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#1

$Old$ April 22nd, 2007, 04:52 PM

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where do you start with this problem

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#2

$Old$ April 22nd, 2007, 05:17 PM

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where do you start with this problem

Here

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#3

$Old$ April 22nd, 2007, 06:09 PM

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Yes, the equation does not has solution, that is all

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Last edited by Krizalid; June 25th, 2007 at 04:41 AM.

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#4

$Old$ June 24th, 2007, 11:14 PM

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in retrospect, i'm thinking we could divide by z - 9 to get the equation z - 7 = 2 and avoid that annoying quadratic. apparently when i did this problem i was concerned about z - 9 being 0, but now i see that that can't be the case, since the original fractions wouldn't be valid in the first place

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$Old$ June 25th, 2007, 06:43 AM

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Yeah i agree, it simply doesnt have a solution.

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#6

$Old$ June 25th, 2007, 09:08 AM

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Yeah i agree, it simply doesnt have a solution.

right, because z can't be 9

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#7

$Old$ June 25th, 2007, 09:26 AM

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$z\neq 9$ .
Then $\displaystyle \frac{z-7}{z-9}=\frac{2}{z-9}\Leftrightarrow \frac{z-7}{z-9}-\frac{2}{z-9}=0\Leftrightarrow \frac{z-9}{z-9}=0\Leftrightarrow 1=0$ , false!

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