Solve $\begin{cases}x\equiv-4\pmod {17}\\ x\equiv 3\pmod{23} \end{cases}$ 
Solve $$\begin{cases}x\equiv-4\pmod {17}\\
x\equiv 3\pmod{23}
\end{cases}$$

My attempt:
$$\gcd (17,23)=1$$
so using the Chinese remainder theorem
 there is a solution modulo $17\times 23=391$
$$x=-4+17t\\
\Longrightarrow-4+17t\equiv 3\pmod{23}\\
\Longrightarrow -4+17t\equiv 3+23k\\
k,t\in \mathbb Z$$
I am stuck here

Once you obtain the equivalence $-4 + 17t \equiv 3 \pmod{23}$, you can add $4$ to each side of the equivalence to obtain 
$$17t \equiv 7 \pmod{23}$$
Since $\gcd(17, 23) = 1$, this equivalence has a solution. To solve it, we can apply the extended Euclidean algorithm to find the multiplicative inverse of $17$ modulo $23$, then multiply both sides of the equivalence $17t \equiv 7 \pmod{23}$ by that inverse.
\begin{align*}
23 & = 1 \cdot 17 + 6\\
17 & = 2 \cdot 6 + 5\\
6 & = 1 \cdot 5 + 1\\
5 & = 5 \cdot 1
\end{align*}
Now, we work backwards to solve for $1$ in terms of $17$ and $23$.
\begin{align*}
1 & = 6 - 5\\
  & = 6 - (17 - 2 \cdot 6)\\
  & = 3 \cdot 6 - 17\\
  & = 3(23 - 17) - 17\\
  & = 3 \cdot 23 - 4 \cdot 17
\end{align*}
Thus, $-4 \cdot 17 = 1 - 3 \cdot 23 \implies -4 \cdot 17 \equiv 1 \pmod{23}$.  Multiplying both sides of the equivalence 
$$17t \equiv 7 \pmod{23}$$
by $-4$ yields
\begin{align*}
-4 \cdot 17t & \equiv -4 \cdot 7 \pmod{23}\\
t & \equiv -28 \pmod{23}\\
t & \equiv -28 + 2 \cdot 23 \pmod{23}\\
t & \equiv 18 \pmod{23}
\end{align*}
Hence, $t = 18 + 23n$ for some integer $n$.  Substituting this value for $t$ into the expression $x = -4 + 17t$ yields
\begin{align*} 
x & = -4 + 17(18 + 23n)\\ 
  & = -4 + 306 + 17 \cdot 23n\\ 
  & = 302 + 391n
\end{align*} 
Thus, $x \equiv 302 \pmod{391}$, the result Andre Nicolas obtained in the comments by making an astute observation.
Check:  Observe that 
\begin{align*}
302 & = 17 \cdot 18 - 4 \implies 302 \equiv -4 \pmod{17}\\
302 & = 13 \cdot 23 + 3 \implies 302 \equiv 3 \pmod{23}
\end{align*} 
