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Thread: X variable on both side- need help pls!

  1. #1
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    Question X variable on both side- need help pls!

    OK...
    New to this forum....
    I'm a CAD geek by nature & would think Algebra would be a snap...
    But it ain't as I remember it...

    So here's my problem...
    Please provide the step by step solutions... so I can learn this... please...
    Thanks in advance....

    Problem:
    (x-5)/3 -3/2=(4x+7)/6

    Soultion:
    ?????
    Last edited by ThePerfectHacker; Nov 22nd 2006 at 07:40 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by architech View Post
    OK...
    New to this forum....
    I'm a CAD geek by nature & would think Algebra would be a snap...
    But it ain't as I remember it...

    So here's my problem...
    Please provide the step by step solutions... so I can learn this... please...
    Thanks in advance....

    Problem:
    (x-5)/3 -3/2=(4x+7)/6

    Soultion:
    ?????
    $\displaystyle
    (x-5)/3 -3/2=(4x+7)/6
    $

    Expand all the brackets:

    $\displaystyle
    x/3 -5/3-3/2=x/3-19/6=2x/3 + 7/6
    $

    Subtract $\displaystyle x/3$ from both sides:

    $\displaystyle
    x/3 - 19/6 -x/3=-19/6 = 2x/3 +7/6 -x/3=x/3+7/6
    $

    or:

    $\displaystyle
    -19/6=x/3+7/6
    $

    subtract $\displaystyle 7/6$ form both sides:

    $\displaystyle x/3=-13/3$

    Multiply by $\displaystyle 3$:

    $\displaystyle x=-13$

    RonL
    Last edited by ThePerfectHacker; Nov 22nd 2006 at 07:41 PM.
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  3. #3
    Super Member

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    Hello, architech!

    Welcome aboard!


    $\displaystyle \frac{x-5}{3} -\frac{3}{2}\;=\;\frac{4x+7}{6}$

    With an equation, we can eliminate the fractions immediately . . .


    Multiply both sides by the LCD, 6:

    . . . . . . .$\displaystyle 6\left(\frac{x-5}{3} - \frac{3}{2}\right) \;= \;6\left(\frac{4x+7}{6}\right)$

    .$\displaystyle \not{6}^{^2}\left(\frac{x-5}{\not{3}}\right) - \not{6}^{^3}\left(\frac{3}{\not{2}}\right) \;= \;\not{6}\left(\frac{4x+7}{\not{6}}\right) $

    . . . . . . .$\displaystyle 2(x - 5) - 3(3) \;= \;4x+7$

    . . . . . . . . .$\displaystyle 2x - 10 - 9 \;=\;4x+7$

    . . . . . . . . . . . . . $\displaystyle -2x \;= \;26$

    . . . . . . . . . . . . . . . $\displaystyle \boxed{x \;= \;-13}$

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