Solve for $x, \;23+35 x ≡ 4$ (mod $ 13)$ I can simplify the equation, but I get stuck at the last step:   $\;x≡ 9^{−1} (4-10)$ (mod $13$). 

My issue is that I am unsure how to compute this by hand.
You are looking for $9^{-1} \pmod {13}$.  You can find that using the Euclidean algorithm:

$$13 - \color{red}{9} = \color{violet}4$$
and
$$\color{red}9 - 2\times \color{violet}4=1.$$
Therefore,

$$\color{red}9 - 2 \times {(13 - 9)} = 1$$

i.e.,
$$3\times 9 -  2 \times 13  = 1.$$

Therefore $3 \times 9 \equiv 1 \pmod {13}.$  

Addendum to answer comment:

$$\color{green}{9^{-1}}\times\color{orange}{(4-10)}\equiv\color{green}3\times\color{orange}{-6}\equiv\color{green}3\times\color{orange}7=21\equiv8 \pmod {13}.$$
We have 23 +35x =4(mod13)
It gives 35x =-19 =7(mod13)
It further gives 9x =7(mod13)
Implies 90x=70 (mod13) gives -x=5(mod13) implies x=8(mod13)