# Rearranging Equations?

• Apr 6th 2011, 06:34 PM
jebus197
Rearranging Equations?
Hi, please can someone watch this video here?

I can follow it right up to the part just before she takes the square root of both sides of the equation. What I don't get is the whole of the video before this point goes on about what you do to the left of the equation you have to do to the other side of the equation too. But in this section she has $mu^2/m$ and $mv^ - deltaEk/m$ but she only removes the m from the left hand side. Surely she should get rid of m from both sides to be consistent before going on to take the square roots of both sides of the equation?

Basically the question is why do the two m's on the right side of the equation not cancel out? It's rather amusing how they laugh heartily but make no effort to explain this.
• Apr 6th 2011, 07:29 PM
DivisionByZero
Well, you have something of the form:

$
\frac{m+k}{m}
$

The m's don't cancel out because m AND k are divided by m. But you could rearrange this to:

$
\frac{m}{m}+\frac{k}{m}=1+\frac{k}{m}
$
• Apr 7th 2011, 03:44 AM
jebus197
Quote:

Originally Posted by DivisionByZero

The m's don't cancel out because m AND k are divided by m.

OK I get that.

Quote:

Well, you have something of the form:

$
\frac{m+k}{m}
$

But you could rearrange this to:

$
\frac{m}{m}+\frac{k}{m}=1+\frac{k}{m}
$

So are you just saying that
$
\frac{m+k}{m}
$
and $
\frac{m}{m}+\frac{k}{m}=1+\frac{k}{m}
$
are exactly equivalent?

I'm slightly lost here. I would have assumed that $
\frac{m+k}{m}
$
would require that you do the 'opposite' from what's in the equation? So since you have $m + k/m$ you would have to somehow subtract $k$ from the top and $k$ from the bottom and then divide them? Can you please break your logic down into smaller steps and explain them in English where possible too?
• Apr 7th 2011, 03:56 AM
topsquark
Let's try this backward. Can you add $1 + \frac{k}{m}$ ? You need to find a common denominator, m in this case.

If that's too many variables for the moment, how do you add $3 + \frac{1}{2}$ ?

-Dan
• Apr 7th 2011, 04:18 AM
jebus197
Quote:

Originally Posted by topsquark
Let's try this backward. Can you add $1 + \frac{k}{m}$ ? You need to find a common denominator, m in this case.

If that's too many variables for the moment, how do you add $3 + \frac{1}{2}$ ?

-Dan

As nuts as it sounds that's not a question I have ever encountered before. (I'm a mature student so am learning a lot of maths for the fist time). To me $3 + \frac{1}{2}$ was always just $3 + \frac{1}{2}$. Just like $3 + \frac{1}{4}$ it was/is just a simple addition. But you have me doubting my own reasoning here, as it seems like you are saying there's a methodology (involving rearranging) to obtain this answer?
• Apr 7th 2011, 04:31 AM
topsquark
Quote:

Originally Posted by jebus197
As nuts as it sounds that's not a question I have ever encountered before. (I'm a mature student so am learning a lot of maths for the fist time). To me $3 + \frac{1}{2}$ was always just $3 + \frac{1}{2}$. Just like $3 + \frac{1}{4}$ it was/is just a simple addition. But you have me doubting my own reasoning here, as it seems like you are saying there's a methodology (involving rearranging) to obtain this answer?

Think of it this way... if we have two fractions:
$\frac{3}{4} + \frac{1}{5}$
then we can add them by getting a common denominator. The common denominator in this case will be 4 x 5 = 20. So we need to multiply the fraction on the right to put it in the form (something)/20.

Now, we can't change the value of 1/5, but we must find a way to multiply the denominator by a factor of 4. How can we do this? Well if we multiply the fraction 1/5 by 1, then we still have 1/5. So I'm going to choose a particular kind of form of 1 to multiply the 1/5 by.

$\frac{1}{5} \times 1 = \frac{1}{5} \times \frac{4}{4} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$.

We do the same thing with 3/4, but in this case we need to multiply the denominator by 5, thus
$\frac{3}{4} \times 1 = \frac{3}{4} \times \frac{5}{5} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

Why do we do all this? In order to add two fractions we need to have a common denominator, in this case 20.
$\frac{3}{4} + \frac{1}{5} = \frac{15}{20} + \frac{4}{20} = \frac{15 + 4}{20} = \frac{19}{20}$.

The problem you are being asked to do is very similar, except in this case I'm going to change the 1 into 1/1, so we can have the number 1 in fractional form. So how do you add
$1 + \frac{1}{2} = \frac{1}{1} + \frac{1}{2}$

-Dan

The eventual point of all this is to show that $1 + \frac{k}{m} = \frac{m}{m} + \frac{k}{m} = \frac{m + k}{m}$. Do you see the similarities?
• Apr 7th 2011, 10:41 AM
jebus197
Quote:

Originally Posted by topsquark
Think of it this way... if we have two fractions:
$\frac{3}{4} + \frac{1}{5}$
then we can add them by getting a common denominator. The common denominator in this case will be 4 x 5 = 20. So we need to multiply the fraction on the right to put it in the form (something)/20.

Now, we can't change the value of 1/5, but we must find a way to multiply the denominator by a factor of 4. How can we do this? Well if we multiply the fraction 1/5 by 1, then we still have 1/5. So I'm going to choose a particular kind of form of 1 to multiply the 1/5 by.

$\frac{1}{5} \times 1 = \frac{1}{5} \times \frac{4}{4} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$.

We do the same thing with 3/4, but in this case we need to multiply the denominator by 5, thus
$\frac{3}{4} \times 1 = \frac{3}{4} \times \frac{5}{5} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

Why do we do all this? In order to add two fractions we need to have a common denominator, in this case 20.
$\frac{3}{4} + \frac{1}{5} = \frac{15}{20} + \frac{4}{20} = \frac{15 + 4}{20} = \frac{19}{20}$.

The problem you are being asked to do is very similar, except in this case I'm going to change the 1 into 1/1, so we can have the number 1 in fractional form. So how do you add
$1 + \frac{1}{2} = \frac{1}{1} + \frac{1}{2}$

Yeah. I can dig that I think you last part just equates to the top heavy fraction $\frac{3}{2} = 1 \frac{1}{2}$

Quote:

The eventual point of all this is to show that $1 + \frac{k}{m} = \frac{m}{m} + \frac{k}{m} = \frac{m + k}{m}$. Do you see the similarities?
But I think I've lost the thread a little on how this relates to the original question? In the case of $\frac{m + k}{m}$ everything on either side of the plus sign is being divided by m, so the m's don't cancel? But in the case of $\frac{mk}{m}$ which is closer to my original example, they do seem to cancel?
• Apr 7th 2011, 12:11 PM
TheChaz
Yes, mk/m reduces to k.
• Apr 7th 2011, 12:21 PM
jebus197
But why is $\frac{m + k}{m}$ different from $\frac{mk}{m}$? Why does one cancel and the other doesn't? (Or are we just going round in circles now?)
• Apr 7th 2011, 12:23 PM
TheChaz
"canceling" is division. You can cancel FACTORS, not TERMS
(the capitalized words are important vocabulary in math, roughly meaning "things that are multplied together" and "things that are added together", respectively)
• Apr 7th 2011, 01:03 PM
jebus197
Quote:

Originally Posted by TheChaz
"canceling" is division. You can cancel FACTORS, not TERMS
(the capitalized words are important vocabulary in math, roughly meaning "things that are multplied together" and "things that are added together", respectively)

At last I see the beginnings of some logic, or at least a rule I can work with until I get a better understanding of everything. Going back to my original equation: YouTube - Rearranging.flv the reason (or rule) why both m's don't cancel is because one side of the equation contains factors and the other contains terms? So only calculations like this with factors as the numerator (above the denominator) would tend to cancel, while those (as on the right side of the equation here) that use terms as the numerator would not?
• Apr 7th 2011, 01:59 PM
TheChaz
More or less!
Notice that in mk/m, you can factor as $\frac{mk}{m*1} = \frac{(m)k}{(m)1}$

and reduce the m's to get k/1 = k

For another example of this, consider
$\frac{2b^2 + 6b}{4b + 12} = \frac{2b(b + 3)}{4(b + 3)}$

You can cancel/reduce the (b + 3)'s...
then you'll have:
$\frac{2b}{4} = \frac{(2)b}{(2)2} = \frac{b}{2}$
After canceling 2's
• Apr 7th 2011, 02:17 PM
jebus197
Cheers dude. I'll give some proper thought to your answer in the morning. (I have my head buried in more books/study at the minute.) But it's definitely beginning to make more sense to me. ;-)
• Apr 7th 2011, 02:22 PM
TheChaz
That's the idea!
Ps love the name - reminds me of when Homer became a missionary :)