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Math Help - Algebra question

  1. #1
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    Algebra question

    Hi I'm on this question where I hacve to find x:

    3x+5 - 2x-1 = 3
    4 6

    and i ended up with 9x + 15 - 4x - 2 = 36
    12 = 12

    so i removed the 12s to leave me with 9x + 15 - 4x - 2 = 36

    Now for my question:

    When I am simplifying the left hand side; does it become 5x + 13 or 5x + 17 because if the equation should be written as (9x + 15) - (4x - 2) it would mean a -- which is a plus to make 5x + 17...

    edit: for some reason the denominators havent come up where i wanted them to be. but its 3x + 5 all over 4 and 2x - 1 all over 6
    and the second error is that both sides should have 12 as their denominators
    Last edited by david18; June 3rd 2007 at 10:02 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by david18 View Post
    Hi I'm on this question where I hacve to find x:

    3x+5 - 2x-1 = 3
    4 6

    and i ended up with 9x + 15 - 4x - 2 = 36
    12 = 12

    so i removed the 12s to leave me with 9x + 15 - 4x - 2 = 36

    Now for my question:

    When I am simplifying the left hand side; does it become 5x + 13 or 5x + 17 because if the equation should be written as (9x + 15) - (4x - 2) it would mean a -- which is a plus to make 5x + 17...

    edit: for some reason the denominators havent come up where i wanted them to be. but its 3x + 5 all over 4 and 2x - 1 all over 6
    and the second error is that both sides should have 12 as their denominators
    \frac {3x + 5}{4} - \frac {2x - 1}{6} = 3 .....the LCD is 12

    \Rightarrow \frac {3(3x + 5) - 2(2x - 1)}{12} = 3

    \Rightarrow \frac {9x + 15 - 4x + 2}{12} = 3

    \Rightarrow 9x + 15 - 4x + 2 = 36 ........multiplied both sides by 12

    \Rightarrow 5x + 17 = 36

    Now continue
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  3. #3
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    Thanks that was really helpful... you used a better method than mine which involved making the 3 top heavy and silly things like that.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by david18 View Post
    Thanks that was really helpful... you used a better method than mine which involved making the 3 top heavy and silly things like that.
    "making the 3 top heavy"? what's that?

    did you get my method? we could have also muliplied the first fraction by \frac {3}{3} and the second by \frac {2}{2} that would make the denominators the same, and we just add the tops from there. there is also the cross multiplication method for combining fractions, but in this case, that was more trouble than it's worth
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