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

Figure 6


Example 3

Solving a System of Nonlinear Equations Representing a Circle and an Ellipse

Solve the system of nonlinear equations.

x^2+y^2=26 (1)

3x^2+25y^2=100 (2)

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

(−3)(x^2+y^2)=(−3)(26)

−3x^2−3y^2=−78

3x2+25y^2=100

22y^2=22

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

y^2=1

y=±\sqrt{1}=±1

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

x^2+(1)^2=26

x^2+1=26

x^2=25

x=±\sqrt{25}=±5

x^2+(−1)^2=26

 x^2+1=26

 x^2=25=±5

There are four solutions: (5,1),(−5,1),(5,−1), and (−5,−1). See Figure 7.

Figure 7

Figure 7


Try It #3

Find the solution set for the given system of nonlinear equations.
4x2+y2x2+y2==1310