# 11.3 Systems of nonlinear equations and inequalities: two variables  (Page 3/9)

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## Solving a system of nonlinear equations representing a circle and an ellipse

Solve the system of nonlinear equations.

$\begin{array}{rr}\hfill {x}^{2}+{y}^{2}=26\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \hfill \left(1\right)\\ \hfill 3{x}^{2}+25{y}^{2}=100& \hfill \left(2\right)\end{array}$

Let’s begin by multiplying equation (1) by $\text{\hspace{0.17em}}-3,\text{}$ and adding it to equation (2).

After we add the two equations together, we solve for $\text{\hspace{0.17em}}y.$

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

Substitute $\text{\hspace{0.17em}}y=±1\text{\hspace{0.17em}}$ into one of the equations and solve for $\text{\hspace{0.17em}}x.$

There are four solutions: $\text{\hspace{0.17em}}\left(5,1\right),\left(-5,1\right),\left(5,-1\right),\text{and}\text{\hspace{0.17em}}\left(-5,-1\right).\text{\hspace{0.17em}}$ See [link] .

Find the solution set for the given system of nonlinear equations.

$\begin{array}{c}4{x}^{2}+{y}^{2}=13\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+{y}^{2}=10\end{array}$

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

## Graphing a nonlinear inequality

All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality    is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality.

Recall that when the inequality is greater than, $\text{\hspace{0.17em}}y>a,$ or less than, $\text{\hspace{0.17em}}y the graph is drawn with a dashed line. When the inequality is greater than or equal to, $\text{\hspace{0.17em}}y\ge a,\text{}$ or less than or equal to, $\text{\hspace{0.17em}}y\le a,\text{}$ the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade. See [link] . (a) an example of   y > a ;   (b) an example of   y ≥ a ;   (c) an example of   y < a ;   (d) an example of   y ≤ a

Given an inequality bounded by a parabola, sketch a graph.

1. Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set.
2. If the boundary is included in the region (the operator is $\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\ge$ ), the parabola is graphed as a solid line.
3. If the boundary is not included in the region (the operator is<or>), the parabola is graphed as a dashed line.
4. Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line.
5. Shade the region representing the solution set.

## Graphing an inequality for a parabola

Graph the inequality $\text{\hspace{0.17em}}y>{x}^{2}+1.$

First, graph the corresponding equation $\text{\hspace{0.17em}}y={x}^{2}+1.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}y>{x}^{2}+1\text{\hspace{0.17em}}$ has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points
$\left(0,2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,0\right).\text{\hspace{0.17em}}$ One point is clearly inside the parabola and the other point is clearly outside.

$\begin{array}{ll}y>{x}^{2}+1\hfill & \hfill \\ 2>{\left(0\right)}^{2}+1\hfill & \hfill \\ 2>1\hfill & \text{True}\hfill \\ \hfill & \hfill \\ \hfill & \hfill \\ \hfill & \hfill \\ 0>{\left(2\right)}^{2}+1\hfill & \hfill \\ 0>5\hfill & \text{False}\hfill \end{array}$

The graph is shown in [link] . We can see that the solution set consists of all points inside the parabola, but not on the graph itself.

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