Why Can We Divide by Variables? I was taught that a graph is simply a visual representation of numbers; I was told that graphing is a nicer way for mathematicians to express very large to infinite solution sets without having to write them all out (or try to at least). If this is true, then every equation has a graphical solution, right?
Okay, so given: $$ y = {-3x^2-5x \over 5x} $$
normally, I would divide out by $x$ and get: $$ y = {-3 \over 5 } x - 1 $$
However, the graphs of these two expressions are different even though algebraically, they seem equal. For example, in the first function, the same $x$--$0$--yields two different solution sets:{$0, undefined$} and {$0, -1$}, respectively.
What confuses me is that if graphs are visual representations of equations, then what allows us to divide through by variables in algebraic expressions to simplify if simplification yields a different graph than the original?
 A: We can divide by variables PROVIDED the variable is not standing for $0$.  So,  when you go to do a division step, you always have to check for the case you are dividing by $0$, and make that a special case.
A: $$y = {-3x^2-5x \over 5x}$$
Note the domain...  $x \in (-\infty,0)\cup(0,\infty)$. If we simplify algebraically, we still carry over the information of this domain. In our original function, $x \ne 0$, so we're allowed to divide through by $x$.
$$y = {-3 \over 5 } x - 1,\quad x\ne0$$
Our domain remains the same. The functions remain the same. It's good form to make a note, such as the one above, of any excluded values. In this case, the excluded point creates a discontinuity in the graph, a hole.

This is different than the function
$$y = {-3 \over 5 } x - 1$$
which is defined over the entire real line.
A: 
I was taught that a graph is simply a visual representation of
  numbers (..) If this is true, then every
  equation has a graphical solution, right?

If the equation has a solution and the mapping between algebra and geometry is proper, then it should show up in both.

However, the graphs of these two expressions are different even though
  algebraically, they seem equal. 

The graphs are different and the algebra is different as well, the first equation does not permit $x = 0$.
For your first equation you have to add a $x \ne 0$ to the graphing to be precise.

The above diagram uses a red circle to indicate that $(0,-1)$ is not part of the graph of the function $y$.
