Results 1 to 3 of 3

Math Help - How do I solve these?

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    14

    How do I solve these?

    x+17/x^2-6x+8 + x-2/x-4 = x-4/x-2

    32/t^2+3t+2 - 3 = t-3/t+1

    r(3-r) + (2r+2)^2 = 0

    Can somebody help me with these please our midterm is tomorrow and if questions like these turn up (which is very likely) then my goose is cooked.

    EDIT: I grouped them together to make things clearer:
    (x+17)/(x^2-6x+8) + (x-2)/(x-4) = (x-4)/(x-2)

    32/(t^2+3t+2) - 3 = (t-3)/(t+1)

    r(3-r) + (2r+2)^2 = 0
    Last edited by strigy; September 2nd 2009 at 03:29 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,238
    Thanks
    1795
    Quote Originally Posted by strigy View Post
    x+17/x^2-6x+8 + x-2/x-4 = x-4/x-2
    If I read this correctly (Parentheses would help. I assume you mean (x+17)/(x^2- 6x+ 8) rather than x+ (17/x^2)- 6x+ 8)
    x^2- 6x+ 8= (x- 4)(x- 2) so this is
    \frac{x+ 17}{(x- 4)(x- 2)}+ \frac{x- 2}{x- 4}= \frac{x- 4}{x- 2}

    Multiplying each term by (x- 4)(x- 2) gets rid of all the fractions.

    32/t^2+3t+2 - 3 = t-3/t+1
    Again, I am going to assume that you really mean \frac{32}{t^2+ 3t+ 3}- 3= \frac{t- 3}{t+ 1}. Once again, [tex]t^3+ 3t+ 2= (t+1)(t+ 2) so multiplying each term by (t+1)(t+2) gets rid of the fractions.

    r(3-r) + (2r+2)^2 = 0
    Just go ahead and multiply it out: 3r- r^2+ 4r^2+ 8r+ 4= 0 which is the same as 3r^2+ 11r+ 4= 0. Complete the square or use the quadratic formula.

    Can somebody help me with these please our midterm is tomorrow and if questions like these turn up (which is very likely) then my goose is cooked.

    EDIT: I grouped them together to make things clearer:
    (x+17)/(x^2-6x+8) + (x-2)/(x-4) = (x-4)/(x-2)

    32/(t^2+3t+2) - 3 = (t-3)/(t+1)

    r(3-r) + (2r+2)^2 = 0
    Why didn't you do that to begin with?!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    14
    Ok thanks HallsofIvy and sorry for the confusion. I got number 1.

    I seem to be stuck at 2 and 3, though.

    I'm stuck here 4t^2-11t+32=0 for number 2, and at 3r^2+11r+4=0 for number 3.

    I tried quadratic but that didn't work. What do I do next?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 11:29 PM
  2. how do i solve this IVP: y'=y^2 -4
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 24th 2010, 12:14 PM
  3. Replies: 1
    Last Post: June 9th 2009, 11:37 PM
  4. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 03:58 PM
  5. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 09:17 AM

Search Tags


/mathhelpforum @mathhelpforum