I want to do inverse modulo to solve the equation
$$23x \equiv 1 \mod 120$$
And to do that I used the extended Euclidean Algorithm. $$120=23\times5+5$$ $$23=5 \times4+3$$ $$5=3\times1+2$$ $$3=2\times1+1$$ Which makes $1 \equiv3-2\times1=3-2$. I then substituted the other values to get the following
$$5 \equiv120-23 \times5$$ $$1 \equiv (23-((120-23\times5)\times4)-(120-23\times5(23-((120-23\times5)\times4))$$
But now I'm not sure how much I need to simplify this down in order to find the answer.