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anonymous
 4 years ago
If (a/b)=(c/d), prove (a+b)/(ab)=(c+d)/(cd).
anonymous
 4 years ago
If (a/b)=(c/d), prove (a+b)/(ab)=(c+d)/(cd).

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hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1hint : 1.Add 1 on both sides>(1) 2.subtract 1 from both sides >(2) 3. divide : (1)/(2)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1@Mello did u get it or should i show detailed steps??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, please. My brain isn't getting it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Im a bit lost from there.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1so u must have got : \(\frac{a+b}{b}=\frac{c+d}{d}\)>(1) right? Step 2: subtract 1 from both sides: \(\frac{a}{b}1=\frac{c}{d}1\) \(\frac{ab}{b}=\frac{cd}{d}\)>(2) got step 2 ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@hartnn Right. Oh subtract from the original. OK.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1sorry i didn't mention that, now divide: (1) / (2) did u see how u get your final result after dividing??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How would you divide an equation by an equation?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1ofcourse we can: if x=y and w=z after dividing we get (x/w)=(y/z) try dividing here. if u don't get, i'll show that step.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, right. I remember learning something like that. I think I should be able to manage from here

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1good,let me know if u still have doubts or could not get final answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alrighty, Thank you so much for the help!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Got it! Thanks again.
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