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 Post subject: Honors Algebra 2 Solving rational equationsPosted: Fri, 6 May 2011 02:56:13 UTC

Joined: Thu, 5 May 2011 00:39:23 UTC
Posts: 6

(1/x^2)-x^2
___________= 3/2

(1/x)+x

The teacher gave us a hint, but I still do not understand how to solve this problem.
(Hint: You may need to use factoring by division to solve this.)

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 Post subject: Re: Honors Algebra 2 Solving rational equationsPosted: Fri, 6 May 2011 02:58:04 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12098
Location: Austin, TX
Bloodprincess15 wrote:

(1/x^2)-x^2
___________= 3/2

(1/x)+x

The teacher gave us a hint, but I still do not understand how to solve this problem.
(Hint: You may need to use factoring by division to solve this.)

Clear denominators then solve the resulting equation which is a polynomial.

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 Post subject: Posted: Fri, 6 May 2011 03:04:09 UTC

Joined: Thu, 5 May 2011 00:39:23 UTC
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what do you multiply the equation by to clear the denominators?

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 Post subject: Posted: Fri, 6 May 2011 03:05:38 UTC
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Bloodprincess15 wrote:
what do you multiply the equation by to clear the denominators?

x.

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 Post subject: Posted: Fri, 6 May 2011 03:07:08 UTC
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Joined: Wed, 1 Oct 2003 04:45:43 UTC
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x^2 actually.

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 Post subject: Posted: Fri, 6 May 2011 03:09:55 UTC
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Matt wrote:
x^2 actually.

Ah yes, silly numerator fractions.

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 Post subject: Posted: Fri, 6 May 2011 03:12:22 UTC

Joined: Thu, 5 May 2011 00:39:23 UTC
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So then would you get:

1-x^4
_____=3x^2/2x^2
x+x^3

Or did I multiply wrong?

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 Post subject: Posted: Fri, 6 May 2011 03:15:12 UTC
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Bloodprincess15 wrote:
So then would you get:

1-x^4
_____=3x^2/2x^2
x+x^3

Or did I multiply wrong?

You didn't multiply the right side by x^2, you multiplied it by x^2/x^2, which is not the same thing.

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 Post subject: Posted: Fri, 6 May 2011 03:21:13 UTC

Joined: Thu, 5 May 2011 00:39:23 UTC
Posts: 6
so then it would be:

....=3x^2/2

Right?

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 Post subject: Posted: Fri, 6 May 2011 03:35:52 UTC
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Multiplying by x^2 on the left-hand side creates an equivalent fraction.
This has nothing to do with the right-hand side. The result is:

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 Post subject: Posted: Fri, 6 May 2011 10:56:15 UTC
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Joined: Sun, 24 Jul 2005 20:12:39 UTC
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Location: Ottawa Ontario
Princess, the "left side multiplication by x^2" means the numerator
AND denominator are multiplied; so that's really a multiplication by
x^2 / x^2 which equals 1, so no need to adjust right side...OK?

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 Post subject: Re: Honors Algebra 2 Solving rational equationsPosted: Fri, 6 May 2011 14:02:31 UTC
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Location: Lexington, MA
Hello, Bloodprincess15!

Another approach . . .

Quote:
Solve: .

Note that the numerator is a difference-of-squares.

Reduce: . . . (Note that: for real )

Multiply by

Therefore: .

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 Post subject: Posted: Wed, 11 May 2011 19:05:43 UTC
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I would have first cross multiplied, and then multiplied both sides by x^2.

Just another way to accomplish the same result.

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 Post subject: Posted: Thu, 12 May 2011 07:37:29 UTC
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Mr. Mathturo wrote:
I would have first cross multiplied, .........

YA!
Take that, TKHunny

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