I assume your students have seen Linear Algebra.
Remember in Linear Algebra how you sometimes have to solve $Ax = b$ for a matrix $A$ with more columns than rows? You usually get free variables, right? So after reducing to echelon form you have something like:
$$\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$$
You know from Linear Algebra that there are infinitely many solutions. How do you find them? Well, you can freely choose $z$ and solve for the remaining variables.
For instance if $z=0$ then $x = 3$ and $y = 3$.
Or, if $z=1$ then $x = 1$ and $y = 4$.
Indeed, $z$ can be any value and the system can still be solved (we call $z$ a free variable). In other words, once $z$ is nailed down, the other two variables are determined uniquely. And another way of saying that is to say there is a unique function $\vec{g}(z)$ such that the equation
$$\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} \vec{g}(z) \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$$
is satisfied for any value $z$ (forgive the abuse of notation in $\vec{g}(z)$).
Well the existence of this function $\vec{g}(z)$ is the conclusion of the Implicit Function Theorem. Whenever you have a situation with more outputs than inputs, like you have a function $f : \mathbb{R}^n \to \mathbb{R}^m$ with $m>n$ and you're solving $f(\vec{x}) = \vec{b}$, there will be a unique function $g$ of the free variables such that the equation $$f(\; g(\text{free variables}), \text{free variables} \;) = \vec{b}$$ is satisfied for any value of the free variables within reason (and "within reason" means within a small neighborhood of the base point where linear reasoning like this is going to be valid).
Thinking back to our matrix example, we can see that the number of free variables is $\#\text{outputs} - \#\text{inputs}$, or $m-n$.
Well, there is one catch which is there is some condition that the total derivative has to have rank $m$ or something, but that makes sense because you only get free variables when your matrix has the right rank.