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anonymous
 one year ago
a,b,c, and d are positive integers such that a/b < c/d
Which one is greater?
a. a+c/b+d
b. c/d
anonymous
 one year ago
a,b,c, and d are positive integers such that a/b < c/d Which one is greater? a. a+c/b+d b. c/d

This Question is Closed

AaronAndyson
 one year ago
Best ResponseYou've already chosen the best response.0< means lessthan > means greaterthan

hybrik
 one year ago
Best ResponseYou've already chosen the best response.0Lets say A and B is 1 and C and D is >1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437751183877:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0we have the subsequent implication: \[\frac{a}{b} < \frac{c}{d} \Rightarrow \frac{a}{c} < \frac{b}{d}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0no, I have multiplied your original inequality by b/c: \[\Large \frac{a}{b} \times \frac{b}{c} < \frac{c}{d} \times \frac{b}{c} \Rightarrow \frac{a}{c} < \frac{b}{d}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay whats the nexxt step

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0we can write the subsequent steps: \[\Large \frac{{\frac{{a + c}}{{b + d}}}}{{\frac{c}{d}}} = \frac{{a + c}}{{b + d}} \times \frac{d}{c} = \frac{{\frac{{a + c}}{c}}}{{\frac{{b + d}}{d}}} = \frac{{\frac{a}{c} + 1}}{{\frac{b}{d} + 1}} < 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this..... is....... super........ confusing can you use that actual numbers please....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0since we have: \[\Large \frac{a}{c} < \frac{b}{d}\] then we can write: \[\Large \frac{a}{c} + 1 < \frac{b}{d} + 1\] and dividing bot sides by (b/d)+1, we get: \[\Large \frac{{\frac{a}{c} + 1}}{{\frac{b}{d} + 1}} < 1\] am I right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0....i have absolutely no idea what all this is i dont understand why what i did up there is wrong...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437753458344:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please note that if I use numbers I'm not giving you a proof, since I'm giving you a check only

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so the subsequent step: \[\Large \frac{a}{b} < \frac{c}{d} \Rightarrow \frac{{\frac{a}{c} + 1}}{{\frac{b}{d} + 1}} < 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0is right for you?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0ok then from this inequality: \[\Large \frac{{\frac{{a + c}}{{b + d}}}}{{\frac{c}{d}}} = \frac{{a + c}}{{b + d}} \times \frac{d}{c} = \frac{{\frac{{a + c}}{c}}}{{\frac{{b + d}}{d}}} = \frac{{\frac{a}{c} + 1}}{{\frac{b}{d} + 1}} < 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0we can write this one: \[\Large \frac{{\frac{{a + c}}{{b + d}}}}{{\frac{c}{d}}} = < 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0oops.. \[\Large \frac{{\frac{{a + c}}{{b + d}}}}{{\frac{c}{d}}} < 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now I multiply both sides by c/d, so I can write: \[\Large \frac{{a + c}}{{b + d}} < \frac{c}{d}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so, what can you conclude?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can conclude that if there is a question like this on my test, I'M DEFINITELY FAILING because I don't understand any of this. This was way too complicated.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please study all my steps carefully, and you will become an expert on this type of question

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, nevertheless my procedure above, is the unique way to solve questions like yours
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