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Math Help - 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?

    \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?

    \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 3x+2 and 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?

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

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

    \frac{{18 - 5x}}<br />
{{3x + 2}}=\frac{7}<br />
{3} \hfill

    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?

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

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

     <br />
\frac{3(18-5x)-7(3x+2)}{3(3x+2)}=0<br />
    Because this still gives you

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

    Which is the same as cross multiplying and dividing which gives

     <br />
3(18-5x)=7(3x+2)<br />

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