Solve Systems of Nonlinear Equations

Solve a System of Nonlinear Equations Using Graphing

We learned how to solve systems of linear equations with two variables by graphing, substitution and elimination. We will be using these same methods as we look at nonlinear systems of equations with two equations and two variables. A system of nonlinear equations is a system where at least one of the equations is not linear.

For example each of the following systems is a system of nonlinear equations.

\begin{matrix} {\begin{matrix} x^{2} + y^{2} = 9 \\ x^{2} - y = 9 \end{matrix} & {\begin{matrix} 9 x^{2} + y^{2} = 9 \\ y = 3 x - 3 \end{matrix} & {\begin{matrix} x + y = 4 \\ y = x^{2} + 2 \end{matrix} \end{matrix}

Just as with systems of linear equations, a solution of a nonlinear system is an ordered pair that makes both equations true. In a nonlinear system, there may be more than one solution. We will see this as we solve a system of nonlinear equations by graphing.

When we solved systems of linear equations, the solution of the system was the point of intersection of the two lines. With systems of nonlinear equations, the graphs may be circles, parabolas or hyperbolas and there may be several points of intersection, and so several solutions. Once you identify the graphs, visualize the different ways the graphs could intersect and so how many solutions there might be.

To solve systems of nonlinear equations by graphing, we use basically the same steps as with systems of linear equations modified slightly for nonlinear equations. The steps are listed below for reference.

Solve the system by graphing: ${\begin{matrix} x - y = −2 \\ y = x^{2} \end{matrix} .$

Solution

Identify each graph.	${\begin{matrix} x - y = −2 & line \\ y = x^{2} & parabola \end{matrix}$
Sketch the possible options for intersection of a parabola and a line.
Graph the line, $x - y = −2 .$ Slope-intercept form $y = x + 2 .$ Graph the parabola, $y = x^{2} .$
Identify the points of intersection.	The points of intersection appear to be $(2, 4)$ and $(−1, 1) .$
Check to make sure each solution makes both equations true. $(2, 4)$ $\begin{matrix} x - y & = & −2 & y & = & x^{2} \\ 2 - 4 & \overset{?}{=} & −2 & 4 & \overset{?}{=} & 2^{2} \\ −2 & = & −2 ✓ & 4 & = & 4 ✓ \end{matrix}$ $(−1, 1)$ $\begin{matrix} x - y & = & −2 & y & = & x^{2} \\ −1 - 1 & \overset{?}{=} & −2 & 1 & \overset{?}{=} & {(−1)}^{2} \\ −2 & = & −2 ✓ & 1 & = & 1 ✓ \end{matrix}$
	The solutions are $(2, 4)$ and $(−1, 1) .$

To identify the graph of each equation, keep in mind the characteristics of the $x^{2}$ and $y^{2}$ terms of each conic.

Solve the system by graphing: ${\begin{matrix} y = −1 \\ {(x - 2)}^{2} + {(y + 3)}^{2} = 4 \end{matrix} .$

Solution

Identify each graph.	${\begin{matrix} y = −1 & line \\ {(x - 2)}^{2} + {(y + 3)}^{2} = 4 & circle \end{matrix}$
Sketch the possible options for the intersection of a circle and a line.
Graph the circle, ${(x - 2)}^{2} + {(y + 3)}^{2} = 4$ Center: $(2, −3)$ radius: 2 Graph the line, $y = −1 .$ It is a horizontal line.
Identify the points of intersection.	The point of intersection appears to be $(2, −1) .$
Check to make sure the solution makes both equations true. $(2, −1)$ $\begin{matrix} {(x - 2)}^{2} + {(y + 3)}^{2} & = & 4 & y & = & −1 \\ {(2 - 2)}^{2} + {(−1 + 3)}^{2} & \overset{?}{=} & 4 & −1 & = & −1 ✓ \\ {(0)}^{2} + {(2)}^{2} & \overset{?}{=} & 4 \\ 4 & = & 4 ✓ \end{matrix}$
	The solution is $(2, −1) .$

Solve a System of Nonlinear Equations Using Substitution

The graphing method works well when the points of intersection are integers and so easy to read off the graph. But more often it is difficult to read the coordinates of the points of intersection. The substitution method is an algebraic method that will work well in many situations. It works especially well when it is easy to solve one of the equations for one of the variables.

The substitution method is very similar to the substitution method that we used for systems of linear equations. The steps are listed below for reference.

Solve the system by using substitution: ${\begin{matrix} 9 x^{2} + y^{2} = 9 \\ y = 3 x - 3 \end{matrix} .$

Solution

Identify each graph.	${\begin{matrix} 9 x^{2} + y^{2} = 9 & ellipse \\ y = 3 x - 3 & line \end{matrix}$
Sketch the possible options for intersection of an ellipse and a line.
The equation $y = 3 x - 3$ is solved for y.

Substitute $3 x - 3$ for y in the first equation.
Solve the equation for x.

Substitute $x = 0$ and $x = 1$ into $y = 3 x - 3$ to find y.

	The ordered pairs are $(0, −3),$ $(1, 0) .$
Check both ordered pairs in both equations. $(0, −3)$ $\begin{matrix} 9 x^{2} + y^{2} & = & 9 & y & = & 3 x - 3 \\ 9 \cdot 0^{2} + {(−3)}^{2} & \overset{?}{=} & 9 & −3 & \overset{?}{=} & 3 \cdot 0 - 3 \\ 0 + 9 & \overset{?}{=} & 9 & −3 & \overset{?}{=} & 0 - 3 \\ 9 & = & 9 ✓ & −3 & = & −3 ✓ \end{matrix}$ $(1, 0)$ $\begin{matrix} 9 x^{2} + y^{2} & = & 9 & y & = & 3 x - 3 \\ 9 \cdot 1^{2} + 0^{2} & \overset{?}{=} & 9 & 0 & \overset{?}{=} & 3 \cdot 1 - 3 \\ 9 + 0 & \overset{?}{=} & 9 & 0 & \overset{?}{=} & 3 - 3 \\ 9 & = & 9 ✓ & 0 & = & 0 ✓ \end{matrix}$
	The solutions are $(0, −3), (1, 0) .$

So far, each system of nonlinear equations has had at least one solution. The next example will show another option.

Solve the system by using substitution: ${\begin{matrix} x^{2} - y = 0 \\ y = x - 2 \end{matrix} .$

Solution

Identify each graph.	${\begin{matrix} x^{2} - y = 0 & parabola \\ y = x - 2 & line \end{matrix}$
Sketch the possible options for intersection of a parabola and a line
The equation $y = x - 2$ is solved for y.

Substitute $x - 2$ for y in the first equation.
Solve the equation for x.
This doesn’t factor easily, so we can check the discriminant.
$\begin{matrix} b^{2} - 4 a c \\ {(−1)}^{2} - 4 \cdot 1 \cdot 2 \\ - 7 \end{matrix}$	The discriminant is negative, so there is no real solution. The system has no solution.

Solve a System of Nonlinear Equations Using Elimination

When we studied systems of linear equations, we used the method of elimination to solve the system. We can also use elimination to solve systems of nonlinear equations. It works well when the equations have both variables squared. When using elimination, we try to make the coefficients of one variable to be opposites, so when we add the equations together, that variable is eliminated.

The elimination method is very similar to the elimination method that we used for systems of linear equations. The steps are listed for reference.

Solve the system by elimination: ${\begin{matrix} x^{2} + y^{2} = 4 \\ x^{2} - y = 4 \end{matrix} .$

Solution

Identify each graph.
Sketch the possible options for intersection of a circle and a parabola.
Both equations are in standard form.
To get opposite coefficients of $x^{2},$ we will multiply the second equation by $−1 .$
Simplify.
Add the two equations to eliminate $x^{2} .$
Solve for y.

Substitute $y = 0$ and $y = −1$ into one of the original equations. Then solve for x.

Write each solution as an ordered pair.	The ordered pairs are $(−2, 0)$ $(2, 0) .$ $(\sqrt{3}, −1) (− \sqrt{3}, −1)$
Check that each ordered pair is a solution to both original equations.
We will leave the checks for each of the four solutions to you.	The solutions are $(−2, 0),$ $(2, 0),$ $(\sqrt{3}, −1),$ and $(− \sqrt{3}, −1) .$

There are also four options when we consider a circle and a hyperbola.

Solve the system by elimination: ${\begin{matrix} x^{2} + y^{2} = 7 \\ x^{2} - y^{2} = 1 \end{matrix} .$

Solution

Identify each graph.	${\begin{matrix} x^{2} + y^{2} = 7 & circle \\ x^{2} - y^{2} = 1 & hyperbola \end{matrix}$
Sketch the possible options for intersection of a circle and hyperbola.
Both equations are in standard form.	${\begin{matrix} x^{2} + y^{2} = 7 \\ x^{2} - y^{2} = 1 \end{matrix}$
The coefficients of $y^{2}$ are opposite, so we will add the equations.	$\begin{matrix} \underset{__________}{{\begin{matrix} x^{2} + y^{2} = 7 \\ x^{2} - y^{2} = 1 \end{matrix}} \\ 2 x^{2} = 8 \end{matrix}$
Simplify.	$\begin{matrix} x^{2} & = & 4 \\ x & = & ± 2 \end{matrix}$ $x = 2 x = −2$
Substitute $x = 2$ and $x = −2$ into one of the original equations. Then solve for y.	$\begin{matrix} x^{2} + y^{2} & = & 7 & x^{2} + y^{2} & = & 7 \\ 2^{2} + y^{2} & = & 7 & {(−2)}^{2} + y^{2} & = & 7 \\ 4 + y^{2} & = & 7 & 4 + y^{2} & = & 7 \\ y^{2} & = & 3 & y^{2} & = & 3 \\ y & = & ± \sqrt{3} & y & = & ± \sqrt{3} \end{matrix}$
Write each solution as an ordered pair.	The ordered pairs are $(−2, \sqrt{3}),$ $(−2, − \sqrt{3}),$ $(2, \sqrt{3}),$ and $(2, − \sqrt{3}) .$
Check that the ordered pair is a solution to both original equations.
We will leave the checks for each of the four solutions to you.	The solutions are $(−2, \sqrt{3}),$ $(−2, − \sqrt{3}),$ $(2, \sqrt{3}),$ and $(2, − \sqrt{3}) .$

Use a System of Nonlinear Equations to Solve Applications

Systems of nonlinear equations can be used to model and solve many applications. We will look at an everyday geometric situation as our example.

The difference of the squares of two numbers is 15. The sum of the numbers is 5. Find the numbers.

Solution

Identify what we are looking for.	Two different numbers.
Define the variables.	$x =$ first number $y =$ second number
Translate the information into a system of equations.
First sentence.	The difference of the squares of two numbers is 15.

Second sentence.	The sum of the numbers is 5.

Solve the system by substitution
Solve the second equation for x.
Substitute x into the first equation.

Expand and simplify.

Solve for y.

Substitute back into the second equation.

	The numbers are 1 and 4.

Myra purchased a small 25” TV for her kitchen. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 300 square inches. What are the length and width of the TV screen?

Solution

Identify what we are looking for.	The length and width of the rectangle
Define the variables.	Let $x =$ width of the rectangle $y =$ length of the rectangle
Draw a diagram to help visualize the situation.
	Area is 300 square inches.
Translate the information into a system of equations.	The diagonal of the right triangle is 25 inches.

	The area of the rectangle is 300 square inches.

Solve the system using substitution.
Solve the second equation for x.
Substitute x into the first equation.

Simplify.
Multiply by $y^{2}$ to clear the fractions.
Put in standard form.
Solve by factoring.


Since y is a side of the rectangle, we discard the negative values.
Substitute back into the second equation.

	If the length is 15 inches, the width is 20 inches.
	If the length is 20 inches, the width is 15 inches.

Key Concepts

How to solve a system of nonlinear equations by graphing.
1. Identify the graph of each equation. Sketch the possible options for intersection.
2. Graph the first equation.
3. Graph the second equation on the same rectangular coordinate system.
4. Determine whether the graphs intersect.
5. Identify the points of intersection.
6. Check that each ordered pair is a solution to both original equations.
How to solve a system of nonlinear equations by substitution.
1. Identify the graph of each equation. Sketch the possible options for intersection.
2. Solve one of the equations for either variable.
3. Substitute the expression from Step 2 into the other equation.
4. Solve the resulting equation.
5. Substitute each solution in Step 4 into one of the original equations to find the other variable.
6. Write each solution as an ordered pair.
7. Check that each ordered pair is a solution to both original equations.
How to solve a system of equations by elimination.
1. Identify the graph of each equation. Sketch the possible options for intersection.
2. Write both equations in standard form.
3. Make the coefficients of one variable opposites.
  Decide which variable you will eliminate.
  Multiply one or both equations so that the coefficients of that variable are opposites.
4. Add the equations resulting from Step 3 to eliminate one variable.
5. Solve for the remaining variable.
6. Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.
7. Write each solution as an ordered pair.
8. Check that each ordered pair is a solution to both original equations.

Section Exercises

Practice Makes Perfect

Solve a System of Nonlinear Equations Using Graphing

In the following exercises, solve the system of equations by using graphing.

${\begin{matrix} y = 2 x + 2 \\ y = − x^{2} + 2 \end{matrix}$

${\begin{matrix} y = 6 x - 4 \\ y = 2 x^{2} \end{matrix}$

Solution

This graph shows the equations of a system, y is equal to 6 x minus 4 which is a line and y is equal to 2 x squared which is a parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 0) and the parabola opens upward. The line has a slope of 6. The line and parabola intersect at the points (1, 2) and (2, 8), which are labeled. The solutions are (1, 2) and (2, 8).

${\begin{matrix} x + y = 2 \\ x = y^{2} \end{matrix}$

${\begin{matrix} x - y = −2 \\ x = y^{2} \end{matrix}$

Solution

${\begin{matrix} y = \frac{3}{2} x + 3 \\ y = − x^{2} + 2 \end{matrix}$

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

Solution

${\begin{matrix} x = −2 \\ x^{2} + y^{2} = 4 \end{matrix}$

${\begin{matrix} y = −4 \\ x^{2} + y^{2} = 16 \end{matrix}$

Solution

This graph shows the equations of a system, x is equal to negative 2 which is a line and x squared plus y squared is equal to 16 which is a circle, on the x y-coordinate plane. The line is horizontal. The center of the circle is (0, 0) and the radius of the circle is 4. The line and circle intersect at (negative 2, 0), so the solution of the system is (negative 2, 0).

${\begin{matrix} x = 2 \\ {(x + 2)}^{2} + {(y + 3)}^{2} = 16 \end{matrix}$

${\begin{matrix} y = −1 \\ {(x - 2)}^{2} + {(y - 4)}^{2} = 25 \end{matrix}$

Solution

${\begin{matrix} y = −2 x + 4 \\ y = \sqrt[]{x} + 1 \end{matrix}$

${\begin{matrix} y = - \frac{1}{2} x + 2 \\ y = \sqrt[]{x} - 2 \end{matrix}$

Solution

Solve a System of Nonlinear Equations Using Substitution

In the following exercises, solve the system of equations by using substitution.

${\begin{matrix} x^{2} + 4 y^{2} = 4 \\ y = \frac{1}{2} x - 1 \end{matrix}$

${\begin{matrix} 9 x^{2} + y^{2} = 9 \\ y = 3 x + 3 \end{matrix}$

Solution

$(−1, 0), (0, 3)$

${\begin{matrix} 9 x^{2} + y^{2} = 9 \\ y = x + 3 \end{matrix}$

${\begin{matrix} 9 x^{2} + 4 y^{2} = 36 \\ x = 2 \end{matrix}$

Solution

$(2, 0)$

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

${\begin{matrix} x^{2} + y^{2} = 169 \\ x = 12 \end{matrix}$

Solution

$(12, −5), (12, 5)$

${\begin{matrix} 3 x^{2} - y = 0 \\ y = 2 x - 1 \end{matrix}$

${\begin{matrix} 2 y^{2} - x = 0 \\ y = x + 1 \end{matrix}$

Solution

No solution

${\begin{matrix} y = x^{2} + 3 \\ y = x + 3 \end{matrix}$

${\begin{matrix} y = x^{2} - 4 \\ y = x - 4 \end{matrix}$

Solution

$(0, −4), (1, −3)$

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

${\begin{matrix} x^{2} + y^{2} = 25 \\ 2 x + y = 10 \end{matrix}$

Solution

$(3, 4), (5, 0)$

Solve a System of Nonlinear Equations Using Elimination

In the following exercises, solve the system of equations by using elimination.

${\begin{matrix} x^{2} + y^{2} = 16 \\ x^{2} - 2 y = 8 \end{matrix}$

${\begin{matrix} x^{2} + y^{2} = 16 \\ x^{2} - y = 4 \end{matrix}$

Solution

$(0, −4), (− \sqrt{7}, 3), (\sqrt{7}, 3)$

${\begin{matrix} x^{2} + y^{2} = 4 \\ x^{2} + 2 y = 1 \end{matrix}$

${\begin{matrix} x^{2} + y^{2} = 4 \\ x^{2} - y = 2 \end{matrix}$

Solution

$(0, −2), (− \sqrt{3}, 1), (\sqrt{3}, 1)$

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

${\begin{matrix} x^{2} + y^{2} = 4 \\ y^{2} - x = 2 \end{matrix}$

Solution

$(−2, 0), (1, − \sqrt{3}), (1, \sqrt{3})$

${\begin{matrix} x^{2} + y^{2} = 25 \\ 2 x^{2} - 3 y^{2} = 5 \end{matrix}$

${\begin{matrix} x^{2} + y^{2} = 20 \\ x^{2} - y^{2} = −12 \end{matrix}$

Solution

$(−2, −4), (−2, 4), (2, −4), (2, 4)$

${\begin{matrix} x^{2} + y^{2} = 13 \\ x^{2} - y^{2} = 5 \end{matrix}$

${\begin{matrix} x^{2} + y^{2} = 16 \\ x^{2} - y^{2} = 16 \end{matrix}$

Solution

$(−4, 0), (4, 0)$

${\begin{matrix} 4 x^{2} + 9 y^{2} = 36 \\ 2 x^{2} - 9 y^{2} = 18 \end{matrix}$

${\begin{matrix} x^{2} - y^{2} = 3 \\ 2 x^{2} + y^{2} = 6 \end{matrix}$

Solution

$(− \sqrt{3}, 0), (\sqrt{3}, 0)$

${\begin{matrix} 4 x^{2} - y^{2} = 4 \\ 4 x^{2} + y^{2} = 4 \end{matrix}$

${\begin{matrix} x^{2} - y^{2} = −5 \\ 3 x^{2} + 2 y^{2} = 30 \end{matrix}$

Solution

$(−2, −3), (−2, 3), (2, −3), (2, 3)$

${\begin{matrix} x^{2} - y^{2} = 1 \\ x^{2} - 2 y = 4 \end{matrix}$

${\begin{matrix} 2 x^{2} + y^{2} = 11 \\ x^{2} + 3 y^{2} = 28 \end{matrix}$

Solution

$(−1, −3), (−1, 3), (1, −3), (1, 3)$

Use a System of Nonlinear Equations to Solve Applications

In the following exercises, solve the problem using a system of equations.

The sum of two numbers is $−6$ and the product is 8. Find the numbers.

The sum of two numbers is 11 and the product is $−42 .$ Find the numbers.

Solution

$−3$ and 14

The sum of the squares of two numbers is 65. The difference of the numbers is 3. Find the numbers.

The sum of the squares of two numbers is 113. The difference of the numbers is 1. Find the numbers.

Solution

$−7$ and $−8$ or 8 and 7

The difference of the squares of two numbers is 15. The difference of twice the square of the first number and the square of the second number is 30. Find the numbers.

The difference of the squares of two numbers is 20. The difference of the square of the first number and twice the square of the second number is 4. Find the numbers.

Solution

$−6$ and $−4$ or $−6$ and 4 or 6 and $−4$ or 6 and 4

The perimeter of a rectangle is 32 inches and its area is 63 square inches. Find the length and width of the rectangle.

The perimeter of a rectangle is 52 cm and its area is 165 ${cm}^{2} .$ Find the length and width of the rectangle.

Solution

If the length is 11 cm, the width is 15 cm. If the length is 15 cm, the width is 11 cm.

Dion purchased a new microwave. The diagonal of the door measures 17 inches. The door also has an area of 120 square inches. What are the length and width of the microwave door?

Jules purchased a microwave for his kitchen. The diagonal of the front of the microwave measures 26 inches. The front also has an area of 240 square inches. What are the length and width of the microwave?

Solution

If the length is 10 inches, the width is 24 inches. If the length is 24 inches, the width is 10 inches.

Roman found a widescreen TV on sale, but isn’t sure if it will fit his entertainment center. The TV is 60”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1728 square inches. His entertainment center has an insert for the TV with a length of 50 inches and width of 40 inches. What are the length and width of the TV screen and will it fit Roman’s entertainment center?

Donnette found a widescreen TV at a garage sale, but isn’t sure if it will fit her entertainment center. The TV is 50”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1200 square inches. Her entertainment center has an insert for the TV with a length of 38 inches and width of 27 inches. What are the length and width of the TV screen and will it fit Donnette’s entertainment center?

Solution

The length is 40 inches and the width is 30 inches. The TV will not fit Donnette’s entertainment center.

Writing Exercises

In your own words, explain the advantages and disadvantages of solving a system of equations by graphing.

Explain in your own words how to solve a system of equations using substitution.

Solution

Answers will vary.

Explain in your own words how to solve a system of equations using elimination.

A circle and a parabola can intersect in ways that would result in 0, 1, 2, 3, or 4 solutions. Draw a sketch of each of the possibilities.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve a system of nonlinear equations using graphing. In row 3, the I can solve a system of nonlinear equations using substitution. In row 4, the I can was solve a system of a nonlinear equations using the elimination. In row 5, the I can was use a system of nonlinear equations to solve applications.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Distance and Midpoint Formulas; Circles

Use the Distance Formula

In the following exercises, find the distance between the points. Round to the nearest tenth if needed.

$(−5, 1)$ and $(−1, 4)$

$(−2, 5)$ and $(1, 5)$

Solution

$d = 3$

$(8, 2)$ and $(−7, −3)$

$(1, −4)$ and $(5, −5)$

Solution

$d = \sqrt{17}, d \approx 4.1$

Use the Midpoint Formula

In the following exercises, find the midpoint of the line segment whose endpoints are given.

$(−2, −6)$ and $(−4, −2)$

$(3, 7)$ and $(5, 1)$

Solution

$(4, 4)$

$(−8, −10)$ and $(9, 5)$

$(−3, 2)$ and $(6, −9)$

Solution

$(\frac{3}{2}, - \frac{7}{2})$

Write the Equation of a Circle in Standard Form

In the following exercises, write the standard form of the equation of the circle with the given information.

radius is 15 and center is $(0, 0)$

radius is $\sqrt{7}$ and center is $(0, 0)$

Solution

$x^{2} + y^{2} = 7$

radius is 9 and center is $(−3, 5)$

radius is 7 and center is $(−2, −5)$

Solution

${(x + 2)}^{2} + {(y + 5)}^{2} = 49$

center is $(3, 6)$ and a point on the circle is $(3, −2)$

center is $(2, 2)$ and a point on the circle is $(4, 4)$

Solution

${(x - 2)}^{2} + {(y - 2)}^{2} = 8$

Graph a Circle

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

$2 x^{2} + 2 y^{2} = 450$

$3 x^{2} + 3 y^{2} = 432$

Solution

ⓐ radius: 12, center: $(0, 0)$
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The center of the circle is (0, 0) and the radius of the circle is 12.

${(x + 3)}^{2} + {(y - 5)}^{2} = 81$

${(x + 2)}^{2} + {(y + 5)}^{2} = 49$

Solution

ⓐ radius: 7, center: $(−2, −5)$
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The center of the circle is (negative 2, negative 5) and the radius of the circle is 7.

$x^{2} + y^{2} - 6 x - 12 y - 19 = 0$

$x^{2} + y^{2} - 4 y - 60 = 0$

Solution

ⓐ radius: 8, center: $(0, 2)$
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The center of the circle is (0, 2) and the radius of the circle is 8.

Parabolas

Graph Vertical Parabolas

In the following exercises, graph each equation by using its properties.

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

$y = 2 x^{2} + 10 x + 7$

Solution

The figure shows an upward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 7 to 7. The vertex is (negative five-halves, negative eleven-halves) and the parabola passes through the points (negative 4, negative 1) and (negative 1, negative 1).

$y = −6 x^{2} + 12 x - 1$

$y = − x^{2} + 10 x$

Solution

The figure shows a downward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 36 to 36. The y-axis of the plane runs from negative 26 to 26. The vertex is (5, 25) and the parabola passes through the points (2, 16) and (8, 16).

In the following exercises, ⓐ write the equation in standard form, then ⓑ use properties of the standard form to graph the equation.

$y = x^{2} + 4 x + 7$

$y = 2 x^{2} - 4 x - 2$

Solution

ⓐ $y = 2 {(x - 1)}^{2} - 4$
ⓑ
The figure shows an upward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 22 to 22. The y-axis of the plane runs from negative 16 to 16. The vertex is (1, negative 4) and the parabola passes through the points (0, negative 2) and (2, negative 2).

$y = −3 x^{2} - 18 x - 29$

$y = − x^{2} + 12 x - 35$

Solution

ⓐ $y = − {(x - 6)}^{2} + 1$
ⓑ
The figure shows a downward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 60 to 60. The y-axis of the plane runs from negative 46 to 46. The vertex is (6, 1) and the parabola passes through the points (5, 0) and (7, 0).

Graph Horizontal Parabolas

In the following exercises, graph each equation by using its properties.

$x = 2 y^{2}$

$x = 2 y^{2} + 4 y + 6$

Solution

The figure shows a rightward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (4, negative 1) and the parabola passes through the points (6, 0) and (6, negative 2).

$x = − y^{2} + 2 y - 4$

$x = −3 y^{2}$

Solution

The figure shows a leftward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (0, 0) and the parabola passes through the points (negative 3, 1) and (negative 3, negative 1).

In the following exercises, ⓐ write the equation in standard form, then ⓑ use properties of the standard form to graph the equation.

$x = 4 y^{2} + 8 y$

$x = y^{2} + 4 y + 5$

Solution

ⓐ $x = {(y + 2)}^{2} + 1$
ⓑ
The figure shows a rightward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (1, negative 2) and the parabola passes through the points (5, 0) and (5, negative 4).

$x = − y^{2} - 6 y - 7$

$x = −2 y^{2} + 4 y$

Solution

ⓐ $x = −2 {(y - 1)}^{2} + 2$
ⓑ
The figure shows a leftward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (2, negative 3) and the parabola passes through the points (0, 2) and (0, 0).

Solve Applications with Parabolas

In the following exercises, create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in standard form.

The figure shows a parabolic arch formed in the foundation of the bridge. The arch is 5 feet high and 20 feet wide.

The figure shows a parabolic arch formed in the foundation of the bridge. The arch is 25 feet high and 30 feet wide.

Solution

$y = - \frac{1}{9} x^{2} + \frac{10}{3} x$

Ellipses

Graph an Ellipse with Center at the Origin

In the following exercises, graph each ellipse.

$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$

$\frac{x^{2}}{4} + \frac{y^{2}}{81} = 1$

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 9), and co-vertices at (plus or minus 2, 0).

$49 x^{2} + 64 y^{2} = 3136$

$9 x^{2} + y^{2} = 9$

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 9 to 9. The y-axis of the plane runs from negative 7 to 7. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 3), and co-vertices at (plus or minus 1, 0).

Find the Equation of an Ellipse with Center at the Origin

In the following exercises, find the equation of the ellipse shown in the graph.

The figure shows an ellipse graphed on the x y coordinate plane. The ellipse has a center at (0, 0), a horizontal major axis, vertices at (plus or minus 10, 0), and co-vertices at (0, plus or minus 4).

The figure shows an ellipse graphed on the x y coordinate plane. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 8), and co-vertices at (plus or minus 6, 0).

Solution

$\frac{x^{2}}{36} + \frac{y^{2}}{64} = 1$

Graph an Ellipse with Center Not at the Origin

In the following exercises, graph each ellipse.

$\frac{{(x - 1)}^{2}}{25} + \frac{{(y - 6)}^{2}}{4} = 1$

$\frac{{(x + 4)}^{2}}{16} + \frac{{(y + 1)}^{2}}{9} = 1$

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (negative 4, negative 1), a horizontal major axis, vertices at (negative 8, negative 1) and (0, negative 1) and co-vertices at (negative 4, 2) and (negative 4, negative 4).

$\frac{{(x - 5)}^{2}}{16} + \frac{{(y + 3)}^{2}}{36} = 1$

$\frac{{(x + 3)}^{2}}{9} + \frac{{(y - 2)}^{2}}{25} = 1$

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (negative 3, 2), a vertical major axis, vertices at (negative 3, 7) and (negative 3, negative 3) and co-vertices at (negative 6, 2) and (0, 2).

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

$4 x^{2} + 16 y^{2} + 48 x + 160 y + 480 = 0$

$25 x^{2} + 4 y^{2} - 150 x - 56 y + 321 = 0$

Solution

ⓐ $\frac{{(x - 3)}^{2}}{4} + \frac{{(y - 7)}^{2}}{25} = 1$
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 18 to 18. The y-axis of the plane runs from negative 14 to 14. The ellipse has a center at (3, 7), a vertical major axis, vertices at (3, 2) and (3, 12) and co-vertices at (negative 1, 7) and (5, 7).

$25 x^{2} + 4 y^{2} + 150 x + 125 = 0$

$4 x^{2} + 9 y^{2} - 126 y + 405 = 0$

Solution

ⓐ $\frac{x^{2}}{9} + \frac{{(y - 7)}^{2}}{4} = 1$
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 15 to 15. The y-axis of the plane runs from negative 11 to 11. The ellipse has a center at (0, 7), a horizontal major axis, vertices at (3, 7) and (negative 3, 7) and co-vertices at (0, 5) and (0, 9).

Solve Applications with Ellipses

In the following exercises, write the equation of the ellipse described.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 10 AU and the furthest is approximately 90 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

The figure shows a model of an elliptical orbit around the sun on the x y coordinate plane. The ellipse has a center at (0, 0), a horizontal major axis, vertices marked at (plus or minus 50, 0), the sun marked as a foci and labeled (50, 0), the closest distance the comet is from the sun marked as 10 A U, and the farthest a comet is from the sun marked as 90 A U.

Hyperbolas

Graph a Hyperbola with Center at $(0, 0)$

In the following exercises, graph.

$\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1$

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 9 to 9. The hyperbola has a center at (0, 0) and branches that pass through the vertices (plus or minus 5, 0), and that open left and right.

$\frac{y^{2}}{49} - \frac{x^{2}}{16} = 1$

$9 y^{2} - 16 x^{2} = 144$

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 19 to 19. The y-axis of the plane runs from negative 15 to 15. The hyperbola has a center at (0, 0) and branches that pass through the vertices (0, plus or minus 4), and that open up and down.

$16 x^{2} - 4 y^{2} = 64$

Graph a Hyperbola with Center at $(h, k)$

In the following exercises, graph.

$\frac{{(x + 1)}^{2}}{4} - \frac{{(y + 1)}^{2}}{9} = 1$

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (negative 1, negative 1) and branches that pass through the vertices (negative 3, negative 1) and (1, negative 1), and that open left and right.

$\frac{{(x - 2)}^{2}}{4} - \frac{{(y - 3)}^{2}}{16} = 1$

$\frac{{(y + 2)}^{2}}{9} - \frac{{(x + 1)}^{2}}{9} = 1$

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (negative 1, negative 2) and branches that pass through the vertices (negative 1, 1) and (negative 1, negative 5), and that open up and down.

$\frac{{(y - 1)}^{2}}{25} - \frac{{(x - 2)}^{2}}{9} = 1$

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

$4 x^{2} - 16 y^{2} + 8 x + 96 y - 204 = 0$

Solution

ⓐ $\frac{{(x + 1)}^{2}}{16} - \frac{{(y - 3)}^{2}}{4} = 1$
ⓑ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (negative 1, 3) and branches that pass through the vertices (negative 5, 3) and (3, 3), and that open left and right.

$16 x^{2} - 4 y^{2} - 64 x - 24 y - 36 = 0$

$4 y^{2} - 16 x^{2} + 32 x - 8 y - 76 = 0$

Solution

ⓐ $\frac{{(y - 1)}^{2}}{16} - \frac{{(x - 1)}^{2}}{4} = 1$
ⓑ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (1, 1) and branches that pass through the vertices (1, negative 3) and (1, 5), and that open up and down.

$36 y^{2} - 16 x^{2} - 96 x + 216 y - 396 = 0$

Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola

In the following exercises, identify the type of graph.

ⓐ $16 y^{2} - 9 x^{2} - 36 x - 96 y - 36 = 0$
ⓑ $x^{2} + y^{2} - 4 x + 10 y - 7 = 0$
ⓒ $y = x^{2} - 2 x + 3$
ⓓ $25 x^{2} + 9 y^{2} = 225$

Solution

ⓐ $x^{2} + y^{2} + 4 x - 10 y + 25 = 0$
ⓑ $y^{2} - x^{2} - 4 y + 2 x - 6 = 0$
ⓒ $x = - y^{2} - 2 y + 3$
ⓓ $16 x^{2} + 9 y^{2} = 144$

Solve Systems of Nonlinear Equations

Solve a System of Nonlinear Equations Using Graphing

In the following exercises, solve the system of equations by using graphing.

${\begin{matrix} 3 x^{2} - y = 0 \\ y = 2 x - 1 \end{matrix}$

Solution

The figure shows a parabola and line graphed on the x y coordinate plane. The x-axis of the plane runs from negative 5 to 5. The y-axis of the plane runs from negative 4 to 4. The parabola has a vertex at (0, 0) and opens upward. The line has a slope of 2 with a y-intercept at negative 1. The parabola and line do not intersect, so the system has no solution.

${\begin{matrix} y = x^{2} - 4 \\ y = x - 4 \end{matrix}$

${\begin{matrix} x^{2} + y^{2} = 169 \\ x = 12 \end{matrix}$

Solution

${\begin{matrix} x^{2} + y^{2} = 25 \\ y = −5 \end{matrix}$

Solve a System of Nonlinear Equations Using Substitution

In the following exercises, solve the system of equations by using substitution.

${\begin{matrix} y = x^{2} + 3 \\ y = −2 x + 2 \end{matrix}$

Solution

$(−1, 4)$

${\begin{matrix} x^{2} + y^{2} = 4 \\ x - y = 4 \end{matrix}$

${\begin{matrix} 9 x^{2} + 4 y^{2} = 36 \\ y - x = 5 \end{matrix}$

Solution

No solution

${\begin{matrix} x^{2} + 4 y^{2} = 4 \\ 2 x - y = 1 \end{matrix}$

Solve a System of Nonlinear Equations Using Elimination

In the following exercises, solve the system of equations by using elimination.

${\begin{matrix} x^{2} + y^{2} = 16 \\ x^{2} - 2 y - 1 = 0 \end{matrix}$

Solution

$(− \sqrt{7}, 3), (\sqrt{7}, 3)$

${\begin{matrix} x^{2} - y^{2} = 5 \\ - 2 x^{2} - 3 y^{2} = −30 \end{matrix}$

${\begin{matrix} 4 x^{2} + 9 y^{2} = 36 \\ 3 y^{2} - 4 x = 12 \end{matrix}$

Solution

$(−3, 0), (0, −2), (0, 2)$

${\begin{matrix} x^{2} + y^{2} = 14 \\ x^{2} - y^{2} = 16 \end{matrix}$

Use a System of Nonlinear Equations to Solve Applications

In the following exercises, solve the problem using a system of equations.

The sum of the squares of two numbers is 25. The difference of the numbers is 1. Find the numbers.

Solution

$−3$ and $−4$ or 4 and 3

The difference of the squares of two numbers is 45. The difference of the square of the first number and twice the square of the second number is 9. Find the numbers.

The perimeter of a rectangle is 58 meters and its area is 210 square meters. Find the length and width of the rectangle.

Solution

If the length is 14 inches, the width is 15 inches. If the length is 15 inches, the width is 14 inches.

Colton purchased a larger microwave for his kitchen. The diagonal of the front of the microwave measures 34 inches. The front also has an area of 480 square inches. What are the length and width of the microwave?

Practice Test

In the following exercises, find the distance between the points and the midpoint of the line segment with the given endpoints. Round to the nearest tenth as needed.

$(−4, −3)$ and $(−10, −11)$

Solution

distance: 10, midpoint: $(−7, −7)$

$(6, 8)$ and $(−5, −3)$

In the following exercises, write the standard form of the equation of the circle with the given information.

radius is 11 and center is $(0, 0)$

Solution

$x^{2} + y^{2} = 121$

radius is 12 and center is $(10, −2)$

center is $(−2, 3)$ and a point on the circle is $(2, −3)$

Solution

${(x + 2)}^{2} + {(y - 3)}^{2} = 52$

Find the equation of the ellipse shown in the graph.

The figure shows an ellipse graphed on the x y coordinate plane. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 10), and co-vertices at (plus or minus 6, 0).

In the following exercises, ⓐ identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, and ⓑ graph the equation.

$4 x^{2} + 49 y^{2} = 196$

Solution

ⓐ ellipse
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The ellipse has a center at (0, 0), a horizontal major axis, vertices at (plus or minus 7, 0) and co-vertices at (0, plus or minus 2).

$y = 3 {(x - 2)}^{2} - 2$

$3 x^{2} + 3 y^{2} = 27$

Solution

ⓐ circle
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The parabola circle has a center at (0, 0) and a radius of 3.

$\frac{y^{2}}{100} - \frac{x^{2}}{36} = 1$

$\frac{x^{2}}{16} + \frac{y^{2}}{81} = 1$

Solution

ⓐ ellipse
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 9) and co-vertices at (plus or minus 4, 0).

$x = 2 y^{2} + 10 y + 7$

$64 x^{2} - 9 y^{2} = 576$

Solution

ⓐ hyperbola
ⓑ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The hyperbola has a center at (0, 0) and branches that pass through the vertices (plus or minus 3, 0) and that open left and right.

In the following exercises, ⓐ identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, ⓑ write the equation in standard form, and ⓒ graph the equation.

$25 x^{2} + 64 y^{2} + 200 x - 256 y - 944 = 0$

$x^{2} + y^{2} + 10 x + 6 y + 30 = 0$

Solution

ⓐ circle
ⓑ ${(x + 5)}^{2} + {(y + 3)}^{2} = 4$
ⓒ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The circle has a center at (negative 5, negative 3) and a radius 2.

$x = − y^{2} + 2 y - 4$

$9 x^{2} - 25 y^{2} - 36 x - 50 y - 214 = 0$

Solution

ⓐ hyperbola
ⓑ $\frac{{(x - 2)}^{2}}{25} - \frac{{(y + 1)}^{2}}{9} = 1$
ⓒ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (2, negative 1) and branches that pass through the vertices (negative 3, negative 1) and (7, negative 1) that open left and right.

$y = x^{2} + 6 x + 8$

Solve the nonlinear system of equations by graphing:
${\begin{matrix} 3 y^{2} - x = 0 \\ y = −2 x - 1 \end{matrix} .$

Solution

No solution
A graph displays a downward-sloping line and a parabola opening to the right, both intersecting at the origin (0,0) on a Cartesian coordinate system.

Solve the nonlinear system of equations using substitution:
${\begin{matrix} x^{2} + y^{2} = 8 \\ y = − x - 4 \end{matrix} .$

Solve the nonlinear system of equations using elimination:
${\begin{matrix} x^{2} + 9 y^{2} = 9 \\ 2 x^{2} - 9 y^{2} = 18 \end{matrix} .$

Solution

$(3, 0), (−3, 0)$

Create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in $y = a x^{2} + b x + c$ form.

The figure shows a parabolic arch formed in the foundation of the bridge. The arch is 10 feet high and 30 feet wide.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 20 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

Solution

$\frac{x^{2}}{2025} + \frac{y^{2}}{1400} = 1$

The sum of two numbers is 22 and the product is $−240 .$ Find the numbers.

For her birthday, Olive’s grandparents bought her a new widescreen TV. Before opening it she wants to make sure it will fit her entertainment center. The TV is 55”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1452 square inches. Her entertainment center has an insert for the TV with a length of 50 inches and width of 40 inches. What are the length and width of the TV screen and will it fit Olive’s entertainment center?

Solution

The length is 44 inches and the width is 33 inches. The TV will fit Olive’s entertainment center.

system of nonlinear equations

A system of nonlinear equations is a system where at least one of the equations is not linear.