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Math Help - Need help with an algebra problem

  1. #1
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    Need help with an algebra problem

    Hello folks, glad I found this site. I wonder if anyone could help me with my son's math problem, I'm kind of a retard when it comes to this stuff. Say I have a group of ten batteries, their are 6volt and 12volt batteries with the total sum of 84 volts, how do I write the equation? I know we need 4 12v and 6 6v batteries for a total 10 batteries and 84volts, but not sure how to write it. Thanks for any help, and glad to be here. Scott
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  2. #2
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    Well, you'll end up with one equation:

    12 times the number of 12v batteries + 6 times the number of 6v batteries = 84

    Since you've only got one equation, you've got to write it so it only has one variable. So, if you've got x batteries in one group, then there must be 10-x batteries left. So ...

     12x + 6(10-x) = 84

    Note, it doesn't matter which battery type goes where, so this gets the same result ...

     6x + 12(10-x) = 84

    -Scott
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  3. #3
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    Thanks, my son picked up on it after seeing your post.
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  4. #4
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    Quote Originally Posted by ScottO View Post
    Well, you'll end up with one equation:

    12 times the number of 12v batteries + 6 times the number of 6v batteries = 84

    Since you've only got one equation, you've got to write it so it only has one variable. So, if you've got x batteries in one group, then there must be 10-x batteries left. So ...

     12x + 6(10-x) = 84

    Note, it doesn't matter which battery type goes where, so this gets the same result ...

     6x + 12(10-x) = 84

    -Scott

    For what it's worth you do actually have 2 equations not 1.

    The first equation is 12x + 6y = 84.

    The second equation is x + y = 10.

    Those are your two simultaneous equations which are easily solved.
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  5. #5
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    Quote Originally Posted by oaksoft View Post
    For what it's worth you do actually have 2 equations not 1.

    The first equation is 12x + 6y = 84.

    The second equation is x + y = 10.

    Those are your two simultaneous equations which are easily solved.
    Good point oaksoft!! Thanks.

    So essentially, I used the second equation without thinking about it. Solved for y, and then substituted. OK for me solving, but not good for explaining to someone else.

    -Scott
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