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.

A General Note: 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.
Image described in main body

Figure 6

Example 3: Solving a System of Nonlinear Equations Representing a Circle and an Ellipse

Solve the system of nonlinear equations.

x2+y2=26(1)3x2+25y2=100(2)

Solution

Let’s begin by multiplying equation (1) by 3, and adding it to equation (2).

(3)(x2+y2)=(3)(26) 3x23y2=78 3x2+25y2=100 22y2=22

After we add the two equations together, we solve for y.

y2=1y=±1=±1

Substitute y=±1 into one of the equations and solve for x.

 x2+(1)2=26 x2+1=26 x2=25 x=±25=±5x2+(1)2=26 x2+1=26 x2=25=±5

There are four solutions: (5,1),(5,1),(5,1),and(5,1).

Circle intersected by ellipse at four points. Those points are negative five, one; five, one; five, negative one; and negative five, negative one.

Figure 7

Try It 3

Find the solution set for the given system of nonlinear equations.

4x2+y2=13x2+y2=10

Solution