Systems of Nonlinear Equations and Inequalities: Two Variables - College Algebra with Corequisite Support

7.3 Systems of Nonlinear Equations and Inequalities: Two Variables

Learning Objectives

In this section, you will:

Solve a system of nonlinear equations using substitution.
Solve a system of nonlinear equations using elimination.
Graph a nonlinear inequality.
Graph a system of nonlinear inequalities.

Corequisite Skills

Learning Objectives

Graph a parabola (IA 11.2.1)
Graph a circle (IA 11.1.4)

Objective 1: Graph a parabola (IA 11.2.1)

A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Properties of parabolas

This figure shows a parabola opening upwards. Below the parabola is a horizontal line labeled directrix. A vertical dashed line through the center of the parabola is labeled axis of symmetry. The point where the axis intersects the parabola is labeled vertex. A point on the axis, within the parabola is labeled focus. A line perpendicular to the directrix connects the directrix to a point on the parabola and another line connects this point to the focus. Both these lines are of the same length.

Vertical Parabolas
	General form $y = a x^{2} + b x + c$	Standard form $y = a {(x - h)}^{2} + k$
Orientation	$a > 0$ up; $a < 0$ down	$a > 0$ up; $a < 0$ down
Axis of symmetry	$x = - \frac{b}{2 a}$	$x = h$
Vertex	Substitute $x = - \frac{b}{2 a}$ and solve for y.	$(h, k)$
y-intercept	Let $x = 0$	Let $x = 0$
x-intercepts	Let $y = 0$	Let $y = 0$

Graphs of parabolas

This figure shows two parabolas with axis x equals h and vertex h, k. The one on the left opens up and A is greater than 0. The one on the right opens down. Here A is less than 0.

How To

Graphing parabolas using properties.

Step 1. Determine whether the parabola opens upward or downward.
Step 2. Find the axis of symmetry.
Step 3. Find the vertex.
Step 4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
Step 5. Find the x-intercepts.
Step 6. Graph the parabola.

Example 1

Graph the parabola $y = - x^{2} + 4 x - 3$

Solution

Standard form	$y = - x^{2} + 4 x - 3 a = - 1, b = 4, c = - 3$
Step 1	Since $a = - 1$ the parabola opens downward
Step 2	The axis of symmetry is given by $x = - \frac{b}{2 a}$ , $x = - \frac{4}{2 (- 1)} = 2$ The axis of symmetry is $x = 2$
Step 3	The vertex is on the line $x = 2$ Let's substitute $x = 2$ into $y = - x^{2} + 4 x - 3$ $\begin{matrix} y = - {(2)}^{2} + 4 (2) - 3 \\ y = - 4 + 8 - 3 \\ y = 1 \end{matrix}$ The vertex is the point (2, 1)
Step 4	To find y-intercept, substitute $x = 0$ into $y = - x^{2} + 4 x - 3$ $\begin{matrix} y = - {(0)}^{2} + 4 (0) - 3 \\ y = - 3 \end{matrix}$ The y-intercept is the point (0, -3) and is it 2 units to the left of the vertex. The symmetric point is 2 units to the right of the vertex and is (4, -3)
Step 5	To find the x-intercepts, substitute $y = 0$ into $y = - x^{2} + 4 x - 3$ and solve for $x$ $\begin{matrix} 0 = - x^{2} + 4 x - 3 \\ 0 = - (x^{2} - 4 x + 3) \\ 0 = x^{2} - 4 x + 3 \\ 0 = (x - 3) (x - 1) \\ x = 3 o r x = 1 \end{matrix}$ The x-intercepts are (3, 0) and (1, 0)
Step 6	Graph the parabola Axis of symmetry $x = 2$ Vertex: (2,1) y-intercept: (0, -3), symmetric point: (4, -3) x-intercepts: (3, 0) and (1, 0)

Practice Makes Perfect

Graph the parabola $y = 2 x^{2} + 4 x + 6$

A blank coordinate plot ranging from -10 to 10

Objective 2: Graph a circle (IA 11.1.4)

Any equation of the form ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is the standard form of the equation of a circle with center, (h,k) and radius. We can then graph the circle on a rectangular coordinate system using the center and radius.

Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.

Example 2

Graph a circle

ⓐ Find the center and radius and then graph the circle ${(x + 3)}^{2} + {(y + 4)}^{2} = 4$
ⓑ Find the center and radius and then graph the circle $x^{2} + y^{2} - 6 x - 8 y + 9 = 0$

Solution

ⓐ

Use the standard form of the equation of a circle. Identify the center, (h,k) and radius, r.	$\begin{matrix} {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ {(x + 3)}^{2} + {(y + 4)}^{2} = 4 \\ (x - {(- 3))}^{2} + (y - {(- 4))}^{2} = 2^{2} \end{matrix}$ Center: (-3, -4) Radius: 2
Graph the circle

ⓑ

We need to rewrite this general form into standard form in order to find the center and radius.

Step 1 Group the x-terms and y-terms. Collect the constants on the right side.	$\begin{matrix} x^{2} + y^{2} - 6 x - 8 y + 9 = 0 \\ x^{2} - 6 x + y^{2} - 8 y = - 9 \end{matrix}$
Step 2 Complete the squares	$\begin{matrix} x^{2} - 6 x + {(\frac{6}{2})}^{2} + y^{2} - 8 y + {(\frac{8}{2})}^{2} = - 9 + {(\frac{6}{2})}^{2} + {(\frac{8}{2})}^{2} \\ x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = - 9 + 9 + 16 \end{matrix}$
Step 3 Rewrite as binomial squares.	${(x - 3)}^{2} + {(y - 4)}^{2} = 16$
Step 4 Identify the center and radius.	Center: (3,4) Radius: 4
Step 5 Graph the circle.

Practice Makes Perfect

Graph a circle.

Find the center and radius and then graph the circle ${(x - 3)}^{2} + {(y + 4)}^{2} = 25$

Use the standard form of the equation of a circle. Identify the center, (h, k) and radius, r.
Graph the circle

Find the center and radius and then graph the circle $x^{2} + y^{2} + 12 x - 14 y + 21 = 0$

Step 1 Group the x-terms and y-terms. Collect the constants on the right side.
Step 2 Complete the squares	________________________________________
Step 3 Rewrite as binomial squares.	________________________________________
Step 4 Identify the center and radius.	Center: ________ Radius: ________
Step 5 Graph the circle.

Halley’s Comet (Figure 1) orbits the sun about once every 75 years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear.

A striking image of a brilliant comet streaking across a deep blue, star-studded sky. The comet's nucleus is intensely bright, fanning out into a long, glowing tail that gradually diffuses into the darkness, with numerous small white stars scattered throughout the celestial backdrop.

Figure 1 Halley’s Comet (credit: "NASA Blueshift"/Flickr)

In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations.

Solving a System of Nonlinear Equations Using Substitution

A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form $A x + B y + C = 0.$ Any equation that cannot be written in this form is nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.

Intersection of a Parabola and a Line

There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.

Possible Types of Solutions for Points of Intersection of a Parabola and a Line

Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.

No solution. The line will never intersect the parabola.
One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.
Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.

Three graphs illustrate solutions for a system of a linear and a quadratic equation: (a) no solutions, (b) one solution (tangent), and (c) two distinct solutions.

Figure 2

How To

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

Solve the linear equation for one of the variables.
Substitute the expression obtained in step one into the parabola equation.
Solve for the remaining variable.
Check your solutions in both equations.

Example 1

Solving a System of Nonlinear Equations Representing a Parabola and a Line

Solve the system of equations.

\begin{array}{l} x - y = −1 \\ y = x^{2} + 1 \end{array}

Solution

Solve the first equation for $x$ and then substitute the resulting expression into the second equation.

\begin{array}{l} x - y = −1 \\ x = y −1 & Solve for x . \\ y = x^{2} + 1 \\ y = {(y −1)}^{2} + 1 & Substitute expression for x . \end{array}

Expand the equation and set it equal to zero.

\begin{array}{l} y = {(y −1)}^{2} + 1 \\ = (y^{2} −2 y + 1) + 1 \\ = y^{2} −2 y + 2 \\ 0 = y^{2} −3 y + 2 \\ = (y −2) (y −1) \end{array}

Solving for $y$ gives $y = 2$ and $y = 1.$ Next, substitute each value for $y$ into the first equation to solve for $x .$ Always substitute the value into the linear equation to check for extraneous solutions.

\begin{array}{l} x - y = −1 \\ x - (2) = −1 \\ x = 1 \\ x - (1) = −1 \\ x = 0 \end{array}

The solutions are $(1, 2)$ and $(0, 1),$ which can be verified by substituting these $(x, y)$ values into both of the original equations. See Figure 3.

A graph showing the intersection of the parabola y = x^2 + 1 (red) and the line x - y = -1 (blue) at points (0, 1) and (1, 2).

Figure 3

Q&A

Could we have substituted values for $y$ into the second equation to solve for $x$ in Example 1?

Yes, but because $x$ is squared in the second equation this could give us extraneous solutions for $x .$

For $y = 1$

\begin{array}{l} y = x^{2} + 1 \\ 1 = x^{2} + 1 \\ x^{2} = 0 \\ x = \pm \sqrt{0} = 0 \end{array}

This gives us the same value as in the solution.

For $y = 2$

\begin{array}{l} y = x^{2} + 1 \\ 2 = x^{2} + 1 \\ x^{2} = 1 \\ x = \pm \sqrt{1} = \pm 1 \end{array}

Notice that $−1$ is an extraneous solution.

Try It #1

Solve the given system of equations by substitution.

\begin{matrix} 3 x - y & = & -2 \\ 2 x^{2} - y & = & 0 \end{matrix}

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

Figure 4 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.

An illustration showing the three possible numbers of intersection points between a line and a circle: zero (no solutions), one (a tangent line), or two (a secant line).

Figure 4

How To

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

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

Example 2

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 \\ y = 3 x −5 \end{array}

Solution

One of the equations has already been solved for $y .$ We will substitute $y = 3 x −5$ into the equation for the circle.

\begin{matrix} x^{2} + {(3 x −5)}^{2} = 5 \\ x^{2} + 9 x^{2} −30 x + 25 = 5 \\ 10 x^{2} −30 x + 20 = 0 \end{matrix}

Now, we factor and solve for $x .$

\begin{array}{l} 10 (x^{2} - 3 x + 2) = 0 \\ 10 (x - 2) (x - 1) = 0 \\ x = 2 \\ x = 1 \end{array}

Substitute the two x-values into the original linear equation to solve for $y .$

\begin{array}{l} y = 3 (2) −5 \\ = 1 \\ y = 3 (1) −5 \\ = −2 \end{array}

The line intersects the circle at $(2, 1)$ and $(1, −2),$ which can be verified by substituting these $(x, y)$ values into both of the original equations. See Figure 5.

A red circle with equation x^2 + y^2 = 5 and a blue line with equation y = 3x - 5 on a Cartesian plane. The line intersects the circle at two points: (2, 1) and (1, -2).

Figure 5

Try It #2

Solve the system of nonlinear equations.

\begin{matrix} x^{2} + y^{2} & = & 10 \\ x - 3 y & = & -10 \end{matrix}

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

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.

A diagram illustrates the possible number of solutions (intersections) between an ellipse (blue) and a circle (red), ranging from zero to four, depending on their relative positions and sizes.

Figure 6

Example 3

Solving a System of Nonlinear Equations Representing a Circle and an Ellipse

Solve the system of nonlinear equations.

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

Solution

Let’s begin by multiplying equation (1) by $−3,$ and adding it to equation (2).

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

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

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

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

\begin{array}{l} 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{array}

There are four solutions: $(5, 1), (−5, 1), (5, −1), and (−5, −1) .$ See Figure 7.

A graph displays a blue circle (x^2+y^2=25) and a red horizontal ellipse centered at the origin. The circle has a radius of 5. The ellipse intersects the circle at points (-5,1), (-5,-1), (5,1), and (5,-1).

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}

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, $y > a,$ or less than, $y < a,$ the graph is drawn with a dashed line. When the inequality is greater than or equal to, $y \geq a,$ or less than or equal to, $y \leq a,$ 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 Figure 8.

This figure displays four graphs, each showing an inequality related to the parabola y = x^2 - 4. Graph (a) illustrates y > x^2 - 4 with a dashed parabolic boundary and the region above the parabola shaded in purple. Graph (b) shows y ≥ x^2 - 4 with a solid parabolic boundary and the region above the parabola shaded in orange. Graph (c) represents y < x^2 - 4 with a dashed parabolic boundary and the region below the parabola shaded in teal. Graph (d) depicts y ≤ x^2 - 4 with a solid parabolic boundary and the region below the parabola shaded in red.

Figure 8 (a) an example of

y > a;

(b) an example of

y \geq a;

y < a;

(d) an example of

y \leq a

How To

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

Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set.
If the boundary is included in the region (the operator is $\leq$ or $\geq$ ), the parabola is graphed as a solid line.
If the boundary is not included in the region (the operator is < or >), the parabola is graphed as a dashed line.
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.
Shade the region representing the solution set.

Example 4

Graphing an Inequality for a Parabola

Graph the inequality $y > x^{2} + 1.$

Solution

First, graph the corresponding equation $y = x^{2} + 1.$ Since $y > x^{2} + 1$ 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
$(0, 2)$ and $(2, 0) .$ One point is clearly inside the parabola and the other point is clearly outside.

\begin{array}{l} y > x^{2} + 1 \\ 2 > {(0)}^{2} + 1 \\ 2 > 1 & True \\ 0 > {(2)}^{2} + 1 \\ 0 > 5 & False \end{array}

The graph is shown in Figure 9. We can see that the solution set consists of all points inside the parabola, but not on the graph itself.

A graph illustrating the inequality y > x^2 + 1, where the region above a dashed parabola with vertex (0, 1) is shaded. Points (0, 2) and (2, 0) are labeled on the graph.

Figure 9

Graphing a System of Nonlinear Inequalities

Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region.

How To

Given a system of nonlinear inequalities, sketch a graph.

Find the intersection points by solving the corresponding system of nonlinear equations.
Graph the nonlinear equations.
Find the shaded regions of each inequality.
Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality.

Example 5

Graphing a System of Inequalities

Graph the given system of inequalities.

\begin{array}{r} x^{2} - y \leq 0 \\ 2 x^{2} + y \leq 12 \end{array}

Solution

These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable $y$ will be eliminated. Then we solve for $x .$

\begin{array}{l} \underset{____________}{\begin{matrix} \begin{array}{l} x^{2} - y = 0 \end{array} \\ 2 x^{2} + y = 12 \end{matrix}} \\ 3 x^{2} = 12 \\ x^{2} = 4 \\ x = \pm 2 \end{array}

Substitute the x-values into one of the equations and solve for $y .$

\begin{array}{r} x^{2} - y = 0 \\ {(2)}^{2} - y = 0 \\ 4 - y = 0 \\ y = 4 \\ {(−2)}^{2} - y = 0 \\ 4 - y = 0 \\ y = 4 \end{array}

The two points of intersection are $(2, 4)$ and $(−2, 4) .$ Notice that the equations can be rewritten as follows.

\begin{array}{l} x^{2} - y \leq 0 \\ x^{2} \leq y \\ y \geq x^{2} \\ 2 x^{2} + y \leq 12 \\ y \leq −2 x^{2} + 12 \end{array}

Graph each inequality. See Figure 10. The feasible region is the region between the two equations bounded by $2 x^{2} + y \leq 12$ on the top and $x^{2} - y \leq 0$ on the bottom.

A Cartesian coordinate system displays two parabolas. A red parabola opens upwards from the origin (0,0). A blue parabola opens downwards from its vertex at (0,12). The parabolas intersect at two points, marked with black dots and labeled as (-2,4) and (2,4). Shaded regions indicate the areas defined by the parabolas: the area between the two curves is shaded gray, the region above the red parabola and outside the blue parabola is shaded light red, and the region below the blue parabola and outside the red parabola is shaded light blue.

Figure 10

Try It #4

Graph the given system of inequalities.

\begin{matrix} y & \geq & x^{2} - 1 \\ x - y & \geq & - 1 \end{matrix}

Media

Access these online resources for additional instruction and practice with nonlinear equations.

7.3 Section Exercises

Verbal

Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers.

When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution?

When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region?

If you graph a revenue and cost function, explain how to determine in what regions there is profit.

If you perform your break-even analysis and there is more than one solution, explain how you would determine which x-values are profit and which are not.

Algebraic

For the following exercises, solve the system of nonlinear equations using substitution.

$\begin{array}{l} x + y = 4 \\ x^{2} + y^{2} = 9 \end{array}$

$\begin{array}{l} y = x −3 \\ x^{2} + y^{2} = 9 \end{array}$

$\begin{array}{l} y = x \\ x^{2} + y^{2} = 9 \end{array}$

$\begin{array}{l} y = - x \\ x^{2} + y^{2} = 9 \end{array}$

10.

$\begin{array}{l} x = 2 \\ x^{2} - y^{2} = 9 \end{array}$

For the following exercises, solve the system of nonlinear equations using elimination.

11.

$\begin{array}{l} 4 x^{2} −9 y^{2} = 36 \\ 4 x^{2} + 9 y^{2} = 36 \end{array}$

12.

$\begin{array}{l} x^{2} + y^{2} = 25 \\ x^{2} - y^{2} = 1 \end{array}$

13.

$\begin{array}{l} 2 x^{2} + 4 y^{2} = 4 \\ 2 x^{2} −4 y^{2} = 25 x −10 \end{array}$

14.

$\begin{array}{l} y^{2} - x^{2} = 9 \\ 3 x^{2} + 2 y^{2} = 8 \end{array}$

15.

$\begin{array}{l} x^{2} + y^{2} + \frac{1}{16} = 2500 \\ y = 2 x^{2} \end{array}$

For the following exercises, use any method to solve the system of nonlinear equations.

16.

$\begin{array}{l} −2 x^{2} + y = −5 \\ 6 x - y = 9 \end{array}$

17.

$\begin{array}{l} - x^{2} + y = 2 \\ - x + y = 2 \end{array}$

18.

$\begin{array}{l} x^{2} + y^{2} = 1 \\ y = 20 x^{2} −1 \end{array}$

19.

$\begin{array}{l} x^{2} + y^{2} = 1 \\ y = - x^{2} \end{array}$

20.

$\begin{array}{l} 2 x^{3} - x^{2} = y \\ y = \frac{1}{2} - x \end{array}$

21.

$\begin{array}{l} 9 x^{2} + 25 y^{2} = 225 \\ {(x −6)}^{2} + y^{2} = 1 \end{array}$

22.

$\begin{array}{l} x^{4} - x^{2} = y \\ x^{2} + y = 0 \end{array}$

23.

$\begin{array}{l} 2 x^{3} - x^{2} = y \\ x^{2} + y = 0 \end{array}$

For the following exercises, use any method to solve the nonlinear system.

24.

$\begin{array}{l} x^{2} + y^{2} = 9 \\ y = 3 - x^{2} \end{array}$

25.

$\begin{array}{l} x^{2} - y^{2} = 9 \\ x = 3 \end{array}$

26.

$\begin{array}{l} x^{2} - y^{2} = 9 \\ y = 3 \end{array}$

27.

$\begin{array}{l} x^{2} - y^{2} = 9 \\ x - y = 0 \end{array}$

28.

$\begin{array}{l} - x^{2} + y = 2 \\ −4 x + y = −1 \end{array}$

29.

$\begin{array}{l} - x^{2} + y = 2 \\ 2 y = - x \end{array}$

30.

$\begin{array}{l} x^{2} + y^{2} = 25 \\ x^{2} - y^{2} = 36 \end{array}$

31.

$\begin{array}{l} x^{2} + y^{2} = 1 \\ y^{2} = x^{2} \end{array}$

32.

$\begin{array}{l} 16 x^{2} −9 y^{2} + 144 = 0 \\ y^{2} + x^{2} = 16 \end{array}$

33.

$\begin{array}{l} 3 x^{2} - y^{2} = 12 \\ {(x −1)}^{2} + y^{2} = 1 \end{array}$

34.

$\begin{array}{l} 3 x^{2} - y^{2} = 12 \\ {(x −1)}^{2} + y^{2} = 4 \end{array}$

35.

$\begin{array}{l} 3 x^{2} - y^{2} = 12 \\ x^{2} + y^{2} = 16 \end{array}$

36.

$\begin{array}{l} x^{2} - y^{2} - 6 x - 4 y - 11 = 0 \\ - x^{2} + y^{2} = 5 \end{array}$

37.

$\begin{array}{l} x^{2} + y^{2} −6 y = 7 \\ x^{2} + y = 1 \end{array}$

38.

$\begin{array}{l} x^{2} + y^{2} = 6 \\ x y = 1 \end{array}$

Graphical

For the following exercises, graph the inequality.

39.

$x^{2} + y < 9$

40.

$x^{2} + y^{2} < 4$

For the following exercises, graph the system of inequalities. Label all points of intersection.

41.

$\begin{array}{l} x^{2} + y < 1 \\ y > 2 x \end{array}$

42.

$\begin{array}{l} x^{2} + y < −5 \\ y > 5 x + 10 \end{array}$

43.

$\begin{array}{l} x^{2} + y^{2} < 25 \\ 3 x^{2} - y^{2} > 12 \end{array}$

44.

$\begin{array}{l} x^{2} - y^{2} > −4 \\ x^{2} + y^{2} < 12 \end{array}$

45.

$\begin{array}{l} x^{2} + 3 y^{2} > 16 \\ 3 x^{2} - y^{2} < 1 \end{array}$

Extensions

For the following exercises, graph the inequality.

46.

$\begin{array}{l} y \geq e^{x} \\ y \leq \ln (x) + 5 \end{array}$

47.

$\begin{array}{l} y \leq - \log (x) \\ y \leq e^{x} \end{array}$

For the following exercises, find the solutions to the nonlinear equations with two variables.

48.

$\begin{array}{l} \frac{4}{x^{2}} + \frac{1}{y^{2}} = 24 \\ \frac{5}{x^{2}} - \frac{2}{y^{2}} + 4 = 0 \end{array}$

49.

$\begin{matrix} \frac{6}{x^{2}} - \frac{1}{y^{2}} = 8 \\ \frac{1}{x^{2}} - \frac{6}{y^{2}} = \frac{1}{8} \end{matrix}$

50.

$\begin{array}{l} x^{2} - x y + y^{2} −2 = 0 \\ x + 3 y = 4 \end{array}$

51.

$\begin{array}{l} x^{2} - x y −2 y^{2} −6 = 0 \\ x^{2} + y^{2} = 1 \end{array}$

52.

$\begin{array}{l} x^{2} + 4 x y −2 y^{2} −6 = 0 \\ x = y + 2 \end{array}$

Technology

For the following exercises, solve the system of inequalities. Use a calculator to graph the system to confirm the answer.

53.

$\begin{array}{l} x y < 1 \\ y > \sqrt{x} \end{array}$

54.

$\begin{array}{l} x^{2} + y < 3 \\ y > 2 x \end{array}$

Real-World Applications

For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions.

55.

Two numbers add up to 300. One number is twice the square of the other number. What are the numbers?

56.

The squares of two numbers add to 360. The second number is half the value of the first number squared. What are the numbers?

57.

A laptop company has discovered their cost and revenue functions for each day: $C (x) = 3 x^{2} −10 x + 200$ and $R (x) = −2 x^{2} + 100 x + 50.$ If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit.

58.

A cell phone company has the following cost and revenue functions: $C (x) = 8 x^{2} −600 x + 21,500$ and $R (x) = −3 x^{2} + 480 x .$ What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit.