As expected, solving systems of linear equations can be extended into systems involving non linear equations such as quadratic equations. Examples of quadratic equations include parabolas, circles, ellipses, and hyperbolas. One important distinction between these systems and the previous linear examples is that due to the curved feature of quadratics, there is the possibility of more than one intersection point. Depending on the combination of equations, you can find up to four common solutions.

Graphing Systems of Non Linear Equations to Determine Solutions

A single intersection point can be seen when a circle and a line are tangent to each other.

One example of two intersection points is seen in the right example with the intersection of a cubic equation and a line.

Another example of two intersection points is seen in the intersection of a circle and a line, or the intersection of two circles.

One example of a system of equations with three intersection points is seen to the left with the intersection of a circle and a parabola. Notice, the vertex of the parabola must be tangent to the circle, otherwise there would be either two or four intersection points.

Finally, four intersection points can be observed with the intersection of a circle and an ellipse.

Each of these examples by no means exhausts the possibilities of intersection points. They simply illustrate the possibilities of up to four solutions to a non linear system of equations.

Solving Systems of Non Linear Equations Algebraically

Just as before, a more accurate way to determine common solutions to a system of equations is to algebraically solve the systems. Just as before, we can use the same tools to solve these systems: substitution and elimination. The only distinct difference is working with solving utilizing factoring and roots. These processes are not explained until Lesson 10 and Lesson 11 so be sure you understand those properties prior to continued progress in this section.

When solving a system with one quadratic equation and one linear equation, it is generally easiest to utilize substitution. This is because the variables must match up perfectly, and the presence of exponents can hinder elimination for this reason.

Lets check out an example of solving a system with a circle and a line. We can have either one or two solutions to this system.