Results 1 to 13 of 13

Math Help - Math 10 Pure Fractions! Add/Multiply/Factor

  1. #1
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7

    Exclamation Math 10 Pure Fractions! Add/Multiply/Factor

    alright, my math is easy but theres always a few hard ones. just wondering if anyone can do these, show restrictions and show as much work as possible, or explain the steps.

    4 | 5
    --------- | - -----------
    y^2+5y+6 | y^2-y-12


    4 | 3
    --------- | + -------------
    w^2+5w+4 | w^2+2w+8

    6b^3 |+ 9g^5
    ------- | ---------
    8g^5 | 12b^3


    1 + | 4 |- 6
    -------|------ | -------
    2x^2 | 3x | x^3


    Thanks for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    Hello, SCT!

    I must assume you know about finding the Lowest Common Denominator
    . . and converting the fractions to the LCD.


    \frac{4}{y^2+5y+6} - \frac{5}{y^2-y-12}

    We have: . \frac{4}{(y+2)(y+3)} - \frac{5}{(y+3)(y-4)} \qquad LCD \:=\:(y+2)(y+3)(y-4)

    Convert: . \frac{4}{(y+2)(y+3)}\cdot{\color{blue}\frac{y-4}{y-4}} \;- \;\frac{5}{(y+3)(y-4)} \cdot{\color{blue}\frac{y+2}{y+2}}

    . . . . =\;\frac{4(y-4)}{(y+2)(y+3)(y+4)} - \frac{5(y+2)}{(y+3)(y-4)(y+2)}

    . . . . = \;\frac{4y - 16 - 5y - 10}{(y+2)(y+3)(y-4)}

    . . . . = \;\frac{-y-26}{(y+2)(y+3)(y-4)}




    \frac{6b^3}{8g^5} + \frac{9g^5}{12b^3}

    Reduce the fractions: . \frac{3b^3}{4g^5} + \frac{3g^5}{4b^3} \qquad LCD \:=\:4b^3g^5

    Convert: . \frac{3b^3}{4g^5}\cdot{\color{blue}\frac{b^3}{b^3}  } \;+\; \frac{3g^5}{4b^3}\cdot{\color{blue}\frac{g^5}{g^5}  } \;= \;\frac{3b^6}{4b^3g^5} + \frac{3g^{10}}{4b^3g^5} \;=\;\frac{3b^6 + 3g^{10}}{4b^3g^5}




    \frac{1}{2x^2} + \frac{4}{3x} - \frac{6}{x^3}
    The LCD is: . 6x^3

    Convert: . \frac{1}{2x^2}\cdot{\color{blue}\frac{3x}{3x}} \;+\;\frac{4}{3x}\cdot{\color{blue}\frac{2x^2}{2x^  2}} \;-\;\frac{6}{x^3}\cdot{\color{blue}\frac{6}{6}}

    . . . . = \;\frac{3x}{6x^3} + \frac{8x^2}{6x^3} - \frac{36}{6x^3} \;=\;\frac{8x^2 + 3x - 36}{6x^3}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7
    thanks!
    any idea about

    m - 3
    ------ -----
    2m-4 30-6


    and

    4 3
    ------ + --------
    w^2+5w+4 w^2+2w-8

    just not getting the 2nd cause of the lcd and such and first one i keep getting different answers
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Did you factor the denominators?

    \frac{4}{w^2 + 5w + 4} + \frac{3}{w^2 + 2w - 8} =

    \frac{4}{(w+4)(w+1)} + \frac{3}{(w+4)(w-2)}

    The LCD is (w+4)(w+1)(w-2).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7
    Ah, thank you.
    Figured out that question. (factored +3 +2 into the 5w+4 for some reason. 5, 3+2 instinct!)
    thank you.
    anyone know about the other one?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Is the problem to simplify

    \frac{m}{2m-4} - \frac{3}{30-6}?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7
    no, i need to find the LCD and solve.
    m doesnt need to be evaluated
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    \frac{m}{2m-4} - \frac{3}{30-6} = \frac{m}{2(m-2)} - \frac{1}{8}.

    The LCD is 8(m-2).

    Can you do the rest?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7
    im confused.. the way our teacher taught us (maybe im incorrect)
    but is to go
    m
    ---
    2(m-2)
    and
    3
    -----
    3(m-2)

    making the LCD 6(m-2)
    or cancelling the (m-2)'s out and making it 6?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    You should always reduce the fractions to lowest terms before figuring out the LCD. I'm still not sure if we're looking at the same problem, though.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7
    oops. typed it wrong.

    m over 2m-4 subtract 3 over 3m-6

    can you get the LCD and then what to multiply the numerators by?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    \frac{m}{2m-4} - \frac{3}{3m-6} =

    \frac{m}{2(m-2)} - \frac{3}{3(m-2)} =

    \frac{m}{2(m-2)} - \frac{1}{m-2}

    The LCD is 2(m-2).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    SCT
    SCT is offline
    Newbie
    Joined
    Sep 2008
    Posts
    7
    would that make the answer
    m-2
    -----
    2(m-2) therefore .:. one half?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Math 20 Pure: Chord Properties
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 30th 2009, 02:22 PM
  2. math help plz for pure 30
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: January 25th 2009, 09:03 PM
  3. math 30 pure help please
    Posted in the Math Topics Forum
    Replies: 7
    Last Post: January 16th 2009, 08:56 PM
  4. Physics vs Pure Math
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 14th 2007, 09:16 PM

Search Tags


/mathhelpforum @mathhelpforum