section 8.4 systems of nonlinear equations in two variables › uploads › 1 › 2 › 0 › 3 ›...

Section 8.4 Systems of Nonlinear Equations in Two Variables 827

Preview Exercises Exercises 66–68 will help you prepare for the material covered in the next section.

66. Solve by the substitution method:

b 4x + 3y = 4y = 2x - 7.

67. Solve by the addition method:

b 2x + 4y = -43x + 5y = -3.

68. Graph x - y = 3 and (x - 2)2 + (y + 3)2 = 4 in the same rectangular coordinate system. What are the two intersection points? Show that each of these ordered pairs satisfi es both equations.

Systems of Nonlinear Equations in Two Variables SECTION 8.4

Objectives � Recognize systems of

nonlinear equations in two variables.

� Solve nonlinear systems by substitution.

� Solve nonlinear systems by addition.

� Solve problems using systems of nonlinear equations.

S cientists debate the probability that a “doomsday rock” will

collide with Earth. It has been estimated that an asteroid, a tiny planet that revolves around the sun, crashes

into Earth about once every 250,000 years, and that such

a collision would have disastrous results. In 1908, a small fragment struck Siberia,

leveling thousands of acres of trees. One theory about the extinction of dinosaurs 65 million years ago involves Earth’s collision with a large asteroid and the resulting drastic changes in Earth’s climate.

Understanding the path of Earth and the path of a comet is essential to detecting threatening space debris. Orbits about the sun are not described by linear equations in the form Ax + By = C. The ability to solve systems that contain nonlinear equations provides NASA scientists watching for troublesome asteroids with a way to locate possible collision points with Earth’s orbit.

Systems of Nonlinear Equations and Their Solutions A system of two nonlinear equations in two variables, also called a nonlinear system , contains at least one equation that cannot be expressed in the form Ax + By = C.Here are two examples:

3x-y=9

x2=2y+10

x2+y2=9.

y=x2+3Not in the formAx + By = C.The term x2 is

not linear.

Neither equation is inthe form Ax + By = C.The terms x2 and y2 are

not linear.

b b

A solution of a nonlinear system in two variables is an ordered pair of real numbers that satisfi es both equations in the system. The solution set of the system is the set of all such ordered pairs. As with linear systems in two variables, the solution of a nonlinear system (if there is one) corresponds to the intersection point(s) of the graphs of the equations in the system. Unlike linear systems, the graphs can be circles, parabolas, or anything other than two lines. We will solve nonlinear systems using the substitution method and the addition method.

� Recognize systems of nonlinear equations in two variables.

M14_BLIT7240_06_SE_08-hr.indd 827 13/10/12 12:22 PM

828 Chapter 8 Systems of Equations and Inequalities

Eliminating a Variable Using the Substitution Method The substitution method involves converting a nonlinear system into one equation in one variable by an appropriate substitution. The steps in the solution process are exactly the same as those used to solve a linear system by substitution. However, when you obtain an equation in one variable, this equation may not be linear. In our fi rst example, this equation is quadratic.

EXAMPLE 1 Solving a Nonlinear System by the Substitution Method

Solve by the substitution method:

b x2 = 2y + 103x - y = 9.

(The graph is a parabola.) (The graph is a line.)

SOLUTION Step 1 Solve one of the equations for one variable in terms of the other. We begin by isolating one of the variables raised to the fi rst power in either of the equations. By solving for y in the second equation, which has a coeffi cient of -1, we can avoid fractions.

3x - y = 9 This is the second equation in the given system.

3x = y + 9 Add y to both sides.

3x - 9 = y Subtract 9 from both sides.

Step 2 Substitute the expression from step 1 into the other equation. We substitute 3x - 9 for y in the fi rst equation.

y= 3x-9 x2=2 y +10

This gives us an equation in one variable, namely,

x2 = 2(3x - 9) + 10.

The variable y has been eliminated.

Step 3 Solve the resulting equation containing one variable.

x2 = 2(3x - 9) + 10 This is the equation containing one variable.

x2 = 6x - 18 + 10 Use the distributive property.

x2 = 6x - 8 Combine numerical terms on the right.

x2 - 6x + 8 = 0 Move all terms to one side and set the quadratic equation equal to 0.

(x - 4)(x - 2) = 0 Factor.

x - 4 = 0 or x - 2 = 0 Set each factor equal to 0.

x = 4 x = 2 Solve for x.

Step 4 Back-substitute the obtained values into the equation from step 1. Now that we have the x@coordinates of the solutions, we back-substitute 4 for x and2 for x into the equation y = 3x - 9.

If x is 4, y = 3(4) - 9 = 3, so (4, 3) is a solution.If x is 2, y = 3(2) - 9 = -3, so (2, -3) is a solution.

� Solve nonlinear systems by substitution.



Step 5 Check the proposed solutions in both of the system’s given equations. We begin by checking (4, 3). Replace x with 4 and y with 3.

x2 = 2y + 10 3x - y = 9 These are the given equations.

42 � 2(3) + 10 3(4) - 3 � 9 Let x = 4 and y = 3.

16 � 6 + 10 12 - 3 � 9 Simplify.

16 = 16, true 9 = 9, true True statements result.

The ordered pair (4, 3) satisfi es both equations. Thus, (4, 3) is a solution of the system.Now let’s check (2, -3). Replace x with 2 and y with -3 in both given equations.

x2 = 2y + 10 3x - y = 9 These are the given equations.

22 � 2(-3) + 10 3(2) - (-3) � 9 Let x = 2 and y = -3.

4 � -6 + 10 6 + 3 � 9 Simplify.

4 = 4, true 9 = 9, true True statements result.

The ordered pair (2, -3) also satisfi es both equations and is a solution of the system. The solutions are (4, 3) and (2, -3), and the solution set is {(4, 3), (2, -3)}.

Figure 8.9 shows the graphs of the equations in the system and the solutions as intersection points. ● ● ●

Check Point 1 Solve by the substitution method:

b x2 = y - 14x - y = -1.

EXAMPLE 2 Solving a Nonlinear System by the Substitution Method

Solve by the substitution method:

b x - y = 3(x - 2)2 + (y + 3)2 = 4.

(The graph is a line.) (The graph is a circle.)

SOLUTION Graphically, we are fi nding the intersection of a line and a circle with center (2, -3) and radius 2.

Step 1 Solve one of the equations for one variable in terms of the other. We will solve for x in the linear equation—that is, the fi rst equation. (We could also solve for y. )

x - y = 3 This is the fi rst equation in the given system.

x = y + 3 Add y to both sides.

Step 2 Substitute the expression from step 1 into the other equation. We substitute y + 3 for x in the second equation.

x= y+3 ( x-2)2+(y +3)2=4

This gives an equation in one variable, namely,

(y + 3 - 2)2 + (y + 3)2 = 4.

The variable x has been eliminated.

−1

1234

−2−3−4

−9−8−7−6−5

1 2 3 4 5−1−2−3−4−5

y

x

3x − y = 9

x2 = 2y + 10

(4, 3)

(2, −3)

FIGURE 8.9 Points of intersection illustrate the nonlinear system’s solutions.

GREAT QUESTION! Why is the graph of (x - 2)2 + (y + 3)2 = 4 a circle?

Recall that

(x - h)2 + (y - k)2 = r2

describes a circle with center (h, k) and radius r.



Step 3 Solve the resulting equation containing one variable.

(y + 3 - 2)2 + (y + 3)2 = 4 This is the equation containing one variable.

(y + 1)2 + (y + 3)2 = 4 Combine numerical terms in the fi rst parentheses.

y2 + 2y + 1 + y2 + 6y + 9 = 4 Use the formula (A + B)2 = A2 + 2AB + B2 to square y + 1 and y + 3.

2y2 + 8y + 10 = 4 Combine like terms on the left.

2y2 + 8y + 6 = 0 Subtract 4 from both sides and set the quadratic equation equal to 0.

2(y2 + 4y + 3) = 0 Factor out 2.

2(y + 3)(y + 1) = 0 Factor completely.

y + 3 = 0 or y + 1 = 0 Set each variable factor equal to 0.

y = -3 y = -1 Solve for y .

Step 4 Back-substitute the obtained values into the equation from step 1. Now that we have the y@coordinates of the solutions, we back-substitute -3 for y and -1 for y in the equation x = y + 3.

If y = -3: x = -3 + 3 = 0, so (0, -3) is a solution.

If y = -1: x = -1 + 3 = 2, so (2, -1) is a solution.

Step 5 Check the proposed solutions in both of the system’s given equations. Take a moment to show that each ordered pair satisfi es both given equations, x - y = 3 and (x - 2)2 + (y + 3)2 = 4 . The solutions are (0, -3) and (2, -1), and the solution set of the given system is {(0, -3), (2, -1)}.


Check Point 2 Solve by the substitution method:

b x + 2y = 0(x - 1)2 + (y - 1)2 = 5.

Eliminating a Variable Using the Addition Method In solving linear systems with two variables, we learned that the addition method works well when each equation is in the form Ax + By = C. For nonlinear systems, the addition method can be used when each equation is in the form Ax2 + By2 = C. If necessary, we will multiply either equation or both equations by appropriate numbers so that the coeffi cients of x2 or y2 will have a sum of 0. We then add equations. The sum will be an equation in one variable.

EXAMPLE 3 Solving a Nonlinear System by the Addition Method

Solve the system:

b4x2 + y2 = 13x2 + y2 = 10.

Equation 1 Equation 2

SOLUTION We can use the same steps that we did when we solved linear systems by the addition method.

−1

123

−2

−4

−7−6−5

1 3 4 765−1−2−3

y

x

x − y = 3

(x − 2)2 + (y + 3)2 = 4

(2, −1)

(0, −3)

FIGURE 8.10 Points of intersection illustrate the nonlinear system’s solutions.

� Solve nonlinear systems by addition.



Step 1 Write both equations in the form Ax2 � By2 � C. Both equations are already in this form, so we can skip this step.

Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x2@coefficients or the sum of the y2@coefficients is 0. We can eliminate y2 by multiplying Equation 2 by -1.

b4x2 + y2 = 13x2 + y2 = 10

No change

" b 4x2 + y2 = 13

-x2 - y2 = -10

Steps 3 and 4 Add equations and solve for the remaining variable.

e 4x2 + y2 = 13-x2 - y2 = -10

3x2 = 3 Add equations.

x2 = 1 Divide both sides by 3.

x = {1 Use the square root property: If x 2 = c, then x = {1c .

Step 5 Back-substitute and fi nd the values for the other variable. We must back-substitute each value of x into either one of the original equations. Let’s use x2 + y2 = 10, Equation 2. If x = 1,

12 + y2 = 10 Replace x with 1 in Equation 2.

y2 = 9 Subtract 1 from both sides.

y = {3. Apply the square root property.

(1, 3) and (1, -3) are solutions. If x = -1,

(-1)2 + y2 = 10 Replace x with - 1 in Equation 2.

y2 = 9 The steps are the same as before.

y = {3.

(-1, 3) and (-1, -3) are solutions.

Step 6 Check. Take a moment to show that each of the four ordered pairs satisfi es the given equations, 4x2 + y2 = 13 and x2 + y2 = 10. The solution set of the given system is {(1, 3), (1, -3), (-1, 3), (-1, -3)}.


Check Point 3 Solve the system:

b3x2 + 2y2 = 354x2 + 3y2 = 48.

In solving nonlinear systems, we include only ordered pairs with real numbers in the solution set. We have seen that each of these ordered pairs corresponds to a point of intersection of the system’s graphs.

Multiply by - 1.

"

−1

12

45

−2

−4−5

1 2 3 4 5−1−2−3−4−5

y

x

(−1, −3)

(−1, 3)

(1, −3)

(1, 3)

4x2 + y2 = 13

x2 + y2 = 10

FIGURE 8.11 A system with four solutions



EXAMPLE 4 Solving a Nonlinear System by the Addition Method

Solve the system:

b y = x2 + 3x2 + y2 = 9.

Equation 1 (The graph is a parabola.) Equation 2 (The graph is a circle.)

SOLUTION We could use substitution because Equation 1, y = x2 + 3, has y expressed in terms of x, but substituting x2 + 3 for y in x2 + y2 = 9 would result in a fourth-degree equation. However, we can rewrite Equation 1 by subtracting x2 from both sides and adding the equations to eliminate the x2@terms.

Notice howlike terms

are arrangedin columns.

y+y2=12

= 9 +y2x2

= 3–x2+yb Subtract x 2 from both sides of Equation 1.

This is Equation 2.

Add the equations.

We now solve this quadratic equation.

y + y2 = 12 This is the equation containing one variable.

y2 + y - 12 = 0 Subtract 12 from both sides and set the quadratic equation equal to 0.

(y + 4)(y - 3) = 0 Factor.

y + 4 = 0 or y - 3 = 0 Set each factor equal to 0.

y = -4 y = 3 Solve for y .

To complete the solution, we must back-substitute each value of y into either one of the original equations. We will use y = x2 + 3, Equation 1. First, we substitute -4 for y.

-4 = x2 + 3

-7 = x2 Subtract 3 from both sides.

Because the square of a real number cannot be negative, the equation x2 = -7 does not have real-number solutions. We will not include the imaginary solutions, x = {2-7, or i27 and - i27, in the ordered pairs that make up the solution set. Thus, we move on to our other value for y, 3, and substitute this value into Equation 1.

y = x2 + 3 This is Equation 1.

3 = x2 + 3 Back-substitute 3 for y.

0 = x2 Subtract 3 from both sides.

0 = x Solve for x .

We showed that if y = 3, then x = 0. Thus, (0, 3) is the solution with a real ordered pair. Take a moment to show that (0, 3) satisfi es the given equations, y = x2 + 3 and x2 + y2 = 9. The solution set of the system is {(0, 3)}. Figure 8.12 shows the system’s graphs and the solution as an intersection point. ● ● ●

GREAT QUESTION! When solving nonlinear systems, do I really have to go to the hassle of checking my solutions?

Yes. Extra solutions may be introduced that do not satisfy both equations in the system. Therefore, you should get into the habit of checking all proposed pairs in each of the system’s two equations.

−1

12

4

765

−2−3

1 2 3 4 5−1−2−3−4−5

y

x

y = x2 + 3

x2 + y2 = 9

(0, 3)

FIGURE 8.12 A system with one real solution



Check Point 4 Solve the system:

b y = x2 + 5x2 + y2 = 25.

Applications Many geometric problems can be modeled and solved by the use of systems of nonlinear equations. We will use our step-by-step strategy for solving problems using mathematical models that are created from verbal conditions.

EXAMPLE 5 An Application of a Nonlinear System

You have 36 yards of fencing to build the enclosure in Figure 8.13 . Some of this fencing is to be used to build an internal divider. If you’d like to enclose 54 square yards, what are the dimensions of the enclosure?

SOLUTION Step 1 Use variables to represent unknown quantities. Let x = the enclosure’s length and y = the enclosure’s width. These variables are shown in Figure 8.13 .

Step 2 Write a system of equations modeling the problem’s conditions. The fi rst condition is that you have 36 yards of fencing.

2x + 2y + y = 36

plus plus equalsFencing alongboth lengths

fencing alongboth widths

fencing for theinternal divider

36yards.

Adding like terms, we can express the equation that models the verbal conditions for the fencing as 2x + 3y = 36.

The second condition is that you’d like to enclose 54 square yards. The rectangle’s area, the product of its length and its width, must be 54 square yards.

x � y = 54

times isLength width 54 square yards.

Step 3 Solve the system and answer the problem’s question. We must solve the system

b2x + 3y = 36xy = 54.

Equation 1

Equation 2

We will use substitution. Because Equation 1 has no coeffi cients of 1 or -1, we will work with Equation 2 and solve for y. Dividing both sides of xy = 54 by x, we obtain

y =54x

.

Now we substitute 54x

for y in Equation 1 and solve for x.

2x + 3y = 36 This is Equation 1.

2x + 3 # 54x

= 36 Substitute 54x

for y.

2x +162x

= 36 Multiply.

� Solve problems using systems of nonlinear equations.

x

y

FIGURE 8.13 Building an enclosure



162

xx a b=36 � x2x+

Clear fractions by multiplying both sides by x.

2x2 + 162 = 36x Use the distributive property on the left side. 2x2 - 36x + 162 = 0 Subtract 36x from both sides and set the

quadratic equation equal to 0. 2(x2 - 18x + 81) = 0 Factor out 2.

2(x - 9)2 = 0 Factor completely using A2 - 2AB + B2 = (A - B)2.

x - 9 = 0 Set the repeated factor equal to zero. x = 9 Solve for x .

We back-substitute this value of x into y =54x

.

If x = 9, y =549

= 6.

This means that the dimensions of the enclosure in Figure 8.13 are 9 yards by 6 yards.

Step 4 Check the proposed solution in the original wording of the problem. Take a moment to check that a length of 9 yards and a width of 6 yards results in 36 yards of fencing and an area of 54 square yards. ● ● ●

Check Point 5 Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.

x

y

FIGURE 8.13 (repeated)

Fill in each blank so that the resulting statement is true .

CONCEPT AND VOCABULARY CHECK

1. A system of two equations in two variables that contains at least one equation that cannot be expressed in the form Ax + By = C is called a system of equations.

2. When solving

ex2 - 4y = 4x + y = -1

by the substitution method, we obtain x = -4or x = 0, so the solution set is .

3. When solving

b3x2 + 2y2 = 354x2 + 3y2 = 48

by the addition method, we can eliminate x2 by multiplying the fi rst equation by -4 and the second equation by and then adding the equations.

4. When solving

ex2 + 4y2 = 16x2 - y2 = 1

by the addition method, we obtain y2 = 3, so the solution set is .

5. When solving

ex2 + y2 = 13x2 - y = 7

by the addition method, we can eliminate x2 by multiplying the second equation by and then adding the equations. We obtain , a quadratic equation in y.

6. When solving

ex2 + 4y2 = 20xy = 4

by the substitution method, we can eliminate y by solving the second equation for y. We obtain

y = . Then we substitute for in the fi rst equation.



37. b x2 + (y - 2)2 = 4x2 - 2y = 0

38. b x2 - y2 - 4x + 6y - 4 = 0x2 + y2 - 4x - 6y + 12 = 0

39. b y = (x + 3)2

x + 2y = -2 40. b (x - 1)2 + (y + 1)2 = 5

2x - y = 3

41. b x2 + y2 + 3y = 222x + y = -1

42. b x - 3y = -5x2 + y2 - 25 = 0

EXERCISE SET 8.4

Practice Exercises In Exercises 1–18, solve each system by the substitution method.

1. b x + y = 2y = x2 - 4

2. b x - y = -1y = x2 + 1

3. b x + y = 2y = x2 - 4x + 4

4. b 2x + y = -5y = x2 + 6x + 7

5. b y = x2 - 4x - 10y = -x2 - 2x + 14

6. b y = x2 + 4x + 5y = x2 + 2x - 1

7. b x2 + y2 = 25x - y = 1

8. b x2 + y2 = 53x - y = 5

9. b xy = 62x - y = 1

10. b xy = -12x - 2y + 14 = 0

11. b y2 = x2 - 92y = x - 3

12. b x2 + y = 42x + y = 1

13. b xy = 3x2 + y2 = 10

14. b xy = 4x2 + y2 = 8

15. b x + y = 1x2 + xy - y2 = -5

16. b x + y = -3x2 + 2y2 = 12y + 18

17. b x + y = 1(x - 1)2 + (y + 2)2 = 10

18. b 2x + y = 4(x + 1)2 + (y - 2)2 = 4

In Exercises 19–28, solve each system by the addition method.

19. b x2 + y2 = 13x2 - y2 = 5

20. b 4x2 - y2 = 44x2 + y2 = 4

21. b x2 - 4y2 = -73x2 + y2 = 31

22. b 3x2 - 2y2 = -52x2 - y2 = -2

23. b 3x2 + 4y2 - 16 = 02x2 - 3y2 - 5 = 0

24. b 16x2 - 4y2 - 72 = 0x2 - y2 - 3 = 0

25. b x2 + y2 = 25(x - 8)2 + y2 = 41

26. b x2 + y2 = 5x2 + (y - 8)2 = 41

27. b y2 - x = 4x2 + y2 = 4

28. b x2 - 2y = 8x2 + y2 = 16

In Exercises 29–42, solve each system by the method of your choice.

29. b 3x2 + 4y2 = 162x2 - 3y2 = 5

30. b x + y2 = 4x2 + y2 = 16

31. b 2x2 + y2 = 18xy = 4

32. b x2 + 4y2 = 20xy = 4

33. b x2 + 4y2 = 20x + 2y = 6

34. b 3x2 - 2y2 = 14x - y = 3

35. b x3 + y = 0x2 - y = 0

36. b x3 + y = 02x2 - y = 0

In Exercises 43–46, let x represent one number and let y represent the other number. Use the given conditions to write a system of nonlinear equations. Solve the system and fi nd the numbers.

43. The sum of two numbers is 10 and their product is 24. Find the numbers.

44. The sum of two numbers is 20 and their product is 96. Find the numbers.

45. The difference between the squares of two numbers is 3. Twice the square of the fi rst number increased by the square of the second number is 9. Find the numbers.

46. The difference between the squares of two numbers is 5. Twice the square of the second number subtracted from three times the square of the fi rst number is 19. Find the numbers.

Practice Plus In Exercises 47–52, solve each system by the method of your choice.

47. b 2x2 + xy = 6x2 + 2xy = 0

48. b 4x2 + xy = 30x2 + 3xy = -9

49. b-4x + y = 12y = x3 + 3x2 50. b-9x + y = 45

y = x3 + 5x2

51. d 3

x2 +1

y2 = 7

5

x2 -2

y2 = -3 52. d 2

x2 +1

y2 = 11

4

x2 -2

y2 = -14

In Exercises 53–54, make a rough sketch in a rectangular coordinate system of the graphs representing the equations in each system.

53. The system, whose graphs are a line with positive slope and a parabola whose equation has a positive leading coeffi cient, has two solutions.

54. The system, whose graphs are a line with negative slope and a parabola whose equation has a negative leading coeffi cient, has one solution.

Application Exercises 55. A planet’s orbit follows a path described by 16x2 + 4y2 = 64.

A comet follows the parabolic path y = x2 - 4. Where might the comet intersect the orbiting planet?

56. A system for tracking ships indicates that a ship lies on a path described by 2y2 - x2 = 1. The process is repeated and the ship is found to lie on a path described by 2x2 - y2 = 1. If it is known that the ship is located in the fi rst quadrant of the coordinate system, determine its exact location.



57. Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 77 square feet.

58. Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

Use the formula for the area of a rectangle and the Pythagorean Theorem to solve Exercises 59–60.

59. A small television has a picture with a diagonal measure of 10 inches and a viewing area of 48 square inches. Find the length and width of the screen.

10 inch

es

60. The area of a rug is 108 square feet and the length of its diagonal is 15 feet. Find the length and width of the rug.

LW

A = 108 feet2

15 feet

61. The fi gure shows a square fl oor plan with a smaller square area that will accommodate a combination fountain and pool. The fl oor with the fountain-pool area removed has an area of 21 square meters and a perimeter of 24 meters. Find the dimensions of the fl oor and the dimensions of the square that will accommodate the pool.

x

x

y

y y

62. The area of the rectangular piece of cardboard shown below is 216 square inches. The cardboard is used to make an open box by cutting a 2-inch square from each corner and turning up the sides. If the box is to have a volume of 224 cubic inches, fi nd the length and width of the cardboard that must be used.

W

L

63. The bar graph shows that compared to a century ago, work in the United States now involves mostly white-collar service jobs.

Per

cent

age

of T

otal

Lab

or F

orce

The Changing Pattern of Work in the United States, 1900–2005

Year

40

30

10

0

20

70

80

90

50

60

White Collar Blue Collar Farming

1900

21

41 38

1920

28

44

28

1940

3644

20

1960

51

41

8

1980

63

34

3

2005

76

22

2

Source: U.S. Department of Labor

The data can be modeled by linear and quadratic functions.

0.5x-y=–18White collar

y=–0.004x2+0.23x+41Blue collar

0.4x+y=35Farming

In each function, x represents the number of years after 1900 and y represents the percentage of the total U.S. labor force.

a. Based on the information in the graph, it appears that there was a year when the percentage of white-collar workers in the labor force was the same as the percentage of blue-collar workers in the labor force. According to the graph, between which two decades did this occur?

b. Solve a nonlinear system to determine the year described in part (a). Round to the nearest year. What percentage of the labor force, to the nearest percent, consisted of white-collar workers and what percentage consisted of blue-collar workers?

c. According to the graph, for which year was the percentage of white-collar workers the same as the percentage of farmers? What percentage of U.S. workers were in each of these groups?

d. Solve a linear system to determine the year described in part (c). Round to the nearest year. Use the models to fi nd the percentage of the labor force consisting of white-collar workers and the percentage consisting of farmers. How well do your answers model the actual data specifi ed in part (c)?

Writing in Mathematics 64. What is a system of nonlinear equations? Provide an example

with your description. 65. Explain how to solve a nonlinear system using the substitution

method. Use x2 + y2 = 9 and 2x - y = 3 to illustrate your explanation.

66. Explain how to solve a nonlinear system using the addition method. Use x2 - y2 = 5 and 3x2 - 2y2 = 19 to illustrate your explanation.


Mid-Chapter Check Point 837

Technology Exercises 67. Verify your solutions to any fi ve exercises from Exercises 1–42

by using a graphing utility to graph the two equations in the system in the same viewing rectangle. Then use the intersection feature to verify the solutions.

68. Write a system of equations, one equation whose graph is a line and the other whose graph is a parabola, that has no ordered pairs that are real numbers in its solution set. Graph the equations using a graphing utility and verify that you are correct.

Critical Thinking Exercises Make Sense? In Exercises 69–72, determine whether each statement makes sense or does not make sense, and explain your reasoning.

69. I use the same steps to solve nonlinear systems as I did to solve linear systems, although I don’t obtain linear equations when a variable is eliminated.

70. I graphed a nonlinear system that modeled the orbits of Earth and Mars, and the graphs indicated the system had a solution with a real ordered pair.

71. Without using any algebra, it’s obvious that the nonlinear system consisting of x2 + y2 = 4 and x2 + y2 = 25 does not have real-number solutions.

72. I think that the nonlinear system consisting of x2 + y2 = 36and y = (x - 2)2 - 3 is easier to solve graphically than by using the substitution method or the addition method.

In Exercises 73–76, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

73. A system of two equations in two variables whose graphs are a circle and a line can have four real ordered-pair solutions.

74. A system of two equations in two variables whose graphs are a parabola and a circle can have four real ordered-pair solutions.

75. A system of two equations in two variables whose graphs are two circles must have at least two real ordered-pair solutions.

76. A system of two equations in two variables whose graphs are a parabola and a circle cannot have only one real ordered-pair solution.

77. The points of intersection of the graphs of xy = 20 and x2 + y2 = 41 are joined to form a rectangle. Find the area of the rectangle.

78. Find a and b in this fi gure.

b

a 1017

9

Solve the systems in Exercises 79–80.

79. e logy x = 3logy(4x) = 5

80. e log x2 = y + 3log x = y - 1

Preview Exercises Exercises 81–83 will help you prepare for the material covered in the next section. In each exercise, graph the linear function.

81. 2x - 3y = 6 82. f(x) = - 23 x 83. f(x) = -2

WHAT YOU KNOW: We learned to solve systems of equations. We solved linear and nonlinear systems in two variables by the substitution method and by the addition method. We solved linear systems in three variables by eliminating a variable, reducing the system to two equations in two variables. We saw that some linear systems, called inconsistent systems, have no solution, whereas other linear systems, called dependent systems, have infi nitely many solutions. We applied systems to a variety of situations, including fi nding the break-even point for a business, fi nding a quadratic function from three points on its graph, and fi nding a rational function’s partial fraction decomposition.

In Exercises 1–12, solve each system by the method of your choice.

1. b x = 3y - 74x + 3y = 2

2. b 3x + 4y = -52x - 3y = 8

3. d 2x3

+y

5= 6

x6

-y

2= -4

4. b y = 4x - 58x - 2y = 10

5. b 2x + 5y = 33x - 2y = 1

6. c x

12- y =

14

4x - 48y = 16

7. c 2x - y + 2z = -8x + 2y - 3z = 9

3x - y - 4z = 3 8. c x - 3z = -5

2x - y + 2z = 167x - 3y - 5z = 19

9. b x2 + y2 = 9x + 2y - 3 = 0

10. b 3x2 + 2y2 = 142x2 - y2 = 7

11. b y = x2 - 6x2 + y2 = 8

12. b x - 2y = 42y2 + xy = 8

Mid-Chapter Check Point CHAPTER 8



In Exercises 13–16, write the partial fraction decomposition of each rational expression.

13. x2 - 6x + 3

(x - 2)3 14. 10x2 + 9x - 7

(x + 2)(x2 - 1)

15. x2 + 4x - 23

(x + 3)(x2 + 4) 16.

x3

(x2 + 4)2

17. A company is planning to manufacture PDAs (personal digital assistants). The fi xed cost will be $400,000 and it will cost $20 to produce each PDA. Each PDA will be sold for $100.

a. Write the cost function, C, of producing x PDAs. b. Write the revenue function, R, from the sale of x PDAs. c. Write the profi t function, P, from producing and selling x

PDAs. d. Determine the break-even point. Describe what this

means.

18. Roses sell for $3 each and carnations for $1.50 each. If a mixed bouquet of 20 fl owers consisting of roses and carnations costs $39, how many of each type of fl ower is in the bouquet?

19. Find the measure of each angle whose degree measure is represented with a variable.

y

x3y + 20

20. Find the quadratic function y = ax2 + bx + c whose graph

passes through the points (-1, 0), (1, 4), and (2, 3). 21. Find the length and width of a rectangle whose perimeter is

21 meters and whose area is 20 square meters.


statement 3 + 2 7 1, or 5 7 1. Because there are infi nitely many pairs of numbers that have a sum greater than 1, the inequality x + y 7 1 has infi nitely many solutions. Each ordered-pair solution is said to satisfy the inequality. Thus, (3, 2) satisfi es the inequality x + y 7 1.

Objectives � Graph a linear inequality

in two variables. � Graph a nonlinear

inequality in two variables.

� Use mathematical models involving linear inequalities.

� Graph a system of inequalities.

Systems of Inequalities SECTION 8.5

W e opened the chapter noting that the modern emphasis on thinness as the ideal body shape has been suggested as a major cause of eating disorders. In this section (Example 5), as well as in the Exercise Set (Exercises 77–80), we use systems of linear inequalities in two variables that will enable you to establish a healthy weight range for your height and age.

Linear Inequalities in Two Variables and Their Solutions We have seen that equations in the form Ax + By = C are straight lines when graphed. If we change the symbol = to 7 , 6 , Ú , or … ,we obtain a linear inequality in two variables . Some examples of linear inequalities in two variables are x + y 7 2, 3x - 5y … 15, and 2x - y 6 4.

A solution of an inequality in two variables , x and y, is an ordered pair of real numbers with the following property: When the x@coordinate is substituted for x and the y@coordinate is substituted for y in the inequality, we obtain a true statement. For example, (3, 2) is a solution of the inequality x + y 7 1. When 3 is substituted for xand 2 is substituted for y, we obtain the true


section 8.4 systems of nonlinear equations in two variables › uploads › 1 › 2 › 0 › 3 ›...

Documents