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Systems of Non-Linear Equations:
     Advanced Systems
(page 5 of 6)

  • Solve the system of nonlinear equations:
    • y = x2
      x2 + (y 2)2 = 4

    From the form of the equations, you should know that this system contains a parabola and a circle.

    According to the graph, there should be three solutions to this system:

     

     

    The solution at the origin is a "neat" one, but the other two intersection points may be messy.

    graph of y = x^2 and x^2 + (y - 2)^2 = 4

    From the first equation, I think I'll plug in "y" for "x2" in the second equation, and solve:

      x2 + (y 2)2 = 4
      y + (y 2)2 = 4
      y + (y2 4y + 4) = 4
      y 2 3y = 0
      y(y 3) = 0
      y = 0,  y = 3

    Now I need to find the corresponding x-values. When y = 0:

      y = x2
      0 = x2
      0 = x

    (This is the solution at the origin that we'd been expecting.) When y = 3:

      y = x2
      3 = x2
      sqrt(3) = x

    Then the solutions are the points (-sqrt(3), 3), (0, 0), and (sqrt(3), 3).

  • Solve the following system:
    • 3x2 + 2y2 = 35
      4x2 3y2 = 24

    I can rearrange the first equation to get:

      (x^2)/(35/3) + (y^2)/(35/2) = 1

    This tells me that the first equation is an ellipse. However, rather than graphing this using ellipse formulae, you could also solve to get a "plus-minus" expression that you can graph as two equations:  Copyright 2002-2011 Elizabeth Stapel All Rights Reserved

      y =  sqrt((35 - 3x^2)/2)

    (You would plug this into your graphing calculator as two graphs, one graph for the top "plus" part of the ellipse, and another for the bottom "minus" part.) The second equation rearranges as:

      (x^2)/6 - (y^2)/8 = 1

    ...which is an hyperbola. The second equation also solves (for your graphing calculator) as:

      y =  sqrt((4/3)x^2 - 8)

    Whatever format you use (the ellipse and the hyperbola center-vertex forms, or the "plus-minus" for-calculator forms), this system graphs as:

      graph of ellipse and hyperbola

    As you can see, there appear to be four solutions. To find them algebraically, I will choose to solve the second equation for x2 (rather than just x), and plug the resulting expression into the first equation, which I will then solve for y:. (It's okay that I "only" solve for x2, because neither equation has an x-term. There is no need, in this particular case, to do any more solving.)

      4x2 3y2 = 24
      4x2 = 3y2 + 24
      x2 = ( 3/4 )y2 + 6

    Then, subsituting into the first equation for the x2, I get:

      3x2 + 2y2 = 35
      3(( 3/4 )y2 + 6) + 2y2 = 35
      ( 9/4 )y2 + 18 + 2y2 = 35
      9y2 + 72 + 8y2 = 140
      17y2 = 68
      y2 = 4
      y = 2

       

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    When  y = 2:

      x2 = ( 3/4 )y2 + 6

        = ( 3/4 )(2)2 + 6

        = ( 3/4 )(4) + 6

        = 3 + 6 = 9

      x = 3

    When  y = 2:

      x2 = ( 3/4 )y2 + 6

        = ( 3/4 )(2)2 + 6

        = ( 3/4 )(4) + 6

        = 3 + 6 = 9

      x = 3

    Then the solution is the points (3, 2), (3, 2), (3, 2), and (3, 2), which may also be written as ( 3, 2), since all the "plus-minus" combinations are included.

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Cite this article as:

Stapel, Elizabeth. "Systems of Non-Linear Equations: Advanced Systems." Purplemath. Available
    from http://www.purplemath.com/modules/syseqgen5.htm. Accessed
 

 



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