# 11.3 Systems of nonlinear equations and inequalities: two variables  (Page 2/9)

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Could we have substituted values for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into the second equation to solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in [link] ?

Yes, but because $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is squared in the second equation this could give us extraneous solutions for $\text{\hspace{0.17em}}x.$

For $\text{\hspace{0.17em}}y=1$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={x}^{2}+1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={x}^{2}+1\hfill \\ \text{\hspace{0.17em}}{x}^{2}=0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±\sqrt{0}=0\hfill \end{array}$

This gives us the same value as in the solution.

For $\text{\hspace{0.17em}}y=2$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={x}^{2}+1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2={x}^{2}+1\hfill \\ {x}^{2}=1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±\sqrt{1}=±1\hfill \end{array}$

Notice that $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ is an extraneous solution.

Solve the given system of equations by substitution.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x-y=-2\hfill \\ 2{x}^{2}-y=0\hfill \end{array}$

$\left(-\frac{1}{2},\frac{1}{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,8\right)$

## Intersection of a circle and a line

Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.

## Possible types of solutions for the points of intersection of a circle and a line

[link] illustrates possible solution sets for a system of equations involving a circle and a line.

• No solution. The line does not intersect the circle.
• One solution. The line is tangent to the circle and intersects the circle at exactly one point.
• Two solutions. The line crosses the circle and intersects it at two points.

Given a system of equations containing a line and a circle, find the solution.

1. Solve the linear equation for one of the variables.
2. Substitute the expression obtained in step one into the equation for the circle.
3. Solve for the remaining variable.
4. Check your solutions in both equations.

## Finding the intersection of a circle and a line by substitution

Find the intersection of the given circle and the given line by substitution.

$\begin{array}{l}{x}^{2}+{y}^{2}=5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=3x-5\hfill \end{array}$

One of the equations has already been solved for $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ We will substitute $\text{\hspace{0.17em}}y=3x-5\text{\hspace{0.17em}}$ into the equation for the circle.

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+{\left(3x-5\right)}^{2}=5\\ {x}^{2}+9{x}^{2}-30x+25=5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}10{x}^{2}-30x+20=0\end{array}$

Now, we factor and solve for $\text{\hspace{0.17em}}x.$

$\begin{array}{l}\text{\hspace{0.17em}}10\left({x}^{2}-3x+2\right)=0\hfill \\ 10\left(x-2\right)\left(x-1\right)=0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=2\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=1\hfill \end{array}$

Substitute the two x -values into the original linear equation to solve for $\text{\hspace{0.17em}}y.$

$\begin{array}{l}y=3\left(2\right)-5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1\hfill \\ \hfill \\ \hfill \\ \hfill \\ y=3\left(1\right)-5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\hfill \end{array}$

The line intersects the circle at $\text{\hspace{0.17em}}\left(2,1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,-2\right),$ which can be verified by substituting these $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ values into both of the original equations. See [link] .

Solve the system of nonlinear equations.

$\begin{array}{l}{x}^{2}+{y}^{2}=10\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-3y=-10\hfill \end{array}$

$\left(-1,3\right)$

## 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

[link] 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.

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