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12 Oct 2017, 10:49
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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) ?
Thanks
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) ?
Thanks
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18 Oct 2017, 10:48
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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.
Please give kudos, if u liked my post!
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.
Please give kudos, if u liked my post!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
