anonymous
  • anonymous
((x-2)/(x-3))-((x+1)(x+3))+((x-9(/x^2-9))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I think the best way to do this is to just multiply out the first two terms, and for the third term, factor the denominator to (x-3)(x+3).
anonymous
  • anonymous
can you multiply it? I dont understand this. it looks like subtraction to me
anonymous
  • anonymous
oh sorry, I didn't see the / in the first term. Ok, for the second term, (x+1)(x+3), multiply this out Then for the third term, (x-9)/(x^2-9), factor the denominator

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anonymous
  • anonymous
uh oh i typed this out wrong its supp((x-2)/(x-3))-((x+1)/(x+3))+((x-9)/(x^2-9))osed to be
anonymous
  • anonymous
\[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2 -9}\] Recognize that \(x^2 - 9 = (x+3)(x-3)\) and put them all under a common denominator of \(x^2-9\)
anonymous
  • anonymous
Over a common denominator rather.
anonymous
  • anonymous
so i need to multiply the first two fractions by (x+3) or x-3?
anonymous
  • anonymous
Multiply the first by (x+3) on top and bottom, and the second by x-3 on top and bottom.
anonymous
  • anonymous
Oh, was the second term also a fraction? Remember when you learned how to add fractions with different denominators? Use the same logic. Multiply the first fraction by (x+3)/(x+3) and the second by (x-3)/(x-3) so that they all have a common denominator. Basically what polpak is saying but it is worth noting that this is just like what you do when you are adding normal fractions with different denominators.
anonymous
  • anonymous
Oh. I may have mistyped it. Was the middle term a fraction? If not it'd be \[\frac{x-2}{x-3} +(x+1)(x+3) +\frac{x-9}{x^2-9}\]
anonymous
  • anonymous
But you still want to get them over a common denominator of \(x^2-9\)
anonymous
  • anonymous
polpak, she posted that they were all fractions a few minutes ago. I missed it too.
anonymous
  • anonymous
you were correct the first time i mistyped it... thank you both for your help! its greatly appreciated
anonymous
  • anonymous
so i came up with 0(x+3)(x-3) is this correct? i somehow dont think it is
anonymous
  • anonymous
im sorry 0/(x-3)(x+3)
anonymous
  • anonymous
I got something else. Did you forget that the second fraction is being subtracted? I almost forgot that.
anonymous
  • anonymous
no i subtracted it but it all seemed to cancel out. what am i doing wrong?
anonymous
  • anonymous
\[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2-9} \] \[= \frac{(x-2)(x+3)}{x^2-9} - \frac{(x+1)(x-3)}{x^2-9} + \frac{x-9}{x^2-9} \] \[= \frac{[x^2+x-6] - [x^2-2x -3] + [x-9]}{x^2-9} \] \[= \frac{x^2-x^2+x + 2x + x -6+3 -9}{x^2-9} \] \[= \frac{4x-12}{x^2-9} \] \[= \frac{4(x-3)}{x^2-9} \] \[= \frac{4(x-3)}{(x+3)(x-3)} \] \[= \frac{4}{x+3} \]
anonymous
  • anonymous
Where'd you go wrong?
anonymous
  • anonymous
And hopefully you can follow that mess.
anonymous
  • anonymous
i didnt change the signs in the second fraction AND i can follow it
anonymous
  • anonymous
Thank you so much!
anonymous
  • anonymous
Cool, np =)
anonymous
  • anonymous
4/( x+3)

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