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Homework Help: System of equations

  1. May 29, 2010 #1
    1. The problem statement, all variables and given/known data

    How do we prove the system of the following equations??

    1) [tex]3xy+y^2 =-7[/tex]

    2) [tex] x^2-2xy = 30[/tex]



    2. Relevant equations



    3. The attempt at a solution


    I tried by solving the 1st equation in terms y and then substituting into equation (2) but the problem got more complicated
     
  2. jcsd
  3. May 29, 2010 #2

    rock.freak667

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    Can you please post your working, because substitution seems like the only viable method.
     
  4. May 30, 2010 #3
    [tex]2y^2 +3xy +7 =0[/tex] and solving for y we have:

    [tex]y= \frac{-3x+\sqrt{9x^2-56}}{4}[/tex] or [tex]y=\frac{-3x-\sqrt{9x^2-56}}{4}[/tex]

    Now we substitute for y in the 2nd equation and we get two equations:

    [tex]x^2-2x\frac{-3x+\sqrt{9x^2-56}}{4}-30 =0[/tex]

    ......................or..................................

    [tex]x^2-2x\frac{-3x-\sqrt{9x^2-56}}{4}-30 =0[/tex]
     
  5. May 30, 2010 #4

    ehild

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    This is the hardest way to proceed. First solve the first equation for x, or the second equation for y.

    ehild
     
  6. May 30, 2010 #5
    and then substitute in the 2nd equation??
     
  7. May 30, 2010 #6

    rock.freak667

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    Yes.
     
  8. May 30, 2010 #7
    Solve by substitution, I'll give you the first 1 (MODS PLEASE DON'T INFRACT ME I'M NOT GIVING HIM THE FULL ANSWER!! < IF SOMETHING IS WRONG PLEASE PM ME NOT INFRACT :()

    2(3xy+y2=-7)
    3(-2xy+x2=30)
    =
    6xy+2y2=-14
    -6xy+6y2=90
    Add those 2 together
    Then you'll be left with 1 variable :p
     
  9. May 31, 2010 #8


    Sorry Cause I'm silly gringa....

    Anyway back to the original problem

    if you try to solve this regulary you get som very silly fractions...
     
    Last edited by a moderator: Jun 11, 2010
  10. May 31, 2010 #9

    Borek

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    How is it related to the original problem?
     
  11. May 31, 2010 #10

    Thank ,it works out.
     
  12. May 31, 2010 #11

    HallsofIvy

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    Well, only because you have changed [itex]x^2[/itex] to [itex]y^2[/itex] in the second equation!

    You should have [itex]6xy+ 2y^2= -14[/itex] and [itex]-6xy+ 6x^2= 90[/itex]. Adding those two equations gives you [itex]6x^2+ 2y^2= 76[/itex]. You still have two variables.
     
  13. Jun 1, 2010 #12
    isn't there a mistake there? its [tex]y^2 +3xy +7 =0[/tex]
     
  14. Jun 11, 2010 #13
    I have tried and tried for the last couple of weeks to try to solve this system using pre calculus tools and I always come up with that the two parts of the system do not intersect and there doesn't exist a commen set of fix points

    I tried to draw a phase portrait in Maple and it does look like something interesting happens in and around the origin and that there asome asympotetes, but I can't get a userable solution :(

    Maybe there is a mistake by the original poster?
     
  15. Jun 11, 2010 #14

    vela

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    I plotted the original equations in Mathematica, and there are in fact four solutions.
     
  16. Jun 12, 2010 #15

    ehild

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    Solve the second equation for y:

    [tex]y=\frac{x^2-30}{2x}[/tex]

    Substitute into the first equation and simplify:

    [tex]3x\frac{x^2-30}{2x}+\frac{(x^2-30)^2}{4x^2}=-7[/tex]

    As x is not equal to 0, you can multiply the whole equation with 4x^2.

    [tex]6x^2(x^2-30)+(x^2-30)^2=-28x^2[/tex]

    By simplifying again, you get

    [tex]7x^4-212x^2+900=0[/tex]

    Solve for x^2.

    [tex]x^2=\frac{106\pm \sqrt{4936}}{7}[/tex]

    x1=5.02, y1=-0.48
    x2=-5.02, y2=0.48
    x3=2.26, y3=-5.51
    x4=-2.26, y4=5.51


    ehild
     
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