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 $\dfrac{18\frac{\text{m}}{\text{sec}}}{78}$?

  • 1
    $\begingroup$ You are correct. The book has an error in the middle term of your second displayed equation $\endgroup$ Feb 19, 2012 at 19:12
  • $\begingroup$ Yes. You are correct. $\endgroup$
    – JavaMan
    Feb 19, 2012 at 19:13
  • $\begingroup$ ok, so which should it be? both negative, or both positive? What exactly should be my next step? $\endgroup$ Feb 19, 2012 at 19:14
  • $\begingroup$ Either. It depends on how you solved the equation. Of course, with the two negative signs, you'd cancel them, giving the simplified result ${18\over 78}{m\over sec} ={3\over 13}{m\over sec}$. $\endgroup$ Feb 19, 2012 at 19:15
  • $\begingroup$ so it doesn't matter if I divide 78kg across all terms, or if I subtract the $-18\frac{kg*m}{sec}$ from both sides then divide the 78kg? $\endgroup$ Feb 19, 2012 at 19:22

1 Answer 1


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: $$ v={3\over13}. $$

  • $\begingroup$ thanks for the very helpful answer! +1 $\endgroup$ Feb 19, 2012 at 19:49
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    $\begingroup$ You're welcome; I'm glad to help. $\endgroup$ Feb 19, 2012 at 19:51

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