Why add before dividing in this equation? For the following equation, I know the correct answer is $9$:
$$
x / 3 + 2 = 5
$$
You subtract $2$ from each side, and the multiply each side by $3$...
But why do you subtract the $2$ first?
Doesn't the order of operations say I should do division before addition?
 A: No, you're getting confused. The "order of operations" doesn't matter here, it's only for solving an expression (i.e. $4(5)-\left(\frac52\right)^3$). Here you just do something to two sides that are equal. You can choose whatever you want to do to both sides, but you have to make sure that you are doing it to both sides. You don't have to subtract 2 first, you can also multiply by 3 first:
$$3\left(\frac{x}3+2\right)=3(5) \\ x+6=15 \\ x=9$$
And you will get the same answer as subtracting first:
$$\left(\frac{x}3+2\right)-2=(5)-2\\\frac{x}3=3\\x=9$$
A: Any order of operations is fine, as long as you are consistent in applying the same operation to both sides of the equation in every step. If you want to multiply by $3$ first:
$$
\frac{x}{3} + 2 = 5
$$
Multiply both sides by $3$:
$$
x + 6 = 15
$$
Subtract $6$ from both sides:
$$
x = 15 - 6
$$
Just work out the solution:
$$
x = 9
$$
Look at equations as a scale balance: anything you do will keep the scale in balance as long as you do it to both sides. The trick is to come up with a sequence of operations that eventually leaves the unknown on one side, and only knowns (in this case, numbers) on the other. 
As a bad example, when we have $x+6=15$ we could, if we wanted to, decide to add $100$ to both sides. The equation would still be perfectly valid, but we wouldn't be any closer to finding $x$!
Finding a good sequence of operations will be challenging at first, but after some practice, it will become second nature.

If you were taught to solve equations in terms of "a $+$ when moved across the $=$ becomes a $-$", etc., I suggest that you forget all that and think instead in terms of a scale balance. This was certainly a breakthrough moment for me.
PS: The "rules" you refer to are just a convention for working out 
expressions with non-existent or ambiguous parentheses, and have nothing to do with solving equations.
A: What gets done last gets undone first.  Since you do division first, you undo division last.
A: Order of operations more so tells you how you should read the problem unambiguously. Here we have "$x$ is divided by $3$, THEN added with $2$ to achieve $5$." To figure out what $x$ was originally, we sort of backtrack.
What did the equation look like before we added $2$?
$$x/3=5-2$$
Then what was it before we divided by $3$?
$$x=(5-2)\times3$$
A: As regards correctness of the solution, the order is irrelevant as long as the operations are legal.
$$\frac x3+2=5.$$
First times $3$:
$$x+3\cdot2=3\cdot5,\\x+6=15,$$
followed by minus $6$:
$$x=15-6,\\x=9.$$
Or first minus $2$:
$$\frac x3=5-2,\\\frac x3=3,$$
followed by times $3$:
$$x=3\cdot3,\\x=9.$$
It's a matter of taste, except that in the first case you perform two multiplies and one subtract, vs. one subtract and one multiply in the second.
