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.

Figure 6

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

Solve the system of nonlinear equations.

\begin{matrix} x^{2} + y^{2} = 26 & (1) 3 x^{2} + 25 y^{2} = 100 & (2) \end{matrix}

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

Solution

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

\frac{\begin{matrix} \begin{matrix} (- 3) (x^{2} + y^{2}) = (- 3) (26) - 3 x^{2} - 3 y^{2} = - 78 \end{matrix} 3 x^{2} + 25 y^{2} = 100 \end{matrix}}{22 y^{2} = 22}

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

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

\begin{matrix} y^{2} = 1 y = \pm \sqrt{1} = \pm 1 \end{matrix}

$\begin{array}{l}{y}^{2}=1\hfill \\ y=\pm \sqrt{1}=\pm 1\hfill \end{array}$

Substitute $y=\pm 1$ into one of the equations and solve for $x$ .

\begin{matrix} x^{2} + {(1)}^{2} = 26 x^{2} + 1 = 26 x^{2} = 25 x = \pm \sqrt{25} = \pm 5 x^{2} + {(- 1)}^{2} = 26 x^{2} + 1 = 26 x^{2} = 25 = \pm 5 \end{matrix}

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

There are four solutions: $\left(5,1\right),\left(-5,1\right),\left(5,-1\right),\text{and}\left(-5,-1\right)$ .

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.

\begin{matrix} 4 x^{2} + y^{2} = 13 x^{2} + y^{2} = 10 \end{matrix}

$\begin{array}{c}4{x}^{2}+{y}^{2}=13\\ {x}^{2}+{y}^{2}=10\end{array}$

Solution

Systems of Nonlinear Equations and Inequalities: Two Variables