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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 is a parabola and a circle.  According to the graph:

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

    ...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.

    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

    When y = 0:

      y = x2
      0 = x2
      0 = x

    When y = 3:

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

    Then the solution is 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 © 2006-2008 Elizabeth Stapel All Rights Reserved

      y = ± sqrt((35 - 3x^2)/2)

    You would plug this into your graphing calculator as two graphs, one 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 equation solves as:

      y = ± sqrt((4/3)x^2 - 8)

    Whatever format you use (the ellipse and the hyperbola, or the "plus-minus" equations), 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:

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

    Then:

      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

    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 alse be written as (± 3, ± 2), since all 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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