Systems
of NonLinear Equations: The basic process for solving more complicated systems of nonlinear equations remains the same as for the previous systems; namely, solve one of the equations for one of the variables, plug that information into the other equation, and solve the resulting onevariable equation. In "real life", advanced systems of nonlinear equations probably won't be algebraically solvable. But you can safely expect that the systems they give you in your algebra class will be solvable. However, you may end up needing to use the Quadratic Formula in new ways, and your solutions may involve square roots.
y
= x^{2} From the form of the equations, you should know that this system contains a parabola and a circle.
From the first equation, I think I'll plug in "y" for "x^{2}" in the second equation, and solve: x^{2}
+ (y 2)^{2} = 4 Now I need to find the corresponding xvalues. When y = 0: y = x^{2}0 = x^{2}0 = x (This is the solution at the origin that we'd been expecting.) When y = 3: y = x^{2}3 = x^{2}sqrt(3) = x Then the solutions are the points , (0, 0), and .
3x^{2}
+ 2y^{2} = 35 I can rearrange the first equation to get: 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 "plusminus" expression that you can graph as two equations: Copyright 20022011 Elizabeth Stapel All Rights Reserved (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: ...which is an hyperbola. The second equation also solves (for your graphing calculator) as: Whatever format you use (the ellipse and the hyperbola centervertex forms, or the "plusminus" forcalculator forms), this system graphs as: As you can see, there appear to be four solutions. To find them algebraically, I will choose to solve the second equation for x^{2} (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 x^{2}, because neither equation has an xterm. There is no need, in this particular case, to do any more solving.) 4x^{2}
3y^{2} = 24 Then, subsituting into the first equation for the x^{2}, I get: 3x^{2}
+ 2y^{2} = 35 When y = 2, I get: x^{2} = (^{ 3}/_{4} )y^{2} + 6 = (^{ 3}/_{4} )(2)^{2} + 6 = (^{ 3}/_{4} )(4) + 6 = 3 + 6 = 9 x = 3 When y = 2, I get: x^{2} = (^{ 3}/_{4} )y^{2} + 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 "plusminus" combinations are included. << Previous Top  1  2  3  4  5  6  Return to Index Next >>



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