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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Hi;

Okay, take all the time you need.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

bobbym wrote:

I was interrupted by some errand, boring one indeed.

Bobbym, it seems you mistakenly made a negative sign a positive sign.

The original is the same as the one Anomnystify solved please check the sign at #244. I mean;

3(x-1)/x-9,

Which you made it

3(x+1)/x^2-9.

*Last edited by EbenezerSon (2013-08-08 20:32:27)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Hi;

I wrote, that there is a mistake in line 2 of post #244

EbenezerSon wrote:

anonimnystefy wrote:Hi EbenezerSon

With a careful look at signs I think should be,

x-3x-3-3/x^2-9 =-2x-6/x^2-3^2 =

-2(x+3)/(x-3)(x+3)

= -2/x-3Please am I right there.

Thanks.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

Please, I mean the problem itself.

You made a negative sign a positive sign.

Here ;

3(x-1). Which you made it 3(x+1).

You check it from one of your post.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

The first line is 3(x+1)

The second line changes that to 3x - 3, that is a mistake.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

Please look at #252.

That is the original question.

The first line is -3(x-1).

And changes the second to be,

-3x+3

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

You are not understanding what I am saying. Post #244 is wrong at the second line.

Please post the original problem from the book.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

1/1+3 - 3(x-1)/x^2-9.

That is it.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

And what did you get for an answer?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

I am not at home now, I will post the original when I get there.

Thanks.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Okay, see you then.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

Hi, Bobbym.

This is the one from the book;

1/x-3 - 3(x-1)/x^2-9

I had -2/x+3 as answer.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Hi;

That is correct!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

bobbym wrote:

Hi;

How did you get,

9^n+2 * 3^n+2 out of 27^n+2?

I have edited it.

*Last edited by EbenezerSon (2013-08-11 04:17:33)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Hi;

I did not get that. Where does 23^n come from?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

Please, I double check it I have edited.

Thanks.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

27^a = 9^a * 3^a

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

It's 27^n+2, and not 27^a.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

You can say a = (n+2), same principle.

27^(n+2) = 9^(n+2) * 3^(n+2)

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

I suppose the base must always be the same in each case. So I percieved it to be,

27^n+2=27^n*27^2=3^3n * 3^3+2.

I had thought should be so.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

I am sorry, I can not follow that. Please bracket it off.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

In fact I have not seen an indicial problem being split to get different numbers as the base. like what is in #269.

.Thiş problem is from indices.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

You should learn to latex or to use parentheses better.

That modern notation they are using in that book is not good.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 466

27^n+2=27^n * 27^2

Thıs is how I mean, I will learn parenthesis in it proper way as you say.

For instance, 6^n+3=6^n * 6^3.

Because, for instance, 3^2 * 3^2=3^(2+2).

What do you say.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

All of that is correct but I have to interpret every bit of it.

When you write 27^n+3 in mathematics that means

It is properly written 27^(n+3) this means

Notice they are both very different.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**