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Math Help - Having Trouble Solveing Equalities with Variables on Each Side

  1. #1
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    Question Having Trouble Solveing Equalities with Variables on Each Side

    EDIT: One more problem I'm having with fuctions...

    solve:
    f(2/3) = (x^2+3)/(x-5)

    having a hard time here
    Last edited by emximer; September 18th 2008 at 05:02 PM.
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  2. #2
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    I'll do the second rational equation ... you try the first one after I show you how ...

    \frac{x+10}{x+3} = \frac{x+8}{x+5}

    cross multiply ...

    (x+10)(x+5) = (x+3)(x+8)

    expand (multiply out) both sides ...

    x^2 + 15x + 50 = x^2 + 11x + 24

    combine like terms ...

    4x = -26

    x = -\frac{13}{2}


    here's the first one ... I'll leave the easier one for you.

    \sqrt{x^2 + 16} = 3+x

    square both sides ...

    x^2 + 16 = 9 + 6x + x^2

    combine like terms ...

    7 = 6x

    \frac{7}{6} = x

    important! ... check the solution

    \sqrt{\frac{49}{36} + 16} = 3 + \frac{7}{6}<br />

    \sqrt{\frac{625}{36}} = \frac{25}{6}<br />

    \frac{25}{6} = \frac{25}{6}

    checks good.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    I'll do the second rational equation ... you try the first one after I show you how ...

    \frac{x+10}{x+3} = \frac{x+8}{x+5}

    cross multiply ...

    (x+10)(x+5) = (x+3)(x+8)

    expand (multiply out) both sides ...

    x^2 + 15x + 50 = x^2 + 11x + 24

    combine like terms ...

    4x = -26

    x = -\frac{13}{2}


    here's the first one ... I'll leave the easier one for you.

    \sqrt{x^2 + 16} = 3+x

    square both sides ...

    x^2 + 16 = 9 + 6x + x^2

    combine like terms ...

    7 = 6x

    \frac{7}{6} = x

    important! ... check the solution

    \sqrt{\frac{49}{36} + 16} = 3 + \frac{7}{6}<br />

    \sqrt{\frac{625}{36}} = \frac{25}{6}<br />

    \frac{25}{6} = \frac{25}{6}

    checks good.

    Thanks a bunch! I knew it would be somewhat easy once I got a little help.

    Got one last problem I'm stuck on with my homework so I edited my post rather than make another thread.
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  4. #4
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    Quote Originally Posted by emximer View Post
    EDIT: One more problem I'm having with fuctions...

    solve:
    f(2/3) = (x^2+3)/(x-5)

    having a hard time here
    Is the question to find f(2/3) given that f(x) = \frac{x^2 + 3}{x-5}? Because if it is, all you have to do is substitute 2/3 into that expression for f(x).

    Or is the question to solve the equation \frac{2}{3} = \frac{x^2 + 3}{x-5}?
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    Is the question to find f(2/3) given that f(x) = \frac{x^2 + 3}{x-5}? Because if it is, all you have to do is substitute 2/3 into that expression for f(x).

    Or is the question to solve the equation \frac{2}{3} = \frac{x^2 + 3}{x-5}?
    you have to substitute 2/3 in and solve, for some reason I can't get the right answer though. These problems are for WebWorks, so you submit and answer and it tells you if you're right.
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  6. #6
    Super Member 11rdc11's Avatar
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    \frac{(\frac{2}{3})^2 + 3}{\frac{2}{3}-5}

    \frac{(\frac{4}{9}) + 3}{\frac{2}{3}-5}

    \frac{\frac{4+27}{9}}{\frac{2-15}{3}}

    \frac{\frac{31}{9}}{\frac{-13}{3}}

    \frac{(3)(31)}{(9)(-13)}

    -\frac{31}{39}
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