Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Thread: rational expressions

  1. #1
    it's not fun, don't do it Moon Rabbits's Avatar
    Join Date
    Mar 2005
    Posts
    5,582

    Default rational expressions

    Simplify.

    (7y^2 / y^2 - 9) * (4y + 2 / 14y^2)

  2. #2
    One Hundred Chimneys Recognized Member Tavrobel's Avatar
    Join Date
    Jan 2006
    Location
    Conjunction Junction
    Posts
    10,455
    Articles
    102
    Contributions
    • Former Site Staff

    Default

    Begin by distributing terms to both the numerator and the denominator. Since this is multiplication, there shouldn't be anything wrong with that.

    (7y<sup>2</sup>)(4y) + (7y<sup>2</sup>)(2) / (4y<sup>2</sup>)(y<sup>2</sup>) - (4y<sup>2</sup>)(9)

    Combine like terms, if applicable. Then simplify.

    (28y<sup>3</sup>+14y<sup>2</sup>) / (4y<sup>4</sup>-36y<sup>2</sup>)

    Since we have a Y squared in every term, remove them from the problem.

    (28y+14) / (4y<sup>2</sup>-36)

    Factor 14 from the top (optional step).

    14*(2y+1) / (4y<sup>2</sup>-36)

    Factor out 2 from both the top and the bottom.

    7*(2y+1) / (2y<sup>2</sup>-18)

    Optionally:
    Factor out a 2 from the bottom. That leaves you with a difference of squares.

    7(2y+1) / 2(y<sup>2</sup>-9)
    7(2y+1) / 2(y+3)(y-3)

    If you need to do something else with the problem, now would be the time to do so.


    Math sucks. Your only hope is suicide.
    Last edited by Tavrobel; 02-28-2008 at 03:05 AM.

  3. #3
    Draw the Drapes Recognized Member rubah's Avatar
    Join Date
    Dec 2004
    Location
    Now Destiny is done.
    Posts
    30,655
    Blog Entries
    21
    Contributions
    • Former Administrator
    • Former Cid's Knight

    Default

    It'd be a lot easier if you divided seven y squared by fourteen y squared at the onset, leaving you with a 2 in the denominator :]] Then you just have (2y-1)/(y-3)(y+3) don't you?

  4. #4
    One Hundred Chimneys Recognized Member Tavrobel's Avatar
    Join Date
    Jan 2006
    Location
    Conjunction Junction
    Posts
    10,455
    Articles
    102
    Contributions
    • Former Site Staff

    Default

    Ohh . I misread it as 4y^2. Waah.

  5. #5
    Hypnotising you crono_logical's Avatar
    Join Date
    May 2001
    Location
    Back in Time
    Posts
    9,313
    Contributions
    • Former Administrator
    • Former Cid's Knight

    Default

    Quote Originally Posted by Rubah Lapah View Post
    It'd be a lot easier if you divided seven y squared by fourteen y squared at the onset, leaving you with a 2 in the denominator :]] Then you just have (2y-1)/(y-3)(y+3) don't you?
    Where'd the 2y-1 come from, rubah?
    Problems playing downloaded videos? Try CCCP


  6. #6
    i n v i s i b l e Tech Admin o_O's Avatar
    Join Date
    Jun 2001
    Location
    New Zealand
    Posts
    2,957
    Blog Entries
    1

    FFXIV Character

    Humphrey Squibbles (Sargatanas)

    Default

    Yeah, Allie.

    Here's the working for anyone that wants it:
    <pre>(7y<sup>2</sup> / y<sup>2</sup> - 9) * (4y + 2 / 14y<sup>2</sup>)
    = (7y<sup>2</sup> / 14y<sup>2</sup>) * (4y + 2 / y<sup>2</sup> - 9) Rearranging the terms into a more desirable order, by commutativity.
    = (1 / 2) * (4y + 2 / y<sup>2</sup> - 9) Reducing left multiplicand.
    = (4y + 2 / 2y<sup>2</sup> - 18) Multiplying multiplicands.
    = (2y + 1 / y<sup>2</sup> - 9) Reducing again.
    = (2y + 1) / (y + 3)(y - 3) Factorising.
    </pre>

  7. #7
    That's me! blackmage_nuke's Avatar
    Join Date
    Aug 2005
    Location
    Yes
    Posts
    8,503
    Blog Entries
    3

    Default

    If you take the 2 out of 4y+2 to get 2(2y+1) then use that 2 to multiply with the 7y^2 to get 14y^2 then that cancels out with the 14y^2 on the bottom and your left with what o_O said

    edit: and put your brackets correctly. i guess most people assume you meant
    (7y^2 / (y^2 - 9)) * ((4y + 2) / 14y^2)

    But sometimes that wont be so obvious
    Last edited by blackmage_nuke; 02-28-2008 at 08:45 AM.
    Kefka's coming, look intimidating!
    Have a nice day!!

  8. #8
    Draw the Drapes Recognized Member rubah's Avatar
    Join Date
    Dec 2004
    Location
    Now Destiny is done.
    Posts
    30,655
    Blog Entries
    21
    Contributions
    • Former Administrator
    • Former Cid's Knight

    Default

    Quote Originally Posted by crono_logical View Post
    Quote Originally Posted by Rubah Lapah View Post
    It'd be a lot easier if you divided seven y squared by fourteen y squared at the onset, leaving you with a 2 in the denominator :]] Then you just have (2y-1)/(y-3)(y+3) don't you?
    Where'd the 2y-1 come from, rubah?
    gosh, didn't you see the +2 right afterwards? I must have forgotten it 8)

  9. #9
    Old school, like an old fool. Flying Mullet's Avatar
    Join Date
    Apr 2003
    Location
    Napping in a peach tree.
    Posts
    19,185
    Articles
    6
    Blog Entries
    7
    Contributions
    • Former Administrator
    • Former Cid's Knight
    • Former Senior Site Staff

    Default

    Similar approach to what's above.

    (7y^2 / y^2 - 9) * (4y + 2 / 14y^2)

    = (7y^2 / y^2 - 9) * (2y + 1 / 7y^2) - factor out a 2 from the second fraction

    = (1/ y^2 - 9) * (2y + 1 / 1) = cancel the 7y^2

    = (2y + 1) / (y^2 - 9) = multiply

    = (2y + 1) / (y +/- 3)^2 = factor the denominator if need be
    Figaro Castle

  10. #10
    Hypnotising you crono_logical's Avatar
    Join Date
    May 2001
    Location
    Back in Time
    Posts
    9,313
    Contributions
    • Former Administrator
    • Former Cid's Knight

    Default

    Though once you get good at it, you can do these in your head
    Problems playing downloaded videos? Try CCCP


  11. #11
    One Hundred Chimneys Recognized Member Tavrobel's Avatar
    Join Date
    Jan 2006
    Location
    Conjunction Junction
    Posts
    10,455
    Articles
    102
    Contributions
    • Former Site Staff

    Default

    But only if you copy down the problem correctly.

  12. #12
    Old school, like an old fool. Flying Mullet's Avatar
    Join Date
    Apr 2003
    Location
    Napping in a peach tree.
    Posts
    19,185
    Articles
    6
    Blog Entries
    7
    Contributions
    • Former Administrator
    • Former Cid's Knight
    • Former Senior Site Staff

    Default

    Details, details...
    Figaro Castle

  13. #13
    i n v i s i b l e Tech Admin o_O's Avatar
    Join Date
    Jun 2001
    Location
    New Zealand
    Posts
    2,957
    Blog Entries
    1

    FFXIV Character

    Humphrey Squibbles (Sargatanas)

    Default

    If you're good enough at maths, you can solve things even if you have the wrong problem to begin with.

  14. #14
    Draw the Drapes Recognized Member rubah's Avatar
    Join Date
    Dec 2004
    Location
    Now Destiny is done.
    Posts
    30,655
    Blog Entries
    21
    Contributions
    • Former Administrator
    • Former Cid's Knight

    Default

    *bad enough

  15. #15

    Default

    I was wondering, in this question to simplifying rational expressions ??

    Find (-6x^2 - 4xy + 8x)/(2x)

    do they want me to find out what the variables are, or do they want me
    to do something else with it? That is the exact way they have the
    question written down.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •