194 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.

[latex]\begin{array}{rr}\hfill {x}^{2}+{y}^{2}=26& \hfill \left(1\right)\\ \hfill 3{x}^{2}+25{y}^{2}=100& \hfill \left(2\right)\end{array}[/latex]

Solution

Let’s begin by multiplying equation (1) by [latex]-3,\text{}[/latex] and adding it to equation (2).

[latex]\frac{\begin{array}{l}\begin{array}{l}\hfill \\ \left(-3\right)\left({x}^{2}+{y}^{2}\right)=\left(-3\right)\left(26\right)\hfill \\ \text{ }-3{x}^{2}-3{y}^{2}=-78\hfill \end{array}\hfill \\ \text{ }3{x}^{2}+25{y}^{2}=100\hfill \end{array}}{\text{ }22{y}^{2}=22}[/latex]

After we add the two equations together, we solve for [latex]y[/latex].

[latex]\begin{array}{l}{y}^{2}=1\hfill \\ y=\pm \sqrt{1}=\pm 1\hfill \end{array}[/latex]

Substitute [latex]y=\pm 1[/latex] into one of the equations and solve for [latex]x[/latex].

[latex]\begin{array}{l}\text{ }{x}^{2}+{\left(1\right)}^{2}=26\hfill \\ \text{ }{x}^{2}+1=26\hfill \\ \text{ }{x}^{2}=25\hfill \\ \text{ }x=\pm \sqrt{25}=\pm 5\hfill \\ \hfill \\ {x}^{2}+{\left(-1\right)}^{2}=26\hfill \\ \text{ }{x}^{2}+1=26\hfill \\ \text{ }{x}^{2}=25=\pm 5\hfill \end{array}[/latex]

There are four solutions: [latex]\left(5,1\right),\left(-5,1\right),\left(5,-1\right),\text{and}\left(-5,-1\right)[/latex].

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.

[latex]\begin{array}{c}4{x}^{2}+{y}^{2}=13\\ {x}^{2}+{y}^{2}=10\end{array}[/latex]

Solution

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