Geoff's algrbra simplification and re-arragnement question!

[jazzygeoff]

Hi all,

I wonder if someone could help me out with the following simplification that I have always had trouble with. It is when there is a division in the algebra, and you need to simplify.

e.g

(-2p-2)/(p-1)

The answer is -2, I know, but could someone please show me EXACT working, with as many words as possible as to how that solution fianlly pops out. I have tried putting in a vaule for p, to find out the answer. I remember in my maths GSCE my maths teacher just used to go around putting slashes through terms all over the place to show that they were cancelling eachother out, but I never really got the hang of it! ;-)

Could someone recommend somewhere to look on the net for some examples of this kind of re-arrangement/simplification with answers that I could work through. I don't see any on the front page of this web-site?

Many thanks in advance!

Geoff

 

[Denis]

Re: Geoff's algrbra simplification and re-arragnement questi

(-2p-2)/(p-1)
The answer is -2, .....
Geoff

The answer is not -2
Would be -2 if division was by (p+1):

(-2p - 2) = -2(p + 1)

 

[Unco]

It looks like a trick at first, but after a while it becomes automatic.

\(\displaystyle \L \mbox{ \frac{-2p - 2}{p + 1}}\)

We want to write the numerator in terms of the denominator.

\(\displaystyle \L \mbox{ -2p - 2 = 2(-p - 1) = -2(p + 1)}\)

Not too bad.

Our quotient becomes

\(\displaystyle \L \mbox{ \frac{-2p - 2}{p + 1} = \frac{-2(p + 1)}{p + 1} = \frac{-2\sout{(p + 1)}}{\sout{p + 1}} = -2}\)

Edit: Woops, I originally had (-2p+2)/(p-1) (which works) but changed to Denis's and forget to edit the main part!

 

[jazzygeoff]

Many thanks to both of you! Well noticed on the deliberate mistake Denis!, and many thanks for the thorough explaination Unco - I have gone on to do the rest of the questions in my exercise correctly now, thanks to your help. My book just has answers without any working, so it's sometimes hard to see how to get the right result.

Also note, that the mix up in my spelling of the word "re-arrangement" in the title of this message was not supposed to be ironic!

 

[Mrspi]

It looks like a trick at first, but after a while it becomes automatic.

\(\displaystyle \L \mbox{ \frac{-2p - 2}{p + 1}}\)

We want to write the numerator in terms of the denominator.

\(\displaystyle \L \mbox{ -2p - 2 = 2(-p - 1) = -2(p + 1)}\)

Not too bad.

Our quotient becomes

\(\displaystyle \L \mbox{ \frac{-2p - 2}{p - 1} = \frac{-2(p - 1)}{p - 1} = \frac{-2\sout{(p - 1)}}{\sout{p - 1}} = -2}\)

Shouldn't the quotient be

\(\displaystyle \L \mbox{ \frac{-2p - 2}{p - 1} = \frac{-2(p + 1)}{p - 1}\)

Maybe a case of "wishful thinking." (wishing that the fraction could be reduced??)

Unfortunately, the numerator and denominator contain no like fractions, and the expression is in simplest form already.

 

[Denis]

Re: Geoff's algrbra simplification and re-arragnement questi

(-2p-2)/(p-1)
The answer is -2, I know...

Once more: the answer is -2 IF expression is corrected to: (-2p-2)/(p+1)