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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Okay, Im here. And Bobbym, really haven't you ANY suggestions on how to begin these equations ? I'm feeling that there is no methodology in what Im doing...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

It does take experience to do some of these. You gain that experience by doing them. The beginning stuff is right here:

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Yes, I guess I'll still keep practicing. thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

How about this one here:

Can you get c as the subject?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

c=b+2/a

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

That is correct!

Make b the subject.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Just a question, If I distribute the b on the right, I must also distribute it for 2 ? (Must I distribute to all the right side of the equation ?

1/a=2b ?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi;

I am not following you. What are you trying to do?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Im trying to isolate b on the right of the equation, but must I also distribute the B to the whole right side of the equation ?

*b 1/a=1/b+2 *b (to the 1/b AND to the 2)

which gives :

b/a=2b

?

*Last edited by Al-Allo (2013-03-10 07:15:06)*

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Foget my question, heres my answer b=2a

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi;

That is not correct. Please show me your first step.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

*b 1/a= 1b +2 *b

b/a=+2

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

The left side is correct.

When you multiply the right side by b, that means everything on the right gets multiplied by b. Try it again.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Sorry Bobbym, Ive got other to finish. Il try it on my own, Im feelng that Im lacking knowledge in algebra, maybe I should do a review ^^

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

You only need some practice on that type.

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**