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    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #1

    Feb 28, 2007, 08:16 AM
    An equation
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Feb 28, 2007, 08:28 AM
    I don't understand what you're trying to do? Are the top 2 equations simultaneous?

    PS. You may want to learn how to use the math notation we have installed here, a picture is not particularly friendly :)
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #3

    Feb 28, 2007, 10:37 AM
    What are you talking about?:confused: lol

    Did it in latex now, I tried to do it before, but forgot to add the math tags lol.
    Oh and yeah, they're supposed to be simultaneous.
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #4

    Feb 28, 2007, 11:01 AM
    your mistake was that your equation for x still contained x on the RHS.
    Try this:


    I haven't worked it through from this point but I suspect you may be in better shape using this equation for x.
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #5

    Feb 28, 2007, 11:05 AM
    Quote Originally Posted by asterisk_man
    your mistake was that your equation for x still contained x on the RHS.
    Try this:


    I haven't worked it through from this point but I suspect you may be in better shape using this equation for x.
    Thanks a lot, I think I'll be able to do it now :)
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #6

    Feb 28, 2007, 11:07 AM
    Much better tex, although you need a little more practice it seems ;)
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #7

    Feb 28, 2007, 11:50 AM
    Quote Originally Posted by Capuchin
    Much better tex, although you need a little more practice it seems ;)
    Well, let's keep practicing then :D
    Quote Originally Posted by kristo
    Thanks alot, I think I'll be able to do it now
    Or not.. lol

    I tried it your way asterisk, but got stuck again.
    Got this far:
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #8

    Feb 28, 2007, 12:23 PM
    Well, I can tell you a few things.
    First, your final y^4 expression is correct, you didn't have any errors getting to that point.
    Second, I'm not sure how we're supposed to solve a 4th order polynomial. This is typically a hard problem to work by hand.
    Finally, I do know that the 4 answers to this are:
    y={4, 12, +sqrt(55)-5,-sqrt(55)-5}
    you can figure out the associated x values.
    I don't think there is a nice x result for the last two y values
    I didn't work the answer by hand so I can't take credit for it but I have verified that they are correct.
    the equation factors into (y-4)(y-12)(y^2+10y-30)=0 but I don't know how you're supposed to get there. I know that the possible integer roots are factors of the constant but the factors of 1440 are 1,2,2,2,2,2,3,3,5 so there are more combinations that one would need to try than is reasonable.

    is there nothing else to this problem other than "solve this system of equations"? And what level of school is this?
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #9

    Mar 1, 2007, 07:16 AM
    This is the exercise:
    Find two numbers, if multiplied they equal 3 times their sum; sum of both numbers squared is 160.
    I hope I worded it correctly.

    Oh and I go to 9th grade, so it's primary school I think? If that's what you meant by school levels.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #10

    Mar 1, 2007, 07:31 AM
    Here you go, I rearranged your equations into the form y = f(x) and plotted them both in excel:

    You can see that there are solutions at [12, 4] and [4, 12], the third solution you see at [3,whatever] is excel trying to handle an asymptote, the blue line shouldn't exist there.

    You can also see another solution on the -x, this is around -12.4132, asterisk gives a proper crossing point there. Also, because I used square root, it wouldn't go below zero, but your equations have no problem with that, so the green line should be a circle, and will intersect at one other point, which explains asterisk's final point.

    This is by far the easiest way to solve it.
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    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #11

    Mar 1, 2007, 08:32 AM
    So you're still missing the (-12.4,2.4) and (2.4,-12.4) intersection points. They'd be on your graph if you look to the left and include the -sqrt part of your graph.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #12

    Mar 1, 2007, 08:36 AM
    I fixed the first one of these, the second won't show because of the -sqrt. But it should be findable if I could be bothered :)
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #13

    Mar 1, 2007, 09:48 AM
    Ok, so we agree that my answers are correct. But, other than graphing and getting close (I don't think I could figure out that it's sqrt(55)-5 from the graph ;) ) how would we get these solutions?
    anger336's Avatar
    anger336 Posts: 1, Reputation: 1
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    #14

    Apr 3, 2007, 07:01 PM
    Quote Originally Posted by kristo
    that question was so easy "not"!!

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