pmod 8\iff3x\equiv 1\pmod4\iff x\equiv 3\pmod4$$
$$5x\equiv 5\pmod6\iff x\equiv 1\pmod6$$ 
$x=4n+3$ (from $1^{rst}$ equation)
Putting in second
$$4n+3\equiv 1\pmod6$$ 
$$4n\equiv 4\pmod6$$ 
$$2n\equiv 2\pmod3$$
$$n\equiv1 \pmod3$$
So $n=3m+1$
$x=4(3m+1)+3=12m+7$
So any integer of the form $12m+7$ is a solution where $m\in\mathbb{Z}$
A: Hint: If $$5x\equiv 5\pmod{6}\implies x\equiv1 \pmod{6}\implies x=6k+1$$
A: Hint: write explicately $x = 24k + x'$ with $0\le x' < 24$ and solve for $x'$.
A: $$
\begin{array}{l}
 \left\{ \begin{array}{l}
 6x \equiv 2\left[ 8 \right] \\ 
 5x \equiv 5\left[ 6 \right] \\ 
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 3x \equiv 1\left[ 4 \right] \\ 
 x \equiv 1\left[ 6 \right] \\ 
 \end{array} \right. \\ 
  \Rightarrow \left\{ \begin{array}{l}
 x \equiv  - 1\left[ 4 \right] \\ 
 x \equiv 1\left[ 6 \right] \\ 
 \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
 x \equiv 3\left[ 4 \right] \\ 
 x \equiv 1\left[ 6 \right] \\ 
 \end{array} \right. \\ 
 4k + 3 = 6m + 1 \Leftrightarrow 1 = 3m - 2k \\ 
 \end{array}
$$
A: We have 
$3x\equiv1\mod 4$
$x\equiv1\mod 6$
Write $3x=4k+1$ and $x=6m+1$
hence we get 
$2k-9m=1$
Solve this diophantine equation and you will get answer.
