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Math Help - Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

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    Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    Please Help with this problem (x-1)/3 = (x+5)/(x-1)
    Last edited by skeeter; May 14th 2012 at 07:26 AM. Reason: place the problem within the post, not in the title.
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    Re: Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    Quote Originally Posted by elvinas View Post
    Once again please help me.
    \displaystyle \begin{align*} \frac{x - 1}{3} &= \frac{x + 5}{x - 1} \\ (x - 1)^2 &= 3(x + 5) \\ x^2 - 2x + 1 &= 3x + 15 \\ x^2 - 5x - 14 &= 0 \\ (x + 2)(x - 7) &= 0 \\ x + 2 = 0 \textrm{ or } x - 7 &= 0 \\ x = -2 \textrm{ or }x &= 7 \end{align*}
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    Re: Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    Thank You very much but in Line 3 isnt 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0
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    Re: Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    Thank You very much but in Line 3 isnt 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0
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    Re: Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    Quote Originally Posted by elvinas View Post
    Thank You very much but in Line 3 isnt 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0
    Definitely not. When moving everything to the LHS, we need to subtract 3x from both sides. -2x - 3x = -5x.
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    Re: Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    Thank You very much but in Line 3 isn't 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0
    (Sorry about multiple posts)
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    Re: Please Help with this problem (x-1)/3 = (x+5)/(x-1) ???????

    From x^2- 2x+ 1= 3x+ 15 you subtract 3x from both sides:
    x^2- 2x+ 1- 3x= 3x+ 15- 3x

    x^2- 5x+ 1= 15

    and now you can subtract 15 from both sides:
    x^2- 5x+ 1- 15= 15- 15
    x^2- 5x- 14= 0
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