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**bronxsystem****Member**- Registered: 2013-06-22
- Posts: 63

Hey all i know this is really basic but i have hit a wall

(d / a+b) * (5a + 5b / 2d^2 + 2d)

i can get the answer but im not sure what im actually doing S:

i khow to multiply problem like (1/2) * (1/2) = 1/4 but what confuses me in this problem is the denominator a+b not sure how to multiply the other denominator with it.

My workings

(d / a+b) * (5a + 5b / 2d^2 + 2d)

i just divide by common factors first step

numerator 5a + 5b to 5 ( a+b)

then denominator 2d (d + 1)

then i just divided both numerator and denominator by common factors (a+b) and d

so i got the answer 5 / 2 (d + 1)

how would you guys solve it? sorry if my reasoning bit over the place, any chance you can show me your working so i can aware myself.

thank you.

*Last edited by bronxsystem (2013-07-13 12:25:09)*

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**bronxsystem****Member**- Registered: 2013-06-22
- Posts: 63

had another one that the method works but again not sure what im doing S:

( a + 3 / a - 5) * (2a - 10 / 3a + 9)

= (a + 3 / a - 5) * 2 ( a - 5) / 3 (a + 3)

then just divide numerator and denominators by common factors and i get 2/3

>< whats this called or what rule is it?

*Last edited by bronxsystem (2013-07-13 12:42:23)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Hi;

I would call it cancellation.

**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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**bronxsystem****Member**- Registered: 2013-06-22
- Posts: 63

sorry those other 2 post had to much fluff a better question maybe is do you think the steps i take are acceptable?

eg

(3ab^2 / 4) * ( 12/ 18a^2)

just simply multiply numerators then denominators and then simplify

(36ab^2) / (72a^2)

then divide by common factor 36a

= b^2 / 2a

thats all it is right?

**not sure why this way works i thought your meant to divide both numerator and denominator by same number S: **

(**3**ab^2 / 4) * (**12** / 18a^2)

first divide numerators by common factor 3

4ab^2

then divide denominators by common factor 2

2 (8a^2)

then divide both numerators and denominators by 2

(2ab^2 / 4a^2)

then divide by 2a

( b^2 / 2a)

*Last edited by bronxsystem (2013-07-14 00:47:35)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

The first method is simpler.

**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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**bronxsystem****Member**- Registered: 2013-06-22
- Posts: 63

roger thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

You are welcome.

**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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