Can I really factor a constant into the $\min$ function? Say I have $\min(5x_1,x_2)$ and I multiply the whole function by $10$, i.e. $10\min(5x_1,x_2)$. Does that simplify to $\min(50x_1,10x_1)$? In one of my classes I think my professor did this but I'm not sure (he makes very hard to read and seemingly bad notes), and I'm just trying to put these notes together. Thanks!

The algebra is all fine, but if the professor blazed straight through it, presumably he expected the students to understand what is going on rather than just memorise it.  So, how about the following slightly informal proof/explanation?
I have two numbers, $5x_1$ held in my left hand and $x_2$ in my right.  One or the other hand is holding the smaller number.  It is clear (isn't it?) that if we multiply them both by $10$, then the same hand will be holding the smaller of the two new numbers, and it will be $10$ times the previous smaller number.  Put this into mathematical notation and we have
$$\min(50x_1,10x_2)=10\min(5x_1,x_2)\,,$$
exactly what we wanted.
Pedagogical note 1: now (and not before) is the time to generalise, consider what happens with a negative multiplier, and so forth.
Pedagogical note 2: this is all much easier, and more convincing, when waving my hands with a student sitting beside me: one of the downsides to teaching maths by phone/email/online/etc.
