How would I solve the following congruence? $5x \equiv 118 (\mod 127)$
This is what I have done so far:
$127 = 25*5+2$
$2 = 127-25*5$
$25 = 12*2+1$
$1 = 25-12*2$
$1 = 25-12(127-25*5)$
I am a little stuck on how to continue.
I know I am supposed to write it in a form such that $5v+127w = 1$ but I am not exactly sure how I would do that.
Any help?

\begin{align*}
1&=25-12(127-25\cdot 5)\\
&=25(1+12\cdot 5)-12(127)\\
&=25(61)-12(127)\\
&=5(305)-12(127)
\end{align*}
