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anonymous

  • 4 years ago

How to subtract a fraction with another fraction?

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

  2. anonymous
    • 4 years ago
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    what is x?

  3. anonymous
    • 4 years ago
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    this question is solved yours other subtittle.

  4. anonymous
    • 4 years ago
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    we say x=0

  5. anonymous
    • 4 years ago
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    lcd = 3(x^2 +2x-3) = 3(x-1)(x+3) multiplying through by this gives: 9(x+3) - 3(2x+10) = (x-1)(x+3) 9x +27 - 6x - 30 = x^2 +2x - 3 x^2 +x = 0 x(x+1)=0 x = 0 or x = -1

  6. anonymous
    • 4 years ago
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    yeah cnsu90, thanks for ur help :)

  7. anonymous
    • 4 years ago
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    oops look like an error in last two steps! 9x +27 - 6x - 30 = x^2 +2x - 3 -x^2 +x = 0 x (1 - x) = 0 x = 0 or x = 1

  8. anonymous
    • 4 years ago
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    hmm..

  9. anonymous
    • 4 years ago
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    @jimmy..denominator cant be 0..so u cant say x=1..

  10. anonymous
    • 4 years ago
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    yes - you are right

  11. anonymous
    • 4 years ago
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    jimmy, whats lcd? lowest common denominator?

  12. anonymous
    • 4 years ago
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    yes

  13. anonymous
    • 4 years ago
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    this is given me a real headache - cant understand how i got erroneous root

  14. anonymous
    • 4 years ago
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    the anwser is x=0

  15. anonymous
    • 4 years ago
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    yes - i know thats correct - i just trying to find out why my method also gave x=1

  16. anonymous
    • 4 years ago
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    your solution is true jimmy.but then you must try your root in function.

  17. anonymous
    • 4 years ago
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    there must be some reason for this

  18. anonymous
    • 4 years ago
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    at x=1 function is undefined.

  19. anonymous
    • 4 years ago
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    yea i know

  20. anonymous
    • 4 years ago
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    - i just dont like mysteries...

  21. anonymous
    • 4 years ago
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    ahh - i'd forgot a very important fact about multiplying the whole equation by the lcd!! - this will sometimes give extraneous roots - when x = 1 you are effectively multiplying by zero. you guys solved it differently - you simplified the lhs and left rhs alone - thats why you didn't get the extra root

  22. anonymous
    • 4 years ago
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    its ok when you multiply by a constant - it only applies when you you multiply by a variable

  23. anonymous
    • 4 years ago
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    hey jimmy, so which method is better? :)

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