How to apply Chinese Remainder Theorem for $x$ If:

$$x \equiv 0 \pmod{17}$$ and $$x \equiv -1 \pmod{9}$$ Then how is:

$$x \equiv 17 \pmod{153}$$

I get that since $\gcd(9, 17) = 153 $ the solution will be $\pmod{153}$ but how do you get the $17 \pmod{153}$ ?
$ x \equiv 0\pmod{\!17}\!\iff\! x = 17\color{#c00}n.\,$ $\, {\rm mod}\ 9\!:\ {-}1\equiv x\equiv 17n\equiv -n$ $\!\iff\!$ $n\equiv 1$ $\!\iff\!$ $\color{#c00}{n = 1\!+\!9k}$  

Therefore $\ x = 17\color{#c00}n = 17(\color{#c00}{1\!+\!9k}) = 17+153k$
Hint first solve
$$ 9x\equiv 1\mod 17$$
which gives $x\equiv2\mod17$. Then solve 
$$17x\equiv1\mod 9$$
which gives $x\equiv-1\mod9$. Then you will get
$$ x\equiv 2\times0\times 9+(-1)\times(-1)\times 17=17\mod 153. $$