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Thread: Linear Equations, Fractions

  1. #1
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    Linear Equations, Fractions

    How would I work this problem out if I did not want to cross multiply?

    $\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

    I would think we should start with the LCD 3(3x+2) no?
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  2. #2
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    Quote Originally Posted by allyourbass2212 View Post
    How would I work this problem out if I did not want to cross multiply?

    $\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

    I would think we should start with the LCD 3(3x+2) no?
    I don't know why you would want to do it that way?

    Multiply both sides by $\displaystyle 3x+2$ and $\displaystyle 3$ and you get rid of both denominators and it becomes simple algebra to solve for x.
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  3. #3
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    Quote Originally Posted by allyourbass2212 View Post
    How would I work this problem out if I did not want to cross multiply?

    $\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

    I would think we should start with the LCD 3(3x+2) no?
    $\displaystyle {\text{Cross - Multiply,}} \hfill \\
    $

    $\displaystyle \frac{{18 - 5x}}
    {{3x + 2}}=\frac{7}
    {3} \hfill$

    $\displaystyle 3(18-5x)=7(3x+2)$
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  4. #4
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    Is this what your trying to do with the LCD?

    $\displaystyle \frac{18-5x}{3x+2}=\frac{7}{3}$

    $\displaystyle \frac{18-5x}{3x+2}-\frac{7}{3}=0$

    $\displaystyle
    \frac{3(18-5x)-7(3x+2)}{3(3x+2)}=0
    $
    Because this still gives you

    $\displaystyle 3(18-5x)-7(3x+2)=0$

    Which is the same as cross multiplying and dividing which gives

    $\displaystyle
    3(18-5x)=7(3x+2)
    $

    So you could do it that way, but it's much more work.
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