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Math Help - Solving an equation where the unknown is included in fractions

  1. #1
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    Solving an equation where the unknown is included in fractions

    How do you do this equation if someone can please explain this step by step
    2x+5 -x-4 =1
    x+4 x-5
    Last edited by mr fantastic; November 7th 2009 at 01:12 PM. Reason: Changed post title
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skml02 View Post
    How do you do this equation if someone can please explain this step by step
    2x+5 -x-4 =1
    x+4 x-5
    Step 1

    multiply both sides by th common denominator

    Step b

    Set the equation to zero

    Step c_3

    Let me know when you're done with steb b.
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  3. #3
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    so I got 2x+5 over 1
    and 4x-16 over x-5 = 4
    but I don't get set the equation to zero
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skml02 View Post
    so I got 2x+5 over 1
    and 4x-16 over x-5 = 4
    but I don't get set the equation to zero
    The common denominator is (x+4)(x-5), so multiplying both sides by this gives:

    (x+4)(x-5)\left[\frac{2x+5}{x+4}-\frac{x-4}{x-5}\right]=1(x+4)(x-5)

    (2x+5)(x-5)-(x-4)(x+4)=(x+4)(x-5)

    2x^2-5x-25-x^2+16=x^2-x-20

    Oh, this is easier than I previously thought...The squared terms dissapear...

    -5x-9=-x-20

    4x=11

    x=\frac{11}{4}
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