How to find unique solution for two variables from a single equation with those same two variables? I’m working on a problem - listed below - that supposedly is not solveable. I’m a bit confused because when we divide $40X$ by $3X$ we get the solution for $P$. Why can’t we just plug back P’s value of $40/3$ to get the solution for $X$.
Is it possible to find X when $40X = 3XP$
 A: When you divide both sides by $3x$, you're assuming that $x$ is nonzero - you can't do that if $x = 0$! In fact, if we take $x = 0$, the equation becomes
$$40 \cdot 0 = 3 \cdot 0 \cdot p$$
which is true for every value of $p$. So we already have infinitely many solutions: $x = 0$, $p = $ anything.
Now, if $x \neq 0$, you're right that we can divide both sides by $3x$ and get $p = 40/3$. If we plug that value of $p$ back in, here's how it goes:
$$40x = 3x \cdot \frac{40}{3}$$
$$40x = 40x$$
$$x = x$$
This is true no matter what $x$ is - so again we have infinitely many solutions, this time with $p = 40/3$ and $x = $ anything.
This is what typically happens when you have more unknowns than you have equations - you can sort of "partially" solve it, but you're left with infinitely many solutions anyway. I'm not sure I would call this "unsolvable", though - certainly you can't solve for a unique solution, but we say equations like $x^2 + 3x + 2 = 0$ are "solvable" even though they have two solutions.
