Algebraic Methods for Solving Systems of Non-Linear Equations
Now, we will explore graphical and algebraic methods for finding the solution to a system of equations containing combinations of different equations, including linear, quadratic, and circles.
Solving a System of Nonlinear Equations Using Elimination
We
have seen that substitution is often the preferred method when a system
of equations includes a linear equation and a nonlinear equation.
However, when both equations in the system have like variables of the
second degree, solving them using elimination by addition is often
easier than substitution. Generally, elimination is a far simpler method
when the system involves only two equations in two variables (a
two-by-two system), rather than a three-by-three system, as there are
fewer steps. As an example, we will investigate the possible types of
solutions when solving a system of equations representing a circle and
an ellipse.
Possible Types of Solutions for the Points of Intersection of a Circle and an Ellipse
Figure 6 illustrates possible solution sets for a system of equations involving a circle and an ellipse.
- No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other.
- One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point.
- Two solutions. The circle and the ellipse intersect at two points.
- Three solutions. The circle and the ellipse intersect at three points.
- Four solutions. The circle and the ellipse intersect at four points.
Figure 6
Example 3
Solving a System of Nonlinear Equations Representing a Circle and an Ellipse
Solve the system of nonlinear equations.
Solution
Let's begin by multiplying equation (1) by −3, and adding it to equation (2).
After we add the two equations together, we solve for .
Substitute into one of the equations and solve for .
There are four solutions: , and . See Figure 7.
Figure 7
Try It #3
Find the solution set for the given system of nonlinear equations.
4x2+y2x2+y2==1310