Download 8.4 - Systems of Nonlinear Equations in Two Variables

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Section 8.4 Systems of Nonlinear Equations in Two Variables
Preview Exercises
Exercises 66–68 will help you prepare for the material covered in
the next section.
66. Solve by the substitution method:
4x + 3y = 4
y = 2x - 7.
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.
Recognize systems of nonlinear
equations in two variables.
783
67. Solve by the addition method:
2x + 4y = - 4
3x + 5y = - 3.
68. Graph x - y = 3 and 1x - 222 + 1y + 322 = 4 in the same
rectangular coordinate system. What are the two intersection
points? Show that each of these ordered pairs satisfies both
equations.
Systems of Nonlinear Equations in Two Variables
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:
e
x2=2y+10
3x-y=9
Not in the form
Ax + By = C.
The term x2 is
not linear.
e
y=x2+3
x2+y2=9.
Neither equation is in
the form Ax + By = C.
The terms x2 and y2 are
not linear.
A solution of a nonlinear system in two variables is an ordered pair of real
numbers that satisfies 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.
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784 Chapter 8 Systems of Equations and Inequalities
�
Solve nonlinear systems by
substitution.
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
first example, this equation is quadratic.
Solving a Nonlinear System by the Substitution Method
EXAMPLE 1
Solve by the substitution method:
b
x2 = 2y + 10
3x - 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 first power in either of the equations.
By solving for y in the second equation, which has a coefficient of - 1, we can avoid
fractions.
3x - y = 9
This is the second equation in the given system.
3x = y + 9
3x - 9 = y
Add y to both sides.
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 first equation.
x2=2 y +10
y= 3x-9
This gives us an equation in one variable, namely
x2 = 213x - 92 + 10.
The variable y has been eliminated.
Step 3 Solve the resulting equation containing one variable.
x2 = 213x - 92 + 10
This is the equation containing one variable.
x2 = 6x - 18 + 10
Use the distributive property.
2
x = 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.
1x - 421x - 22 = 0
Factor.
x - 4 = 0 or
x = 4
x - 2 = 0
x = 2
Set each factor equal to 0.
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 and 2 for
x into the equation y = 3x - 9.
If x is 4,
If x is 2,
y = 3142 - 9 = 3,
y = 3122 - 9 = - 3,
so 14, 32 is a solution.
so 12, -32 is a solution.
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Section 8.4 Systems of Nonlinear Equations in Two Variables
785
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
4 2 ⱨ 2132 + 10
These are the given equations.
3142 - 3 ⱨ 9
16 ⱨ 6 + 10
Let x = 4 and y = 3.
12 - 3 ⱨ 9
Simplify.
y
16 = 16,
4
3
2
1
−5 −4 −3 −2 −1−1
x2
−2
−3
−4
−5
−6
−7
−8
−9
= 2y + 10
(4, 3)
1 2 3 4 5
x
x2 = 2y + 10
3x - y = 9
4 ⱨ -6 + 10
3x − y = 9
Figure 8.9 Points of intersection
illustrate the nonlinear system’s
solutions.
4 = 4,
True statements result.
These are the given equations.
3122 - 1-32 ⱨ 9
Let x = 2 and y = - 3.
6 + 3ⱨ9
9 = 9,
true
Simplify.
true
True statements result.
The ordered pair 12, - 32 also satisfies both equations and is a solution of the system.
The solutions are (4, 3) and 12, -32, and the solution set is 514, 32, 12, -326.
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 - 1
4x - y = - 1.
Solving a Nonlinear System by the Substitution Method
EXAMPLE 2
Study Tip
true
The ordered pair (4, 3) satisfies both equations. Thus, (4, 3) is a solution of the system.
Now let’s check 12, -32. Replace x with 2 and y with - 3 in both given equations.
2 2 ⱨ 21-32 + 10
(2, −3)
9 = 9,
true
Solve by the substitution method:
Recall that
1x - h2 + 1y - k2 = r
2
2
2
describes a circle with center 1h, k2
and radius r.
b
x - y = 3
1x - 22 + 1y + 322 = 4.
2
(The graph is a line.)
(The graph is a circle.)
Solution Graphically, we are finding the intersection of a line and a circle with
center 12, -32 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 first equation. (We could also solve for y.)
x - y = 3
This is the first 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
1y + 3 - 222 + 1y + 322 = 4.
The variable x has been eliminated.
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786 Chapter 8 Systems of Equations and Inequalities
Step 3 Solve the resulting equation containing one variable.
1y + 3 - 222 + 1y + 322 = 4
1y + 122 + 1y + 322 = 4
y2 + 2y + 1 + y2 + 6y + 9 = 4
2y2 + 8y + 10 = 4
2y2 + 8y + 6 = 0
21y2 + 4y + 32 = 0
21y + 321y + 12 = 0
y + 3 = 0 or y + 1 = 0
y = -3
y = -1
This is the equation containing one variable.
Combine numerical terms in the first parentheses.
Use the formula 1A + B22 = A2 + 2AB + B2 to
square y + 1 and y + 3.
Combine like terms on the left.
Subtract 4 from both sides and set the quadratic
equation equal to 0.
Factor out 2.
Factor completely.
Set each variable factor equal to 0.
Solve for y.
y
x−y=3
3
2
(2, −1)
1
−3 −2 −1−1
(0, −3)
1
3 4 5 6 7
x
−2
−4
−5
−6
−7
(x −
2)2
+ (y +
3)2
=4
Figure 8.10 Points of intersection
illustrate the nonlinear system’s
solutions.
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,
If y = - 1:
x = - 1 + 3 = 2,
2
Solve by the substitution method:
b
Solve nonlinear systems by
addition.
so 12, -12 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 satisfies both given equations, x - y = 3
and (x - 2)2 + (y + 3)2 = 4. The solutions are 10, - 32 and 12, -12, and the
solution set of the given system is 510, - 32, 12, - 126.
Figure 8.10 shows the graphs of the equations in the system and the solutions
as intersection points.
Check Point
�
so 10, -32 is a solution.
x + 2y = 0
1x - 12 + 1y - 122 = 5.
2
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 coefficients 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:
b
4x2 + y2 = 13 Equation 1
x2 + y2 = 10. Equation 2
Solution We can use the same steps that we did when we solved linear systems
by the addition method.
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Section 8.4 Systems of Nonlinear Equations in Two Variables
787
Step 1 Write both equations in the form Ax 2 ⴙ By 2 ⴝ 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 x 2-coefficients or the sum of the y 2-coefficients is 0.
We can eliminate y2 by multiplying Equation 2 by - 1.
b
4x2 + y2 = 13
x2 + y2 = 10
No change
"
Multiply by - 1.
"
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 find 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 11, - 32 are solutions. If x = - 1,
y
(−1, 3)
5
4
1-122 + y2 = 10
4x2 + y2 = 13
y2 = 9
(1, 3)
2
1
−5 −4 −3 −2 −1−1
−4
−5
1 2 3 4 5
x
x2 + y2 = 10
(1, −3)
Figure 8.11 A system with four
solutions
The steps are the same as before.
y = ; 3.
−2
(−1, −3)
Replace x with -1 in Equation 2.
1- 1, 32 and 1-1, - 32 are solutions.
Step 6 Check. Take a moment to show that each of the four ordered pairs satisfies
the given equations, 4x2 + y2 = 13 and x2 + y2 = 10. The solution set of the given
system is 511, 32, 11, -32, 1-1, 32, 1-1, -326.
Figure 8.11 shows the graphs of the equations in the system and the solutions
as intersection points.
Check Point
3
Solve the system:
b
3x2 + 2y2 = 35
4x2 + 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.
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788 Chapter 8 Systems of Equations and Inequalities
Study Tip
Solving a Nonlinear System by the Addition Method
EXAMPLE 4
When solving nonlinear systems,
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.
Solve the system:
b
y = x2 + 3
x + y2 = 9.
2
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 how
like terms
are arranged
in columns.
e
–x2+y
= 3
x2
+y 2= 9
y+y 2=12
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
1y + 421y - 32 = 0
y + 4 = 0 or y - 3 = 0
y = -4
y = 3
Subtract 12 from both sides and set the
quadratic equation equal to 0.
Factor.
Set each factor equal to 0.
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
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
7
6
5
4
y=
x2
+3
(0, 3)
2
1
−5 −4 −3 −2 −1−1
−2
−3
1 2 3 4 5
x
x2 + y2 = 9
Figure 8.12 A system with one
real solution
Subtract 3 from both sides.
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) satisfies the given equations, y = x2 + 3 and
x2 + y2 = 9. The solution set of the system is 510, 326. Figure 8.12 shows the
system’s graphs and the solution as an intersection point.
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Section 8.4 Systems of Nonlinear Equations in Two Variables
4
Check Point
Solve the system:
b
�
Solve problems using systems of
nonlinear equations.
y
Figure 8.13 Building an enclosure
y = x2 + 5
x2 + 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.
An Application of a Nonlinear System
EXAMPLE 5
x
789
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 first
condition is that you have 36 yards of fencing.
Fencing along
both lengths
2x
plus
fencing along
both widths
plus
+
2y
+
fencing for the
internal divider
equals
y
=
36
yards.
36
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.
Length
x
times
ⴢ
width
is 54 square yards.
y =
54
Step 3 Solve the system and answer the problem’s question. We must solve the system
b
2x + 3y = 36
xy = 54.
Equation 1
Equation 2
We will use substitution. Because Equation 1 has no coefficients 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 =
Now we substitute
54
.
x
54
for y in Equation 1 and solve for x.
x
2x + 3y = 36 This is Equation 1.
54
54
for y.
2x + 3 #
= 36 Substitute
x
x
162
2x +
= 36 Multiply.
x
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790 Chapter 8 Systems of Equations and Inequalities
x a2x+
162
b =36 ⴢ x Clear fractions by multiplying both sides by x.
x
2x2 + 162 = 36x
2x2 - 36x + 162 = 0
21x2 - 18x + 812 = 0
21x - 922 = 0
x - 9 = 0
x = 9
Use the distributive property on the left side.
Subtract 36x from both sides and set the
quadratic equation equal to 0.
Factor out 2.
Factor completely using
A2 - 2AB + B2 = 1A - B22.
Set the repeated factor equal to zero.
Solve for x.
We back-substitute this value of x into y =
54
.
x
x
If x = 9, y =
54
= 6.
9
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.
y
Figure 8.13 (repeated)
Check Point
5
Find the length and width of a rectangle whose perimeter is 20 feet
and whose area is 21 square feet.
Exercise Set 8.4
Practice Exercises
13. b
xy = 3
x2 + y2 = 10
14. b
xy = 4
x2 + y2 = 8
16. b
x + y = -3
x2 + 2y2 = 12y + 18
In Exercises 1–18, solve each system by the substitution method.
1. b
x + y = 2
y = x2 - 4
2. b
x - y = -1
y = x2 + 1
15. b
x + y = 1
x2 + xy - y2 = - 5
3. b
x + y = 2
y = x2 - 4x + 4
4. b
2x + y = - 5
y = x2 + 6x + 7
17. b
5. b
y = x2 - 4x - 10
y = - x2 - 2x + 14
6. b
y = x2 + 4x + 5
y = x2 + 2x - 1
x + y = 1
2x + y = 4
18. b
1x - 122 + 1y + 222 = 10
1x + 122 + 1y - 222 = 4
In Exercises 19–28, solve each system by the addition method.
x2 + y2 = 25
7. b
x - y = 1
x2 + y2 = 5
8. b
3x - y = 5
19. b
x2 + y2 = 13
x2 - y2 = 5
20. b
4x2 - y2 = 4
4x2 + y2 = 4
9. b
xy = 6
2x - y = 1
10. b
xy = - 12
x - 2y + 14 = 0
21. b
x2 - 4y2 = - 7
3x2 + y2 = 31
22. b
3x2 - 2y2 = - 5
2x2 - y2 = - 2
11. b
y2 = x2 - 9
2y = x - 3
12. b
x2 + y = 4
2x + y = 1
23. b
3x2 + 4y2 - 16 = 0
2x2 - 3y2 - 5 = 0
24. b
16x2 - 4y2 - 72 = 0
x2 - y2 - 3 = 0
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Section 8.4 Systems of Nonlinear Equations in Two Variables
25. b
x2 + y2 = 25
1x - 822 + y2 = 41
26. b
x2 + y2 = 5
x2 + 1y - 822 = 41
27. b
y2 - x = 4
x2 + y2 = 4
28. b
x2 - 2y = 8
x2 + y2 = 16
In Exercises 29–42, solve each system by the method of your choice.
29. b
3x2 + 4y2 = 16
2x2 - 3y2 = 5
2x2 + y2 = 18
31. b
xy = 4
30. b
x + y2 = 4
x2 + y2 = 16
x2 + 4y2 = 20
32. b
xy = 4
33. b
x2 + 4y2 = 20
x + 2y = 6
34. b
3x2 - 2y2 = 1
4x - y = 3
35. b
x3 + y = 0
x2 - y = 0
36. b
x3 + y = 0
2x2 - y = 0
37. b
x2 + 1y - 222 = 4
x2 - 2y = 0
38. b
x2 - y2 - 4x + 6y - 4 = 0
x2 + y2 - 4x - 6y + 12 = 0
y = 1x + 322
39. b
x + 2y = - 2
41. b
x2 + y2 + 3y = 22
2x + y = - 1
1x - 122 + 1y + 122 = 5
40. b
2x - y = 3
42. b
54. The system, whose graphs are a line with negative slope and
a parabola whose equation has a negative leading coefficient,
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 2y 2 - 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 first 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.
x - 3y = - 5
x2 + y2 - 25 = 0
44. The sum of two numbers is 20 and their product is 96. Find
the numbers.
s
he
nc
0i
1
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 find the numbers.
43. The sum of two numbers is 10 and their product is 24. Find
the numbers.
791
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.
45. The difference between the squares of two numbers is 3.
Twice the square of the first number increased by the square
of the second number is 9. Find the numbers.
A = 108 feet2
15 feet
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 first number is 19. Find the numbers.
W
L
Practice Plus
In Exercises 47–52, solve each system by the method of your choice.
2
2
47. b
2x + xy = 6
x2 + 2xy = 0
48. b
4x + xy = 30
x2 + 3xy = - 9
49. b
- 4x + y = 12
y = x3 + 3x2
50. b
-9x + y = 45
y = x3 + 5x2
1
3
+ 2 = 7
x2
y
51. d
5
2
- 2 = -3
2
x
y
2
1
+ 2 = 11
x2
y
52. d
4
2
- 2 = - 14
2
x
y
61. The figure shows a square floor plan with a smaller square
area that will accommodate a combination fountain and
pool. The floor with the fountain-pool area removed has an
area of 21 square meters and a perimeter of 24 meters. Find
the dimensions of the floor and the dimensions of the square
that will accommodate the pool.
y
y
x
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 coefficient,
has two solutions.
y
x
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792 Chapter 8 Systems of Equations and Inequalities
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, find the length and width of the cardboard that must
be used.
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 x 2 - y2 = 5 and 3x2 - 2y2 = 19 to illustrate
your explanation.
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.
The Changing Pattern of Work in the United States, 1900–2005
Percentage of Total Labor Force
White Collar
Blue Collar
Farming
76
70
63
60
51
50
30
44
41 38
40
44
41
36
28
34
28
21
8
10
0
22
20
20
1900
1920
67. Verify your solutions to any five 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.
90
80
Technology Exercises
1960
1940
3
1980
Critical Thinking Exercises
2
2005
Year
Source: U.S. Department of Labor
The data can be modeled by linear and quadratic functions.
White collar
0.5x-y=–18
Blue collar
y=–0.004x2+0.23x+41
Farming
0.4x+y=35
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 find
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 specified in part (c)?
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 x 2 + y2 = 36
and y = 1x - 222 - 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 orderedpair solution.
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Mid-Chapter Check Point
77. The points of intersection of the graphs of xy = 20 and
x 2 + y2 = 41 are joined to form a rectangle. Find the area of
the rectangle.
793
Solve the systems in Exercises 79–80.
80. log x2 = y + 3
79. logy x = 3
logy14x2 = 5
78. Find a and b in this figure.
log x = y - 1
Preview Exercises
a
10
Exercises 81–83 will help you prepare for the material covered in
the next section. In each exercise, graph the linear function.
17
82. f1x2 = - 23 x
81. 2x - 3y = 6
b
Chapter
83. f1x2 = - 2
9
8
Mid-Chapter Check Point
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 infinitely many solutions. We applied systems to a
variety of situations, including finding the break-even
point for a business, finding a quadratic function from
three points on its graph, and finding a rational
function’s partial fraction decomposition.
In Exercises 1–12, solve each system by the method of your
choice.
1. b
x = 3y - 7
4x + 3y = 2
y
2x
+
= 6
3
5
3. d
y
x
= -4
6
2
5. b
2x + 5y = 3
3x - 2y = 1
2x - y + 2z = - 8
7. c x + 2y - 3z = 9
3x - y - 4z = 3
9. b
11. b
2. b
4. b
3x + 4y = - 5
2x - 3y = 8
y = 4x - 5
8x - 2y = 10
x
1
- y =
6. c 12
4
4x - 48y = 16
x 3z = - 5
8. c 2x - y + 2z = 16
7x - 3y - 5z = 19
x2 + y2 = 9
x + 2y - 3 = 0
10. b
3x2 + 2y2 = 14
2x2 - y2 = 7
y = x2 - 6
x2 + y2 = 8
12. b
x - 2y = 4
2y2 + xy = 8
In Exercises 13–16, write the partial fraction decomposition of
each rational expression.
13.
x2 - 6x + 3
1x - 223
14.
15.
x2 + 4x - 23
1x + 321x2 + 42
16.
10x2 + 9x - 7
1x + 221x2 - 12
x3
1x2 + 42
2
17. A company is planning to manufacture PDAs (personal
digital assistants).The fixed 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 profit 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 flowers consisting of roses and carnations costs
$39, how many of each type of flower is in the bouquet?
19. Find the measure of each angle whose degree measure is
represented with a variable.
y
3y + 20
x
20. Find the quadratic function y = ax2 + bx + c whose graph
passes through the points 1 -1, 02, (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.