1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

System of equations: help needed

  1. Jan 7, 2007 #1
    can anyone tell me how to solve the system of equations:

    [tex](x-1)(y^2+6) = y(x^2+1)[/tex]

    [tex](y-1)(x^2+6) = x(y^2+1)[/tex]

    note: this is not a homework problem. as far as i know, this is a problem from australia's national math olympiad.
     
  2. jcsd
  3. Jan 8, 2007 #2

    benorin

    User Avatar
    Homework Helper

    I used the http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=advanced [Broken] and put

    Code (Text):
    (x-1)(y^2+6)=y(x^2+1)
    (y-1)(x^2+6)=x(y^2+1)
    into the field marked "EQUATION(S)" to get the solutions

    [tex](x,y)=(2,2),\, (2,3),\, (3,2),\, (3,3),\, \left( \frac{1}{2}\left(1\pm i\sqrt{15}\right) , \frac{1}{2}\left(1\mp i\sqrt{15}\right)\right)[/tex].
     
    Last edited by a moderator: May 2, 2017
  4. Jan 8, 2007 #3

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I do not think that resorting to a software solution would be acceptable on a test.

    Expand both equations, then add them together. You will now have quadratic equations in x any y. Complete the squares and you will easily get the real solutions.
     
  5. Jan 8, 2007 #4
    benorin, i also got those solutions using Maple. but what i want to know is how to solve them without using softwares.

    Integral, i tried doing what you suggested. i expanded and subtracted one from the other and got an equation of a circle. but i don't know what to do next. can you give some more hints?
     
  6. Jan 8, 2007 #5

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Try completing the square
     
  7. Jan 8, 2007 #6
    by adding the 2 equations i got, [tex](x-y)(x+y+2xy+7) = 0[/tex]

    by subtracting one from the other i got, [tex]\left(x-\frac{5}{2}\right)^2 + \left(y-\frac{5}{2}\right)^2 = \frac{1}{2}[/tex]

    but i can't figure out what to do next.
     
  8. Jan 8, 2007 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    That's the eqn of a circle in the Oxy plane with radius 1/2 and located with the center at (5/2,5/2). You'll have to intersect this circle with the 2 line y=x and with the curve x+y+2xy+7=0 to get the possible solutions.

    Daniel.
     
  9. Jan 8, 2007 #8
    the radius is actually [tex]\frac{1}{\sqrt{2}}[/tex]

    but how do i do that? if i take the intersection between the circle and the line y = x, i get the solutions (3,3) and (2,2). but how do i get the intersection of the circle and the curve x+y+2xy+7=0?
     
  10. Jan 8, 2007 #9
    well, i tried the following:
    from x+y+2xy+7=0, we get [tex]y = \frac{-7-x}{1+2x}[/tex]

    substituting this into the equation of the circle and simplifying, i got the quartic equation:

    [tex]x^4 - 4x^3 + 10x^2 + 33x + 24 = 0[/tex]

    now how do i solve this?
     
  11. Jan 8, 2007 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    First of all, you got the first equation wrong, it should read:
    (x-y)(x+y-2xy+7)=0

    Secondly, eliminating the (x-y) factor, ADD the two equations you now have, yielding:
    [tex]x^{2}-5x+y^{2}-5y+\frac{25}{2}+(x+y-2xy+7)=\frac{1}{2}[/tex]
    that is:
    [tex](x-y)^{2}-4(x+y)+19=0[/tex]

    Similarly, subtract the first one from your second one to get:
    [tex](x+y)^{2}-6(x+y)+5=0[/tex]

    Now, introduce the new variables u=x+y, v=x-y to rewrite your system of equations as:
    [tex]u^{2}-6u+5=0[/tex]
    [tex]v^{2}-4u+19=0[/tex]
    This system is readily solvable. :smile:
     
  12. Jan 8, 2007 #11

    Gib Z

    User Avatar
    Homework Helper

    Is it helpful to notice here that these equations are inverses of each other? Then would mean that the intersect on the like y=x correct? And a few other places of course...but thought it might help since if you have one solution, you immediately have another :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: System of equations: help needed
Loading...