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Math Help - Math Help plz!

  1. #1
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    Exclamation Math Help plz!

    1.)rewrite in order to solve

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



    2.)rewrite in order to complete the square. (i hate these)

    y^2 +3y-1=0
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  2. #2
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    in future, please use grouping symbols for the sake of clarity ...

    3/(2x-6) + 5/(x-3) = 1/6

    3/[2(x-3)] + 5/(x-3) = 1/(2*3) note that x cannot equal 3.

    get a common denominator ...

    3*3/[3*2(x-3)] + 5*2*3/[2*3(x-3)] = 1(x-3)/[2*3(x-3)]

    now that you have a common denominator, the numerators form the equation ...

    9 + 30 = x - 3
    now can you find x ?



    y^2 + 3y - 1 = 0

    y^2 + 3y = 1

    half the coefficient of the linear term (that's the 3), square, and add result to both sides ...

    y^2 + 3y + (3/2)^2 = 1 + (3/2)^2

    [y + (3/2)]^2 = 13/4

    y + (3/2) = +/- sqrt(13/4) = +/- sqrt(13)/2

    y = [3 +/- sqrt(13)]/2

    get over your hatred of these ... they are not going to go away.
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  3. #3
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    Quote Originally Posted by Peyton Sawyer View Post
    1.)rewrite in order to solve

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



    2.)rewrite in order to complete the square. (i hate these)

    y^2 +3y-1=0
    for the first one i am trying ,it maybe .. i am not sure
    3/2x-6 + 5/x-3 = 1/6
    3/x(x-3)+5/x-3 = 1/6
    3+5(2)/2(x-3) = 1/6
    13/2(x-3) = 1/6
    2x-6 = 78
    2x = 72
    x = 36
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  4. #4
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    Quote Originally Posted by Peyton Sawyer View Post
    1.)rewrite in order to solve

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



    2.)rewrite in order to complete the square. (i hate these)

    y^2 +3y-1=0
    \frac{3}{2x-6}+\frac{5}{x-3}=\frac{1}{6}
    \frac{3}{2(x-3)}+\frac{5}{x-3}=\frac{1}{6}

    Do you see that the denominator of the second term is half of the first term's denominator? So multiply the top and bottom of the second term by 2 to get the same denominator for both...

    \frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}
    \frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}
    \frac{3}{2x-6}+\frac{10}{2x-6}=\frac{1}{6}
    \frac{3+10}{2x-6}=\frac{1}{6}
    \frac{13}{2x-6}=\frac{1}{6}

    Cross multiplication gets rid of the denominators

    13\times 6 = 2x-6
    78 = 2x-6
    72 = 2x
    36 = x

    So yes you are right.
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  5. #5
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    \frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}
    \frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}
    \frac{3}{2x-6}+\frac{10}{2x-6}=\frac{1}{6}
    \frac{3+10}{2x-6}=\frac{1}{6}
    \frac{13}{2x-6}=\frac{1}{6}

    Cross multiplication gets rid of the denominators

    13\times 6 = 2x-6
    78 = 2x-6 better check going from this step to the next, x = 36 is not correct
    72 = 2x
    36 = x

    So yes you are right.
    x = 42 is the correct solution.
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