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Gmat Math Book - clarificaiton -- cmponendo & dividendo

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Gmat Math Book - clarificaiton -- cmponendo & dividendo [#permalink]

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New post 12 Oct 2017, 09:49
if:
a/b = c/d (a,b,c,d non zero real numbers), it is given that:
1. (a+b)/(a-b) = (c+d)/(c-d)

Clarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)
How did we (algebraically) arrive at this equivalency?

Clarification 2: can we also claim that: (a-b)/(a+b) = (c-d)/(c+d) ?


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Gmat Math Book - clarificaiton -- cmponendo & dividendo [#permalink]

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New post 18 Oct 2017, 09:48
Hi sevenplusplus

Clarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)
How did we (algebraically) arrive at this equivalency? - Here the book only gives the rule that can be applied to ratios. Incase a-b=0, a=b, ratio is 1, so there's no need to go for componendo dividendo. It is useful to apply in some problems directly.
\(\frac{a}{b}\) = \(\frac{c}{d}\) =>Adding and subtracting 1 from both sides;
\(\frac{a}{b}\)+1=\(\frac{c}{d}\)+1 = \(\frac{(a+b)}{b}\)=\(\frac{(c+d)}{d}\) -------(1)
\(\frac{a}{b}\)-1=\(\frac{c}{d}\)-1 = \(\frac{(a-b)}{b}\)=\(\frac{(c-d)}{d}\)--------(2)
Dividing (1) by (2) gives the desired result.


Clarification 2: can we also claim that: \(\frac{(a-b)}{(a+b)}\) = \(\frac{(c-d)}{(c+d)}\) ?
Yes. As it is inverse of the ratio.

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Gmat Math Book - clarificaiton -- cmponendo & dividendo   [#permalink] 18 Oct 2017, 09:48
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