anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AaronAndyson
  • AaronAndyson
< means less-than > means greater-than
hybrik
  • hybrik
Lets say A and B is 1 and C and D is >1
anonymous
  • anonymous
|dw:1437751183877:dw|

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Michele_Laino
  • Michele_Laino
we have the subsequent implication: \[\frac{a}{b} < \frac{c}{d} \Rightarrow \frac{a}{c} < \frac{b}{d}\]
anonymous
  • anonymous
Is that a rule?
Michele_Laino
  • Michele_Laino
no, 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
  • anonymous
okay whats the nexxt step
Michele_Laino
  • Michele_Laino
we 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
  • anonymous
this..... is....... super........ confusing can you use that actual numbers please....
Michele_Laino
  • Michele_Laino
since 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?
Michele_Laino
  • Michele_Laino
oops.. both*
anonymous
  • anonymous
....i have absolutely no idea what all this is i dont understand why what i did up there is wrong...
anonymous
  • anonymous
|dw:1437753458344:dw|
Michele_Laino
  • Michele_Laino
please note that if I use numbers I'm not giving you a proof, since I'm giving you a check only
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
so the subsequent step: \[\Large \frac{a}{b} < \frac{c}{d} \Rightarrow \frac{{\frac{a}{c} + 1}}{{\frac{b}{d} + 1}} < 1\]
Michele_Laino
  • Michele_Laino
is right for you?
anonymous
  • anonymous
yea
Michele_Laino
  • Michele_Laino
ok 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
  • Michele_Laino
we can write this one: \[\Large \frac{{\frac{{a + c}}{{b + d}}}}{{\frac{c}{d}}} = < 1\]
Michele_Laino
  • Michele_Laino
oops.. \[\Large \frac{{\frac{{a + c}}{{b + d}}}}{{\frac{c}{d}}} < 1\]
anonymous
  • anonymous
yea...
Michele_Laino
  • Michele_Laino
now I multiply both sides by c/d, so I can write: \[\Large \frac{{a + c}}{{b + d}} < \frac{c}{d}\]
anonymous
  • anonymous
ok...
Michele_Laino
  • Michele_Laino
so, what can you conclude?
anonymous
  • anonymous
I 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
  • Michele_Laino
please study all my steps carefully, and you will become an expert on this type of question
anonymous
  • anonymous
so..... lost
Michele_Laino
  • Michele_Laino
I'm sorry, nevertheless my procedure above, is the unique way to solve questions like yours

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