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Math Help - Dividing Rational expressions

  1. #1
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    Dividing Rational expressions

    The book says the answer is supposed to be x+2 for this question with the restrictions of x does not equal -2, \frac12

    I couldn't get that

    The question is:

    \frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9}
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by t-dot View Post
    The book says the answer is supposed to be x+2 for this question with the restrictions of x does not equal -2, \frac12

    I couldn't get that

    The question is:

    \frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9}
    \frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9} = \frac{2x^2-8}{6x+3} \times \frac{18x+9}{6x-12}

    = \frac{2x^2-8}{6x+3} \times \frac{3(6x+3)}{6x-12}

    = \frac{3(2x^2-8)}{6x - 12}

    = \frac{6(x^2-4)}{6(x - 2)}

    = \frac{6(x - 2)(x + 2)}{6(x - 2)}

    = (x + 2)
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  3. #3
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    Hello, t-dot!

    \frac{2x^2-8}{6x+3} \div \frac{6x-12}{18x+9}\;\; {\color{blue}=\;\;\frac{2x^2-8}{6x+3} \times \frac{18x+9}{6x-12}}

    Book's answer: . x+2
    Factor all the expressions:

    . . 2x^2-8\:=\:2(x^2-4)\:=\:2(x-2)(x+2)
    . . 6x+3\:=\:3(2x+1)
    . . 6x-12 \:=\:6(x-2)
    . . 18x + 9 \:=\:9(2x+1)

    The problem becomes: . \frac{\not{2}(x-2)(x+2)}{\not3(2x+1)} \times \frac{\not9(2x+1)}{\not6(x-2)} \;\;=\;\;x + 2

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  4. #4
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    thanks guys...
    i just kept doing my math wrong
    that was my problem
    lol
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