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Math Help - Can you tell me if i solved this right

  1. #1
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    Can you tell me if i solved this right

    Got one more question; is this correct? can it be simplified further

    I just uploaded a picture quick because im not so good at writing equations outside of word

    http://img580.imageshack.us/img580/9407/b3ku.png

    The first line is the question. It justs says ' Solve'.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    You got this far:

    \frac {x^2-3x-2}{(x-2)(x-4)} = \frac {2}{(x-2)(x-4)}

    Since the denminators are the same, you can simply multiply both sides by (x-2)(x-4) to get

    x^2 -3x -2 = 2

    Now rearrange and factor to solve for x. One thing to be aware of however: note that because you had factors (x-2) and (x-4) in the denominators to start with, x cannot be equal to either 2 or 4.
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    Re: Can you tell me if i solved this right

    Thank you for bearing with me and my terrible looking equations. If I may ask, what are you using to make your equations so clean?

    x2 -3x -2 -2 = 2 -2

    x2- 3x - 4 = 0

    (x-4) (x+1) ?
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    OK, so far so good. So what values for x do you get?

    As for equation formatting we use a tool called "LaTeX." It uses commands between start and end delimiters of [ tex] and [/tex]. Inside those delimiters the symbol "^" specifies "raised to the value of" and functions with a leading backslash and arguments within pairs of {} squiggly brackets such as \frac {xx}{yy} and \sqrt{abc} yield  \frac {xx}{yy} and  \sqrt {abc}. Here's a good introductory explanation: mimetextutorial.html
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  5. #5
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    Re: Can you tell me if i solved this right

    Ooh that will come in handy thank you.

    I'm not sure how to go about finding the values for x is there some kind of equation I should be following to get the answer?

    Is it: 'The values you get for x are: x= 4. -1.' ?
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    Don't forget what I said about x = 4 back in post #2! As a matter of course you should always double-check your math, in this case by verifying that if x = -1 and x = 4 the original equation is satisfied. So - what happens if you substitute x = 4 into the original equation?
    Last edited by ebaines; July 8th 2013 at 12:50 PM.
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    Re: Can you tell me if i solved this right

    Quote Originally Posted by ebaines View Post
    Don't forget what I said about x = 4 back in post #2! As a matter of course you should always double-check your math, in this case by verifying that if x = -1 and x = 4 the original equation is satisfied. So - what happens if you substitute x = 4 into the original equation?
    Yes yes thats right my mistake. By original equation do you mean the very first one or x2-3x-4?
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  8. #8
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    The original equation is what they gave you to work with:

     \frac x {x-2} + \frac 1 {x-4} = \frac 2 {x^2-6x+8}
    Last edited by ebaines; July 8th 2013 at 01:06 PM.
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  9. #9
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    Re: Can you tell me if i solved this right

    Sorry about the late reply... we had a flood here in Toronto, power went out.

    Im pretty sure I made a mistake

    but


    if I sub the x=4 in the original equation I'm going to get

    1
    __
    (4) -4

    and I can't divide by zero right..?
    Last edited by Urpes; July 9th 2013 at 07:20 AM.
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  10. #10
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    Quote Originally Posted by Urpes View Post
    and I can't divide by zero right..?
    Right. So x=4 is NOT a solution to the original equation.
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    Re: Can you tell me if i solved this right

    Quote Originally Posted by ebaines View Post
    Right. So x=4 is NOT a solution to the original equation.
    Now I'm completely lost so I can't use 2, or 4. What do I do to solve it then? Sub in -1?
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  12. #12
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    Quote Originally Posted by Urpes View Post
    Now I'm completely lost so I can't use 2, or 4. What do I do to solve it then? Sub in -1?
    Back in post#5 you came up with two possible answers: 4 and -1. You have now ruled out 4 as a possibility. That leaves -1 as a possible answer. So - does it work in the original equation?
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  13. #13
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    Re: Can you tell me if i solved this right

    Not when I put it in it doesn't.

    (-1) 1 2
    ____ + ____ = ______
    (-1)-2 (-1)-4 (-1)2-6x+8

    -1 1 2
    ___ + ___ = __
    -3 -5 15
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  14. #14
    MHF Contributor ebaines's Avatar
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    Re: Can you tell me if i solved this right

    Substituting -1 for x, for the left hand side you have:

     \frac {-1}{-3} + \frac 1 {-5} = \frac 5 {15} + \frac {-3}{15} =  \frac 2 {15}

    And for the right hand side:

     \frac 2 {(-1)^2 - 6(-1) + 8} = \frac 2 {1 + 6 + 8} = \frac 2 {15}

    Looks good to me!
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  15. #15
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    Re: Can you tell me if i solved this right

    Quote Originally Posted by ebaines View Post
    Substituting -1 for x, for the left hand side you have:

     \frac {-1}{-3} + \frac 1 {-5} = \frac 5 {15} + \frac {-3}{15} =  \frac 2 {15}

    And for the right hand side:

     \frac 2 {(-1)^2 - 6(-1) + 8} = \frac 2 {1 + 6 + 8} = \frac 2 {15}

    Looks good to me!
    All of that makes sense to me except for the  \frac 5 {15} + \frac {-3}{15}  part

    Where did those two come from?
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