y/x + y/x+n = p is the general form of the annoying equation thing from the exam - knows are y, n and p, so x needs to be found in terms of y, n and p.
I suspect you wrote that wrong because it simplifies to 2y/x + n = p 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).
*sigh* There was a time I'd have been able to do that. (Well, I could probably still have done the bit you went through, but I haven't used the quadratic formula or completed the square since my A Level days, over 7 years ago...)
Actualy she had numbers which gave something that factorised nicely so you wouldn't have needed the knowledge (really poilt the question that, I was quite impressed with it for a while).
Neil, who can't really talk being as he's just discovered he can't do maths.
Well, I only wanted to post the equation in lieu of writing it down and so you could see it, so I didn't feel the need to bore the entire world with it.
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).
Re: Solving the maths
Date: 2001-06-09 02:27 am (UTC)From: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.
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: I have a question ...
Date: 2001-06-10 01:59 am (UTC)From:Im honoured
*hugs*