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September 2nd, 2019, 07:01 PM  #1 
Newbie Joined: Sep 2019 From: New york Posts: 6 Thanks: 0  a simple system with an impossible answer...
first post here... currently an applied math student attending Stonybrook university. The question seems so simple. But I am unable to find an answer. jack can do 3 chemistry problems and 6 math problems an hour. Jill can do 4 chemistry problems and 7 math problems an hour. How long must they both work to finish 11 chemistry problems and 17 math problems? If you solve this system, you get jack's time as (3 hours) and Jill's is 5 hours. Obviously, this is impossible. But the only other way I found to accurately estimate this is just guessing and checking.... getting closest with jack doing 1.4 hours of study and Jill doing 1.7 ... producing 11 completed chemistry problems but 20.3 math questions. Very confused how to approach this question efficiently without guessing. Please help!!!! Last edited by skipjack; September 4th, 2019 at 04:36 PM. 
September 2nd, 2019, 07:10 PM  #2  
Senior Member Joined: Aug 2012 Posts: 2,393 Thanks: 749  Quote:
https://en.wikipedia.org/wiki/H._Blaine_Lawson Last edited by Maschke; September 2nd, 2019 at 07:20 PM.  
September 2nd, 2019, 07:28 PM  #3 
Newbie Joined: Sep 2019 From: New york Posts: 6 Thanks: 0  No!!
No I don’t know him... but I have come across the term “complex variables” when trying to do some research on a problem like this.... how do you approach it???

September 2nd, 2019, 07:45 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,393 Thanks: 749  Complex variables is calculus on the complex numbers. It's got a very different flavor than regular (real number) calculus. I don't think it bears on your problem.

September 2nd, 2019, 07:49 PM  #5 
Newbie Joined: Sep 2019 From: New york Posts: 6 Thanks: 0 
I’m affluent with multivariable calculus... but linear algebra is a lot different... how do I efficiently compute the answer to this? Without putting it into a computer or something?

September 2nd, 2019, 08:41 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 
the thing to remember is rates add $r_{kc}=3 ~prob/hr$ $r_{km}=6~prob/hr$ $r_{lc}=4~prob/hr$ $r_{lm} = 7~prob/hr$ $T = \dfrac{11}{r_{kc}+r_{lc}} + \dfrac{17}{r_{km}+r_{lm}} = \\ \dfrac{11}{7} + \dfrac{17}{13} = \dfrac{262}{91} ~hr \approx 2~hr~53~min$ 
September 3rd, 2019, 05:33 PM  #7 
Newbie Joined: Sep 2019 From: New york Posts: 6 Thanks: 0  Incorrect
This system is asking for a certain number of hours for each students to work.... you need to isolate exactly how much jack needs to work and Jill separately.

September 3rd, 2019, 05:58 PM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  Quote:
$T_{jack} = \dfrac{11}{3} + \dfrac{17}{6} = \dfrac{13}{2} = 6~hr~30~min$ $T_{jill} = \dfrac{11}{4} + \dfrac{17}{7} = \dfrac{77+68}{28}=\dfrac{145}{28} \approx 5~hr~11~min$  
September 3rd, 2019, 08:32 PM  #9 
Newbie Joined: Sep 2019 From: New york Posts: 6 Thanks: 0  Noo
you want to find the MINIMUM amount of time (two variables) for jack and Jill. To complete exactly 11 chemistry problems and 17 math problems. When you solve this system mathematically it yields 3 and 5 ... which doesn’t make any real world sense. I have approximated their times to be somewhere Around 1.7 hours for jack and 1.4 hours for Jill. But this was through no other means than brute force guessing and checking. Mathematically I would like to know how to approach a problem like this more efficiently...

September 4th, 2019, 01:02 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,972 Thanks: 2222 
Please clarify what system you solved, so that we know what you tried. Can jack do 3 chemistry problems and 6 math problems an hour simultaneously?
Last edited by skipjack; September 4th, 2019 at 04:37 PM. 

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answer, impossible, linear algebra, math, simple, simply, system, system of equations 
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