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anonymous

  • 5 years ago

x+2 / x^2 -9 + x-3/x^2-2x-3 = 2x-3/ x^2+4x+3

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  1. anonymous
    • 5 years ago
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    what about it?

  2. anonymous
    • 5 years ago
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    algebra , i dont know how to solve it.

  3. anonymous
    • 5 years ago
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    kathyworld, it is not clear how the equation is formed. Can you please put in some brackets or parentheses to make the equation better?

  4. anonymous
    • 5 years ago
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    [ x+2 / x^2 -9] + [x-3/x^2-2x-3] = 2x-3/ x^2+4x+3

  5. anonymous
    • 5 years ago
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    ^ = power

  6. anonymous
    • 5 years ago
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    I assume it looks like this: [(x+2 )/( x^2 -9)] + [(x-3)/(x^2-2x-3)] = (2x-3)/ (x^2+4x+3)

  7. anonymous
    • 5 years ago
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    yes perfect :)

  8. anonymous
    • 5 years ago
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    ok, thanks

  9. anonymous
    • 5 years ago
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    i just dont understand how to solve it

  10. anonymous
    • 5 years ago
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    [(x+2) / (x-3)(x+3)] + [(x-3) / (x-3)(x+1)] = (2x-3)/(x+3)(x+1)

  11. anonymous
    • 5 years ago
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    is that the answer ?

  12. anonymous
    • 5 years ago
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    [(x+2 )/( x^2 -9)] + [(x-3)/(x^2-2x-3)] = (2x-3)/ (x^2+4x+3) [(x+2 )/{( x-3)(x+3)}] + [(x-3)/{(x-3)(x+1)}] = (2x-3)/ {(x+3)(x+1)} So the monomials appear in the denominator are x-3, x+3 and x+1. Make all the fractions have the same denominator: [(x+2 )/{( x-3)(x+3)}] + [(x-3)/{(x-3)(x+1)}] = (2x-3)/ {(x+3)(x+1)} [{(x+2 )(x+1)}/{( x-3)(x+3)(x+1)}] + [{(x-3)(x+3)}/{(x-3)(x+1)(x+3)}] = {(2x-3)(x-1)}/ {(x+3)(x+1)(x-1)} So the numerator we have (forget for a while the denominator) (x+2)(x+1) + (x-3)(x+3) = (2x-3)(x-1) (x^2 + 3x +2) + (x^2 - 9) = 2x^2 - 5x +3 8x = 10 x = 5/4

  13. anonymous
    • 5 years ago
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    wow you did the whole thing... I would've let vic23 do the basic algebric manipulations...

  14. anonymous
    • 5 years ago
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    THANK YOU VERY MUCH SKLEE ! :)

  15. anonymous
    • 5 years ago
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    Ya, I should leave some part for kathy to complete it ^_^

  16. anonymous
    • 5 years ago
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    sorry vic23 is not the guy who asked the question. My apologies. Let me correct myself. I would've let kathyworld09 do the basic algebric manipulations

  17. anonymous
    • 5 years ago
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    you're welcome kathy. Hope you learn something and understand it. I am sure you can do it next time :)

  18. anonymous
    • 5 years ago
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    so its not fully completed ?

  19. anonymous
    • 5 years ago
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    kathy, i have given the complete solution. :)

  20. anonymous
    • 5 years ago
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    You should learn and understand how it is done, so that you can do it again in other problem next time

  21. anonymous
    • 5 years ago
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    oh thank you , i just am having difficulty in a similar problem though

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