# a simple system with an impossible answer...

#### Bioobird

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.

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#### Maschke

first post here... currently an applied math student attending Stonybrook university. The question seems so simple. But I am unable to find an answer.
Do you know professor Lawson, or know of him? He was my prof for complex variables way back in the day at Berkeley.

https://en.wikipedia.org/wiki/H._Blaine_Lawson

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#### Bioobird

No!!

No I donâ€™t know him... but I have come across the term â€œcomplex variablesâ€ when trying to do some research on a problem like this.... how do you approach it???

#### Maschke

No I donâ€™t know him... but I have come across the term â€œcomplex variablesâ€ when trying to do some research on a problem like this.... how do you approach it???
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.

#### Bioobird

Iâ€™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?

#### romsek

Math Team
the thing to remember is rates add

$r_{kc}=3 ~prob/hr$
$r_{km}=6~prob/hr$

$r_{lc}=4~prob/hr$
$r_{lm} = 7~prob/hr$

$T = \dfrac{11}{r_{kc}+r_{lc}} + \dfrac{17}{r_{km}+r_{lm}} = \\ \dfrac{11}{7} + \dfrac{17}{13} = \dfrac{262}{91} ~hr \approx 2~hr~53~min$

#### Bioobird

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.

#### romsek

Math Team
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.
so you want to treat it as if each works alone?

$T_{jack} = \dfrac{11}{3} + \dfrac{17}{6} = \dfrac{13}{2} = 6~hr~30~min$

$T_{jill} = \dfrac{11}{4} + \dfrac{17}{7} = \dfrac{77+68}{28}=\dfrac{145}{28} \approx 5~hr~11~min$

#### Bioobird

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 doesnâ€™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...

#### skipjack

Forum Staff
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?

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