Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$ Find the smallest positive integer $x$ that solves the following simultaneously.
Note: I haven't been taught the Chinese Remainder Theorem, and have had trouble trying to apply it.
$$
\begin{cases}
2x \equiv 11 \pmod{15}\\
3x \equiv 6 \pmod{8}
\end{cases}
$$
I tried solving each congruence individually. 
The inverse for the first is 8: $x \equiv 8\cdot11 \equiv 88 \equiv 13 \pmod{15}$.
The inverse for the second is 3: $x \equiv 3\cdot6 \equiv 18 \equiv 2 \pmod{8}$.
But I can't figure out how where to go from here.

You must start from Bézout's identity between  the moduli $15$ and $8$: $\,2\cdot 8-1\cdot 15=1$. From the solutions to the individual congruences $\color{red} {13} \pmod{15}$ and $\color{red}2 \pmod{8}$, you deduce the solutions to the system:
$$2\cdot 8\cdot\color{red} {13}-1\cdot 15\cdot\color{red}2=178\equiv 58\pmod{120}.$$ 
