confused about order of operations Ok, I'm confused on which is the next step in solving this equation:
$$0 = (6.0\text{ kg}) (-3.0\frac{\text{m}}{\text{sec}}) + (78\text{ kg})  v$$
I'm supposed to solve for $v$, however, I'm not sure what the next step is. The solution guide to my problem states that I'm supposed to end up with this:
$$v = \frac{-18\frac{\text{m}}{\text{sec}}}{78} = 0.23\frac{\text{m}}{\text{sec}}$$
but how does the $78$ stay positive or the $18$ stay negative. Shouldn't it either be $\dfrac{-18\frac{\text{m}}{\text{sec}}}{-78}$ or $\dfrac{18\frac{\text{m}}{\text{sec}}}{78}$?
 A: You are correct, the book is in error.
To address the concerns in your comments, I'll offer the following:
There are three main procedures you can use to solve an equation:
$\ \ \ $1) Simplify one or both sides of the equation.
$\ \ \ $2) Multiply or divide both sides of the equation by the same non-zero number.
$\ \ \ $ 3) Add or subtract the same number from both sides of the equation. 
Given an equation, performing one of the above three operations to the equation will produce an equation that has exactly the same solutions as the original; that is, the resulting equation is equivalent to the original.   
Solving simple equations such as yours (linear equations, that is) consists of performing the above operations to the equation repeatedly, until you've isolated the quantity of interest. It does not matter in which order you apply those operations.

For example, with your equation (and I'll omit the units if that's ok)
$$\tag{1}
0=6\cdot(-3)+78v,
$$
to solve for $v$,
you could start by doing  the multiplication first (rule 1)):
$$
0=-18+78v;
$$
then add 18 to both sides (rule 3)):
$$18=78 v;$$
then divide both sides by 78 (rule 2)):
$$
v={18\over 78}.
$$

Alternatively, to solve equation $(1)$ for $v$, after doing the initial multiplication (it's usually a good idea to apply rule 1) whenever you can), you could divide both sides by 78 first:
$$
0={-18\over 78}+1\cdot v;
$$
then add ${ 18\over 78}$ to both sides:
$$
v={18\over 78}.
$$
Either way works. You just do what you have to to isolate $v$ on one side.
Note that  it is preferable to simplify the answer: 
$$
v={18\over 78}={6\cdot 3\over 6\cdot 13}={  3\over  13}.
$$

Here's another route to the answer:
You could start by dividing through by 6 first:
$$
0=-3+13v;
$$
then add 3 to both sides and finally divide both sides by 13:
$$
v={3\over13}.
$$
