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    CyCervantes's Avatar
    CyCervantes Posts: 2, Reputation: 1
    New Member
     
    #1

    Mar 4, 2009, 12:00 AM
    Eliminitation Method
    I have these two equations and can't seem to get one exact answer for x and y. I tried to follow the right steps, but don't know what I'm doing wrong. Can anyone tell me how to do this?
    3x-5y=1 (2)
    6x-10y=4 (1)

    6x-10y=2
    6x-10y=4
    -------------
    12x-20=6??
    I don't understand what to do, I'm supposed to be solving for X and Y but I got this... :o
    Clough's Avatar
    Clough Posts: 26,677, Reputation: 1649
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    #2

    Mar 4, 2009, 12:37 AM

    Hi, CyCervantes!

    Greetings and WELCOME to the site! I just moved your question that you had posted in Introductions to this forum topic area so that it will get the most exposure to those who are best able to answer it. Introductions is for people to introduce themselves and we try to not ask questions there.

    It can be a little confusing when first learning how to use this site! Your question will get noticed much more in this forum topic area.

    We would appreciate it if you would return to Introductions sometime to tell us a little about yourself though, if you would be willing to do that.

    Thanks!
    yaode3zy's Avatar
    yaode3zy Posts: 31, Reputation: 2
    Junior Member
     
    #3

    Mar 4, 2009, 03:39 AM

    The purpose of the elimination method is to cancel out terms. Your on a good start by multiplying (2) to 3x-5y=1 ; however, none of the terms cancel out.

    In these kinds of problems, you would pick a term to eliminate. For example, you want to eliminate "x". Since 6x is a multiple of 3x, you can use "2" to multiply. To cancel the terms out, you will need to have one negative.
    So by multiplying by "-2", you will end up with:

    -6x+10y=-2
    6x-10y=4
    0 + 0= -2

    In this case, all the terms canceled out, meaning they are the same (Correct me if I'm wrong; it's been a long time heh... ).

    All in all, this problem set is not a great example to use to learn the concept, but that's how you do it.
    sarnian's Avatar
    sarnian Posts: 462, Reputation: 9
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    #4

    Mar 4, 2009, 04:41 AM
    Hello CyCervantes

    Indeed : it seems that the data provided here is incorrect (at least insufficient to solve).
    yaode3zy's Avatar
    yaode3zy Posts: 31, Reputation: 2
    Junior Member
     
    #5

    Mar 4, 2009, 05:43 AM

    I think I remember now; this would mean they do not intersect, therefore there is no solution. To be the same line, I believe the equations are completely identical.

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