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anonymous

  • 5 years ago

((x-2)/(x-3))-((x+1)(x+3))+((x-9(/x^2-9))

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  1. anonymous
    • 5 years ago
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    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).

  2. anonymous
    • 5 years ago
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    can you multiply it? I dont understand this. it looks like subtraction to me

  3. anonymous
    • 5 years ago
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    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

  4. anonymous
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    \[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2 -9}\] Recognize that \(x^2 - 9 = (x+3)(x-3)\) and put them all under a common denominator of \(x^2-9\)

  6. anonymous
    • 5 years ago
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    Over a common denominator rather.

  7. anonymous
    • 5 years ago
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    so i need to multiply the first two fractions by (x+3) or x-3?

  8. anonymous
    • 5 years ago
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    Multiply the first by (x+3) on top and bottom, and the second by x-3 on top and bottom.

  9. anonymous
    • 5 years ago
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    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.

  10. anonymous
    • 5 years ago
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    Oh. I may have mistyped it. Was the middle term a fraction? If not it'd be \[\frac{x-2}{x-3} +(x+1)(x+3) +\frac{x-9}{x^2-9}\]

  11. anonymous
    • 5 years ago
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    But you still want to get them over a common denominator of \(x^2-9\)

  12. anonymous
    • 5 years ago
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    polpak, she posted that they were all fractions a few minutes ago. I missed it too.

  13. anonymous
    • 5 years ago
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    you were correct the first time i mistyped it... thank you both for your help! its greatly appreciated

  14. anonymous
    • 5 years ago
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    so i came up with 0(x+3)(x-3) is this correct? i somehow dont think it is

  15. anonymous
    • 5 years ago
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    im sorry 0/(x-3)(x+3)

  16. anonymous
    • 5 years ago
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    I got something else. Did you forget that the second fraction is being subtracted? I almost forgot that.

  17. anonymous
    • 5 years ago
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    no i subtracted it but it all seemed to cancel out. what am i doing wrong?

  18. anonymous
    • 5 years ago
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    \[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2-9} \] \[= \frac{(x-2)(x+3)}{x^2-9} - \frac{(x+1)(x-3)}{x^2-9} + \frac{x-9}{x^2-9} \] \[= \frac{[x^2+x-6] - [x^2-2x -3] + [x-9]}{x^2-9} \] \[= \frac{x^2-x^2+x + 2x + x -6+3 -9}{x^2-9} \] \[= \frac{4x-12}{x^2-9} \] \[= \frac{4(x-3)}{x^2-9} \] \[= \frac{4(x-3)}{(x+3)(x-3)} \] \[= \frac{4}{x+3} \]

  19. anonymous
    • 5 years ago
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    Where'd you go wrong?

  20. anonymous
    • 5 years ago
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    And hopefully you can follow that mess.

  21. anonymous
    • 5 years ago
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    i didnt change the signs in the second fraction AND i can follow it

  22. anonymous
    • 5 years ago
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    Thank you so much!

  23. anonymous
    • 5 years ago
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    Cool, np =)

  24. anonymous
    • 5 years ago
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    4/( x+3)

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