anonymous
  • anonymous
Two pipe running together can fill a cistern in 3.5mins and one pipe takes 3 mins more than the other find the time in which each pipe would fill cistern
Mathematics
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anonymous
  • anonymous
Two pipe running together can fill a cistern in 3.5mins and one pipe takes 3 mins more than the other find the time in which each pipe would fill cistern
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
perl
  • perl
Let x be the rate of the first pipe and y is the rate of the second pipe. And t be the time it takes for first pipe to fill cistern. Then we have the following equations, use d = r* t 1 cistern = (x+y) * 3.5 min 1 cistern = x * t 1 cistern = y * (t + 3)
anonymous
  • anonymous
I can't understand the above.

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perl
  • perl
Have you used the formula distance = rate * time
anonymous
  • anonymous
Yes..
perl
  • perl
I am using a similar equation. 1 cistern = rate of pipe * time
anonymous
  • anonymous
What next?
perl
  • perl
Let x be the rate of the first pipe and y is the rate of the second pipe. And t be the time it takes for first pipe to fill cistern. 1 cistern = (x+y) * 3.5 min 1 cistern = x * t 1 cistern = y * (t + 3)
anonymous
  • anonymous
3 cistern?
perl
  • perl
they are three different equations
perl
  • perl
"Two pipe running together can fill a cistern in 3.5mins" That gives us 1 cistern = (x+y) * 3.5 min , where x and y are the rates of the pipes
anonymous
  • anonymous
Help PLEASE?
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
I think that @perl gave you the right explanation
Michele_Laino
  • Michele_Laino
nevertheless I can give you another way to solve your problem. The working rates of the two pipes are: \[\Large\frac{W}{x},\quad \frac{W}{{x + 3}}\]
Michele_Laino
  • Michele_Laino
where W is the work to be done
Michele_Laino
  • Michele_Laino
now, following the text of your problem, we can write: \[\Large \frac{W}{x} + \frac{W}{{x + 3}} = \frac{W}{{3.5}}\] and, simplifying that expression for W, we get: \[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{1}{{3.5}}\] Please solve that equation for x
anonymous
  • anonymous
x+3+x/x^2+3x = 1/3.5?
Michele_Laino
  • Michele_Laino
first step: we have: \[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{2}{7}\]
Michele_Laino
  • Michele_Laino
yes! since 3.5= 7/2
Michele_Laino
  • Michele_Laino
now, the least common multiple, of x, x+3 and 7, is: x*(x+3)*7 am I right?
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
is that equation is equivalent to this one? \[\Large 1 \times 7 \times \left( {x + 3} \right) + 7x = 2x\left( {x + 3} \right)\]
Michele_Laino
  • Michele_Laino
oops..is that equation equivalent to this one?
anonymous
  • anonymous
I guess..
Michele_Laino
  • Michele_Laino
more steps: \[\Large \frac{{1 \times 7\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}} + \frac{{1 \times 7x}}{{7x\left( {x + 3} \right)}} = \frac{{2x\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}}\]
Michele_Laino
  • Michele_Laino
now?
anonymous
  • anonymous
I'm lost...
Michele_Laino
  • Michele_Laino
as denominator, of each fraction, we find the least common multiple
anonymous
  • anonymous
Getting it.. Now?
Michele_Laino
  • Michele_Laino
now, if we consider the first fraction, for example, we have: |dw:1436624598950:dw|
Michele_Laino
  • Michele_Laino
similarly for the other 2 fractions
Michele_Laino
  • Michele_Laino
after that, we have to simplify the last equation, so we get: \[\Large 7x + 21 + 7x = 2{x^2} + 6x\] I have applied the distributive property of multiplication over addition Please rewrite that equation in this form: \[\Large bA{x^2} + Bx + C = 0\]
Michele_Laino
  • Michele_Laino
oops.. in this form: \[\Large A{x^2} + Bx + C = 0\]
Michele_Laino
  • Michele_Laino
what do you get?
anonymous
  • anonymous
It should be 7x^2, right?
Michele_Laino
  • Michele_Laino
hint: here are more steps: \[\Large \begin{gathered} 7x + 21 + 7x = 2{x^2} + 6x \hfill \\ 14x + 21 = 2{x^2} + 6x \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now I subtract 14 x from both sides so I can write: \[\Large 14x + 21 - 14x = 2{x^2} + 6x - 14x\] please simplify
Michele_Laino
  • Michele_Laino
hint: \[\Large \begin{gathered} 14x - 14x = 0 \hfill \\ 6x - 14x = - 8x \hfill \\ \end{gathered} \]
anonymous
  • anonymous
2x^2-8x-21=0
Michele_Laino
  • Michele_Laino
perfect!
Michele_Laino
  • Michele_Laino
now you have to solve that equation, using the standard formula: \[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\] where \[\Large \begin{gathered} a = 2, \hfill \\ b = - 8, \hfill \\ c = - 21. \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Um? I can't use this formula as it's npt stated in the question *not
Michele_Laino
  • Michele_Laino
have you studied that formula in your course of mathematics?
anonymous
  • anonymous
Yep.. But my teacher would add "correct to 2 or 3 decimals" ONLY THEN can I use this formula.
Michele_Laino
  • Michele_Laino
then we can use that formula, here is the next step: \[\Large \begin{gathered} x = \frac{{8 \pm \sqrt {{{\left( { - 8} \right)}^2} - 4 \times 2 \times \left( { - 21} \right)} }}{{2 \times 2}} = \hfill \\ \hfill \\ = \frac{{8 \pm \sqrt {64 + 168} }}{4} = ... \hfill \\ \end{gathered} \] please continue
anonymous
  • anonymous
I would get this marked wrong if I use the formula... Please understand....
Michele_Laino
  • Michele_Laino
ok! I understand
anonymous
  • anonymous
So I'll HAVE TO factor it out..
Michele_Laino
  • Michele_Laino
ok! I will write your next steps
Michele_Laino
  • Michele_Laino
If I divide both sides of your equation by 2, I get: \[\Large {x^2} - 4x - \frac{{21}}{2} = 0\]
Michele_Laino
  • Michele_Laino
Then I add and subtract 4 at left side, so I can write: \[\Large {x^2} - 4x + 4 - 4 - \frac{{21}}{2} = 0\]
Michele_Laino
  • Michele_Laino
is it ok, for you?
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
now we note that: \[\Large {x^2} - 4x + 4 = {\left( {x - 2} \right)^2}\]
Michele_Laino
  • Michele_Laino
so we can rewrite my last expression as follows: \[\Large {\left( {x - 2} \right)^2} - 4 - \frac{{21}}{2} = 0\]
Michele_Laino
  • Michele_Laino
and, being: \[\Large - 4 - \frac{{21}}{2} = \frac{{ - 8 - 21}}{2} = - \frac{{29}}{2}\]
Michele_Laino
  • Michele_Laino
we have: \[\Large {\left( {x - 2} \right)^2} - \frac{{29}}{2} = 0\]
Michele_Laino
  • Michele_Laino
now I add -29/2 to both sides of that equation, so I get: \[\Large \begin{gathered} {\left( {x - 2} \right)^2} - \frac{{29}}{2} + \frac{{29}}{2} = 0 + \frac{{29}}{2} \hfill \\ \hfill \\ {\left( {x - 2} \right)^2} = \frac{{29}}{2} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
ok! now I ask you, what is: \[\Large \sqrt {\frac{{29}}{2}} = ...?\] nearest to tenth?
anonymous
  • anonymous
3.8
anonymous
  • anonymous
wow! @Michele_Laino perfect explanation :)
Michele_Laino
  • Michele_Laino
that's right! Now if I take the square rooth of both sides of my last equation, I get: \[\Large x - 2 = \pm 3.8\] please remember we have 2 square roots
Michele_Laino
  • Michele_Laino
thanks!! @behappy
Michele_Laino
  • Michele_Laino
so we have 2 equations: \[\Large \begin{gathered} x - 2 = 3.8 \hfill \\ \hfill \\ x - 2 = - 3.8 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
5.8 and -1.8
Michele_Laino
  • Michele_Laino
yes! Nevertheless -1.8 can not be acceptable, since x represents a time. So only x=5.8 is the acceptable solution
Michele_Laino
  • Michele_Laino
finally the requested time are: \[\Large \begin{gathered} {t_1} = 5.8\min \hfill \\ {t_2} = 5.8 + 3 = ...\min \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
times*
anonymous
  • anonymous
8.8
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
THANKS!!!!!! CAN YOU PLEASE HELP ME MORE?!
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
thanks again for your appreciation to my work @behappy :)

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