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Posted by on Wednesday, April 10, 2013 at 12:19am.

I'm having trouble understanding how to do complex fractions.. My book doesn't explain it very well..

12.
(x+3/12)/(4x-5/15)

14.
(2/x^2 + 1/x)/(4/x^2 - 1/x)

16.
(1/y + 3/y^2)/(y + 27/y^2)

If you could please explain how to begin the process I would appreciate it.

  • Pre-Calculus - , Wednesday, April 10, 2013 at 2:07am

    12.


    3 / 12 = 3 / ( 3 * 4 ) = 1 / 4

    5 / 15 = 5 / ( 5 * 3 ) = 1 / 3


    ( x + 3 / 12 ) / ( 4 x - 5 / 15 ) =

    ( x + 1 / 4 ) / ( 4 x - 1 / 3 ) =

    ( 4 x / 4 + 1 / 4 ) / ( 4 x * 3 / 3 - 1 / 3 ) =

    [ ( 4 x + 1 ) / 4 ] / [ ( 12 x - 1 ) / 3 ] =

    ( 4 x + 1 ) * 3 / [ 4 * ( 12 x - 1 ) ] =

    ( 3 / 4 ) ( 4 x + 1 ) / ( 12 x - 1 ) =

    3 * ( 4 x + 1 ) / [ 4 * ( 12 x - 1 ) ] =

    ( 12 x + 3 ) / ( 48 x + 4 )


    14.

    ( 2 / x ^ 2 + 1 / x ) / ( 4 / x ^ 2 - 1 / x ) =

    [ 2 / x ^ 2 + 1 * x / ( x * x ) ] / [ 4 / x ^ 2 - 1 * x / ( x * x ) ] =

    ( 2 / x ^ 2 + x / x ^ 2 ) / ( 4 / x ^ 2 - x / x ^ 2 ) =

    [ ( 2 + x ) / x ^ 2 ] / [ ( 4 - x ) / x ^ 2 ] =

    [ x ^ 2 * ( 2 + x ) / x ^ 2 ] / [ x ^ 2 * ( 4 - x ) / x ^ 2 ] =

    ( 2 + x ) / ( 4 - x )


    16.


    ( 1 / y + 3 / y ^ 2 ) / ( y + 27 / y ^ 2 ) =

    [ 1 * y / ( y * y ) + 3 / y ^ 2 ) ] / [ y * y ^ 2 / y ^ 2 + 27 / y ^ 2 ) ] =

    ( y / y ^ 2 + 3 / y ^ 2 ) / ( y ^ 3 / y ^ 2 + 27 / y ^ 2 ) =

    [ ( y + 3 ) / y ^ 2 ] / [ ( y ^ 3 + 27 ) / y ^ 2 ) =

    [ y ^ 2 * ( y + 3 ) / y ^ 2 ] / [ y ^ 2 ( y ^ 3 + 27 ) / y ^ 2 ) =

    ( y + 3 ) / ( y ^ 3 + 27 ) =

    ( y + 3 ) / [ ( y + 3 ) * ( y ^ 2 - 3 y + 9 ) =

    1 / ( y ^ 2 - 3 y + 9 )

  • Pre-Calculus - , Wednesday, April 10, 2013 at 10:14am

    12.
    (x+3/12)/(4x-5/15)
    first reduce those fractions to lowest terms
    = (x + 1/4) / (4x - 1/3)
    multipy top and bottom by 12 , the LCD
    = (12x + 3)/(48x+4) or 3(4x+1)/(4(12x+1) )

    14.
    (2/x^2 + 1/x)/(4/x^2 - 1/x)
    do it the same way, multiply top and bottom by x^2 , the LCD
    = (2 + x)/(4 - x)
    all done!

    try the last one using the same method

  • Pre-Calculus - , Thursday, April 11, 2013 at 12:01am

    Thank you both!

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