Simultaneous Equation problem

- Ahsome

Simultaneous Equation problem

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- chestercat

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- Ahsome

Two 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

Is it 3 and 6 hours .

- Ahsome

They haven't given answers @Pawanyadav. How did you solve it?

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## More answers

- anonymous

this is an equation right

- Ahsome

Yes

- anonymous

ok this is what i found .......
http://www.algebra.com/algebra/homework/equations/Equations.faq.question.616566.html

- anonymous

this should give you a better understanding then what we gave you .. this question does not seem complete

- Ahsome

It doesn't :(

- anonymous

what do you need

- Ahsome

I want to know how one would solve this question

- anonymous

Will i do or you will do @HWBUSTER00

- anonymous

well i tired to help @Catch.me can you do it

- anonymous

OK
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

So, \(A=2-B\) and \(A=B-3\)?
Therefore
\[2-B=B-3\]\[-B=B-5\]\[-2B=-5\]\[B=\dfrac{5}{2}\]

- Ahsome

\[A=2-b\]\[A=-\dfrac{1}{2}\]Which is impossible :(

- Ahsome

@Catch.me?

- anonymous

OK 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

- anonymous

no equal 2 but 1

- anonymous

hole tank

- Ahsome

I'm really confused now. Can you elaborate @Catch.me?

- anonymous

I 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

- Ahsome

ok

- anonymous

now you have 2 equations 2 unknowns solve them
\[\frac{ 4b-6 }{ b ^{2}-3b }= 1 \]
of course when b = 3 is neglected.
so \[B ^{2}-7B+6=0\]
B=6 or 1 is neglected

- anonymous

@Ahsome ?

- Ahsome

How did you go from (2/a)+(2/b)=1 to that @Catch.me?

- anonymous

put a = b-3 and make multiply them

- ParthKohli

OK 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 \(1/2\)\[\Rightarrow \dfrac{1}{x} + \dfrac{1}{x+3} = \dfrac{1}{2}\]\[\Rightarrow x = 3\]

- anonymous

common denominator

- Ahsome

Ahh, I see. So x would be the slower tap, right?

- anonymous

@Ahsome do you understand this question now?

- Ahsome

No, I still don't :(

- anonymous

B 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/(x-3)) = 1

- anonymous

when 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

the 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/(x-3)) = 1
you will get a similar equation to what he had

- anonymous

i hope that helps

- anonymous

i recommend watching some of these videos
https://www.khanacademy.org/math/algebra/ratio-proportion-topic/advanced-ratios/v/another-take-on-the-rate-problem

- Ahsome

Thank you so much @billj5. I'm kinda busy now, but I will definately look at this afterwards :)

- anonymous

yw

- Pawanyadav

The answer was right.

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