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anonymous

  • one year ago

Solve and see whether or not it is extraneous 4/x^2+3x-10 -1/x^2+x-6 = 3/x^2-x-12

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  1. anonymous
    • one year ago
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    @satellite73

  2. anonymous
    • one year ago
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    now it gets real annoying real fast factor the denominators first i suck at factoring so let me know what you get

  3. anonymous
    • one year ago
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    Alright hold on lol:)

  4. anonymous
    • one year ago
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    4/(x-2)(x+5) -1/(x-2)(x+3)= 3(x-4)(x+3) @satellite73 done :)

  5. anonymous
    • one year ago
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    whew ok

  6. anonymous
    • one year ago
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    then you can clear the fractions as before but this time you have to multiply by the least common multiple of the denominators which is \[(x-2)(x+5)(x+3)(x-4)\] ugh

  7. anonymous
    • one year ago
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    oh..oh gosh...headachee

  8. anonymous
    • one year ago
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    actually it is not that bad, since the numerators are all numbers

  9. anonymous
    • one year ago
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    \[4(x+3)(x-4)-1(x+5)(x-4)=3(x+5)(x-2)\] i think check it

  10. anonymous
    • one year ago
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    ok hold on:)

  11. anonymous
    • one year ago
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    it is really just cancelling what you need to cancel

  12. anonymous
    • one year ago
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    yes, thats true

  13. anonymous
    • one year ago
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    then we have a raft of algebra to do multiply all this muck out, combine like terms etc etc i refuse to do it, i would cheat

  14. anonymous
    • one year ago
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    Haha! I totally agree:) Ok, lets move onto another one

  15. anonymous
    • one year ago
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    lol actually all the \(x^2\) terms go bye bye and you end up with only \(14x=2\) so \(x=\frac{1}{7}\)

  16. anonymous
    • one year ago
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    here is how to cheat http://www.wolframalpha.com/input/?i=4%28x%2B3%29%28x-4%29-1%28x%2B5%29%28x-4%29%3D3%28x%2B5%29%28x-2%29

  17. anonymous
    • one year ago
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    and since \(\frac{1}{7}\) does not make any denominator zero, it is NOT extraneous

  18. anonymous
    • one year ago
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    Marry me? Lol jk! Thanks so much :)

  19. anonymous
    • one year ago
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    yw

  20. anonymous
    • one year ago
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    and sure why not? it can be an OS wedding

  21. anonymous
    • one year ago
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    Perfect! I'll start planning :D

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