handicapped_runner
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- Joined
- Feb 11, 2021
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I have the following inequality:
[MATH] (C+Bk)(1-m)+B(m+f)(1-a)>0 [/MATH]I follow the following steps:
[MATH] (C+Bk)(1-m)>-B(m+f)(1-a)\\ C(1-m)>-B(m+f)(1-a)-Bk(1-mm)\\ \frac{C}{B}(1-m)>-(m+f)(1-a)-k(1-mm)\\ \frac{C}{B}>\frac{-(m+f)(1-a)-k(1-m)}{1-m}\\ \frac{-C}{B}<\frac{(m+f)(1-a)+k(1-m)}{(1-m)}\\ [/MATH]Is this correct? But what if I do the opposite, that is:
[MATH] B(m+f)(1-a)>-(C+Bk)(1-m)\\ B(m+f)(1-a)>-C(1-m)-Bk(1-m)\\ B((m+f)(1-a)+k(1-m))>-C(1-m)\\ B((m+f)(1-a)+k(1-m))>-C(1-m)\\ B>-C\frac{(1-m)}{((m+f)(1-a)+k(1-m))}\\ \frac{B}{-C}<\frac{(1-m)}{((m+f)(1-a)+k(1-m))}\\ [/MATH]They are clearly describing different things so I feel like I'm doing something wrong, but I cannot figure out what and which result is the correct one. Sorry, not a mathematician, I hope this isn't a very stupid question. C and B are positive(>0), while the other variables are between 0 and 1.
[MATH] (C+Bk)(1-m)+B(m+f)(1-a)>0 [/MATH]I follow the following steps:
[MATH] (C+Bk)(1-m)>-B(m+f)(1-a)\\ C(1-m)>-B(m+f)(1-a)-Bk(1-mm)\\ \frac{C}{B}(1-m)>-(m+f)(1-a)-k(1-mm)\\ \frac{C}{B}>\frac{-(m+f)(1-a)-k(1-m)}{1-m}\\ \frac{-C}{B}<\frac{(m+f)(1-a)+k(1-m)}{(1-m)}\\ [/MATH]Is this correct? But what if I do the opposite, that is:
[MATH] B(m+f)(1-a)>-(C+Bk)(1-m)\\ B(m+f)(1-a)>-C(1-m)-Bk(1-m)\\ B((m+f)(1-a)+k(1-m))>-C(1-m)\\ B((m+f)(1-a)+k(1-m))>-C(1-m)\\ B>-C\frac{(1-m)}{((m+f)(1-a)+k(1-m))}\\ \frac{B}{-C}<\frac{(1-m)}{((m+f)(1-a)+k(1-m))}\\ [/MATH]They are clearly describing different things so I feel like I'm doing something wrong, but I cannot figure out what and which result is the correct one. Sorry, not a mathematician, I hope this isn't a very stupid question. C and B are positive(>0), while the other variables are between 0 and 1.