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passage.livejournal.com - Solving the maths
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Solving the maths
Date: 2001-06-07 05:54 am (UTC)From:which gives n-p = 2y/x
taking the reciprocal: x/2y =1/(n-p)
To give: x=2y/(n-p)
Which is not hard
did you infact mean y/x + y/(x+n) =p
in which case y(2x+n)/x(x+n)=p (common denomiator)
Multiplying through by x(x+n)/y and exanding the bracket gives: 2x+n = px^2/y + npx/y
=> px^2/y+(npx/y -2)x-n = 0
Which looks nicer as:
px^2 + (npx-2y)x -n =0
Now, if you have numbers sub them in as it may facorise, else you'll have to solve that using quadratic formular (or completing the sqaure). Neith or which are too taxing, can you figure out the rest from here?
Neil
P.S.: If it was the first feel suitably embarressed, if it was the second then that is surprisingly hard for a GCSE paper, I could find A level students who couldn't do that (on the other thadn I know some throughly incometant A level mathematicians).
I have a question ...
Date: 2001-06-07 06:01 am (UTC)From:Neil
Re: I have a question ...
Date: 2001-06-07 11:15 am (UTC)From:Re: I have a question ...
Date: 2001-06-08 06:55 am (UTC)From:Re: I have a question ...
Date: 2001-06-08 12:04 pm (UTC)From:Re: Solving the maths
Date: 2001-06-09 02:27 am (UTC)From:Re: I have a question ...
Date: 2001-06-10 01:59 am (UTC)From:Im honoured
*hugs*
Re: Solving the maths
Date: 2001-06-14 07:14 am (UTC)From:Neil, who can't really talk being as he's just discovered he can't do maths.