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Ahsome
 one year ago
Simultaneous Equation problem
Ahsome
 one year ago
Simultaneous Equation problem

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ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Two taps A and B fill a swimming pool together in two hours. Alone, it takes tap A three haours less than B to fill the same pool. How many hours does it take each tap to fill the pool separately?

Pawanyadav
 one year ago
Best ResponseYou've already chosen the best response.0Is it 3 and 6 hours .

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0They haven't given answers @Pawanyadav. How did you solve it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is an equation right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok this is what i found ....... http://www.algebra.com/algebra/homework/equations/Equations.faq.question.616566.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this should give you a better understanding then what we gave you .. this question does not seem complete

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0I want to know how one would solve this question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Will i do or you will do @HWBUSTER00

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well i tired to help @Catch.me can you do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK the main concept of these problems is to convert words to equation, so just get sentences and write the corresponding equation Two taps A and B fill a swimming pool together in two hours. means \[A + B = 2\] Alone, it takes tap A three hours less than B to fill the same pool. \[A = B  3\] now you have 2 equations 2 unknowns substitute in the first equation using last equation

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0So, \(A=2B\) and \(A=B3\)? Therefore \[2B=B3\]\[B=B5\]\[2B=5\]\[B=\dfrac{5}{2}\]

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0\[A=2b\]\[A=\dfrac{1}{2}\]Which is impossible :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK first equation was wrong A = x to fill 1 liter B = y to fill 1 liter so in 2 hours x : 1 2 : ?? A will fill 2/x B will fill 2/y so 2/a + 2/b = 2

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0I'm really confused now. Can you elaborate @Catch.me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I made a mistake in understanding the first sentence Two taps A and B fill a swimming pool together in two hours means that the amount of A filling in 2 hours + the amount of B filling in 2 hours will make the tank full lets say A fills tank in 7 hours so in 2 hours it fills 2/7 of that tank. so (2/a) + (2/b) = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you have 2 equations 2 unknowns solve them \[\frac{ 4b6 }{ b ^{2}3b }= 1 \] of course when b = 3 is neglected. so \[B ^{2}7B+6=0\] B=6 or 1 is neglected

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0How did you go from (2/a)+(2/b)=1 to that @Catch.me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0put a = b3 and make multiply them

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0OK so let's say that one tap takes \(x\) hours and the other one takes \(x+3\). Then the fraction of work they do in one hour is given by \(1/x + 1/(x+3)\) but since it takes two hours for them to do the work, it is also given by \(1/2\)\[\Rightarrow \dfrac{1}{x} + \dfrac{1}{x+3} = \dfrac{1}{2}\]\[\Rightarrow x = 3\]

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, I see. So x would be the slower tap, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Ahsome do you understand this question now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0B fills 1 pool per x hours > rate of B filling a pool = 1/x A = B  3 hours, so A fills 1 pool per x  3 hours > rate of A filling a pool = 1/(x  3) now we know the rates of each so we have this, both taps fill 1 pool in 2 hours, 2 hours * (rate of B) + 2 hours * (rate of A) = 1 pool filled 2* (1/x) + 2 * (1/(x3)) = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when you solve that for x you get 2 answers for x one of these will not make sense if you plug into this equation A = B  3, will give a negative result, and we know this makes no sense because the number of hours it takes to fill the pool cannot be negative, so the answer that makes sense is 6 hours for B, then you just subtract 3 from that to get 3 hours for A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the method ParthKholi used is the same thing except he used B = A + 3 instead of A = B 3 , if you divide both sides of this equation by 2 2* (1/x) + 2 * (1/(x3)) = 1 you will get a similar equation to what he had

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i recommend watching some of these videos https://www.khanacademy.org/math/algebra/ratioproportiontopic/advancedratios/v/anothertakeontherateproblem

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much @billj5. I'm kinda busy now, but I will definately look at this afterwards :)

Pawanyadav
 one year ago
Best ResponseYou've already chosen the best response.0The answer was right.
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