Results 1 to 2 of 2

Math Help - Solve the system

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    54

    Solve the system

    Solve the system

     <br />
x^2-4\sqrt{3x-2}+6=y
    <br />
y^2-4\sqrt{3y-2}+6=x<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by mathwizard View Post
    Solve the system

     <br />
x^2-4\sqrt{3x-2}+6=y
    <br />
y^2-4\sqrt{3y-2}+6=x<br />
    There is an obvious solution x = y = 2. In fact, this is the only solution.

    To see why, let f(x) =  x^2-4\sqrt{3x-2}+6 (defined for x ≥ 2/3). The function f(x) – x has a minimum value of 0 at x=2 (calculus exercise). Hence f(x) > x whenever x ≠ 2. So if x ≠ 2 and y = f(x) then y > x, and f(y) ≥ y > x. Thus for x ≠ 2 it can never happen that f(x) = y and f(y) = x.
    Last edited by Opalg; June 24th 2008 at 12:16 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solve system
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 3rd 2011, 04:53 PM
  2. Solve the system?
    Posted in the Algebra Forum
    Replies: 0
    Last Post: April 30th 2009, 09:18 PM
  3. solve the system
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 14th 2008, 11:12 AM
  4. solve the system
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 2nd 2007, 09:19 PM
  5. Solve System of Dif EQ
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 31st 2007, 08:42 AM

Search Tags


/mathhelpforum @mathhelpforum