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Brad Baillod
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 Posted: Wed Jul 02, 2008 10:07 pm Post subject: Solve this algebraic equation? |
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Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
I'm ending up with a quadratic for x that's kind of hairy.
Thanks,
Brad |
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porky_pig_jr@my-deja.com
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| Posted: Wed Jul 02, 2008 10:29 pm Post subject: Re: Solve this algebraic equation? |
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On Jul 2, 6:07 pm, Brad Baillod <bradbail...@gmail.com> wrote:
| Quote: |
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
I'm ending up with a quadratic for x that's kind of hairy.
Thanks,
Brad
|
Solving for y, we get y = f(x) a rational function of a form y = P(x)/
Q(x), where
P(x) = (cx-a)(13-5x), Q(x) = 1-x, (x =/= 1, x =/= a/c),
I don't think you can get any simpler than that.
Is this exactly what the problem states? Express y in terms of x, a,
and c? |
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Mensanator
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| Posted: Wed Jul 02, 2008 10:29 pm Post subject: Re: Solve this algebraic equation? |
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On Jul 2, 5:07 pm, Brad Baillod <bradbail...@gmail.com> wrote:
| Quote: |
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
I'm ending up with a quadratic for x that's kind of hairy.
|
y = (cx-a)(5+(8/(1-x)))
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Brad
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| Posted: Thu Jul 03, 2008 1:35 am Post subject: Re: Solve this algebraic equation? |
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Thanks for the replies... I got it backwards. I want to solve for x
in terms of y, c, and a. I expect that is a little more difficult.
| Quote: |
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
I'm ending up with a quadratic for x that's kind of hairy.
y = (cx-a)(5+(8/(1-x)))
Thanks,
Brad |
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Arturo Magidin
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| Posted: Thu Jul 03, 2008 1:59 am Post subject: Re: Solve this algebraic equation? |
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In article <53900117-8547-4e3f-9b9d-247cf5c93386@v26g2000prm.googlegroups.com>,
Brad <bradbaillod@gmail.com> wrote:
Please don't top-post. If you don't know what that means, follow the links:
http://www.caliburn.nl/topposting.html
http://www.html-faq.com/etiquette/?toppost
| Quote: |
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
I'm ending up with a quadratic for x that's kind of hairy.
Thanks for the replies... I got it backwards. I want to solve for x
in terms of y, c, and a. I expect that is a little more difficult.
|
Don't know what is "kind of hairy" about a simple quadratic in x,
which is what you'll get. You have
5(1-x)+8 = y(1-x)/(cx-a)
5(1-x)(cx-a)+8(cx-a) = y(1-x)
5(-cx^2 + (a+c)x - a) + 8cx - 8a -y + yx = 0
(-5c)x^2 + (5(a+c)+8c +y)x + (-5a -8a - y) = 0.
(-5c)x^2 + (5a + 13c + y)x + (-13a - y) = 0.
If c =0, this is a linear equation on x, and you get
x = (13a+y)/(5a+y).
If c is not 0, then you can use the quadratic formula.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org |
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Brad
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| Posted: Thu Jul 03, 2008 2:43 am Post subject: Re: Solve this algebraic equation? |
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On Jul 2, 10:24 pm, Anon <a...@domain.invalid> wrote:
| Quote: |
On Wed, 2 Jul 2008 18:35:53 -0700 (PDT), Brad <bradbail...@gmail.com
wrote:
Thanks for the replies... I got it backwards. I want to solve for x
in terms of y, c, and a. I expect that is a little more difficult.
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
If c = 0, x = (y + 13a) / (y + 5a), otherwise
x = 1/(10 c) [ y + 5a + 13c +- sqrt[(y + 5a + 13c)^2 - 20c(y + 13a)] ]
where +- means 'plus or minus'.
I'm ending up with a quadratic for x that's kind of hairy.
y = (cx-a)(5+(8/(1-x)))
Thanks,
Brad
|
Thanks all for the replies... the quadratic formula had completely
slipped my mind. |
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Anon
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| Posted: Thu Jul 03, 2008 6:55 am Post subject: Re: Solve this algebraic equation? |
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On Wed, 02 Jul 2008 18:07:14 EDT, Brad Baillod <bradbaillod@gmail.com>
wrote:
| Quote: |
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
|
y / (c x - a) = 5 + 8/(1 - x)
= (5 - 5x +  / (1 - x)
= (13 - 5x) / (1 - x)
Multiply both sides by (c x - a), giving
y = (c x - a) (13 - 5x) / (1 - x) or equivalently
y = (5x - 13) (c x - a) / (x - 1)
If you want the numerator in expanded form,
y = [ 5 c x^2 - (5a + 13c)x + 13a ] / (x - 1)
| Quote: |
I'm ending up with a quadratic for x that's kind of hairy.
Thanks,
Brad |
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Anon
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| Posted: Thu Jul 03, 2008 7:24 am Post subject: Re: Solve this algebraic equation? |
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On Wed, 2 Jul 2008 18:35:53 -0700 (PDT), Brad <bradbaillod@gmail.com>
wrote:
| Quote: |
Thanks for the replies... I got it backwards. I want to solve for x
in terms of y, c, and a. I expect that is a little more difficult.
Hi,
I wonder if any of you could help solve this equation--not for values--but to express y in terms of x, c, and a. C and a are constants.
5 + (8/(1-x)) = y / (cx - a)
where 'cx' means c times x
|
If c = 0, x = (y + 13a) / (y + 5a), otherwise
x = 1/(10 c) [ y + 5a + 13c +- sqrt[(y + 5a + 13c)^2 - 20c(y + 13a)] ]
where +- means 'plus or minus'.
| Quote: |
I'm ending up with a quadratic for x that's kind of hairy.
y = (cx-a)(5+(8/(1-x)))
Thanks,
Brad |
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