Download SECTION 6-3 Systems Involving Second

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6 Systems of Equations and Inequalities
SECTION
6-3
Systems Involving Second-Degree Equations
• Solution by Substitution
• Other Solution Methods
If a system of equations contains any equations that are not linear, then the system is
called a nonlinear system. In this section we investigate nonlinear systems involving second-degree terms such as
x 2 y2 5
x 2 2y2 2
x 2 3xy y2 20
3x y 1
xy 2
xy y2 0
It can be shown that such systems have at most four solutions, some of which may
be imaginary. Since we are interested in finding both real and imaginary solutions to
the systems we consider, we now assume that the replacement set for each variable
is the set of complex numbers, rather than the set of real numbers.
• Solution by
Substitution
EXAMPLE 1
The substitution method used to solve linear systems of two equations in two variables is also an effective method for solving nonlinear systems. This process is best
illustrated by examples.
Solving a Nonlinear System by Substitution
Solve the system:
Solution
x 2 y2 5
3x y 1
Solve the second equation for y in terms of x; then substitute for y in the first equation to obtain an equation that involves x alone.
3x y 1
y 1 3x
aEEbEEc
Substitute this expression for y in the first equation.
x 2 y2 5
x 2 (1 3x)2 5
10x 2 6x 4 0
5x 2 3x 2 0
Simplify and write in standard quadratic form.
Divide through by 2 to simplify further.
(x 1)(5x 2) 0
x 1, 25
If we substitute these values back into the equation y 1 3x, we obtain two
solutions to the system:
6-3
y
3x y 1
5
x2 y2 5
5
5
x
Systems Involving Second-Degree Equations
x1
x 25
y 1 3(1) 2
y 1 3( 25) 11
5
447
A check, which you should provide, verifies that (1, 2) and (25, 11
5 ) are both solutions to the system. These solutions are illustrated in Figure 1. However, if we substitute the values of x back into the equation x2 y2 5, we obtain
x1
5
x 25
12 y2 5
FIGURE 1
( 25)2 y2 5
y2 4
y2 121
25
y 2
y 11
5
It appears that we have found two additional solutions, (1, 2) and (52, 11
5 ). But neither of these solutions satisfies the equation 3x y 1, which you should verify.
So, neither is a solution of the original system. We have produced two extraneous
roots, apparent solutions that do not actually satisfy both equations in the system.
This is a common occurrence when solving nonlinear systems.
It is always very important to check the solutions of any nonlinear system to ensure that extraneous roots have not been introduced.
Matched Problem 1
EXPLORE-DISCUSS 1
Solve the system:
x 2 y2 10
2x y 1
In Example 1, we saw that the line 3x y 1 intersected the circle x2 y2 5 in two points.
(A) Consider the system
x 2 y2 5
3x y 10
Graph both equations in the same coordinate system. Are there any real solutions to this system? Are there any complex solutions? Find any real or complex solutions.
(B) Consider the family of lines given by
3x y b
b any real number
What do all these lines have in common? Illustrate graphically the lines in
this family that intersect the circle x2 y2 5 in exactly one point. How
many such lines are there? What are the corresponding value(s) of b? What
are the intersection points? How are these lines related to the circle?
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6 Systems of Equations and Inequalities
EXAMPLE 2
Solving a Nonlinear System by Substitution
Solve:
Solution
x 2 2y2 2
xy 2
Solve the second equation for y, substitute in the first equation, and proceed as before.
xy 2
y
x2 2
2
x
2
2
x
x2 2
8
2
x2
x 4 2x 2 8 0
Multiply both sides by x2 and simplify.
u2 2u 8 0
Substitute u x2 to transform to quadratic form and solve.
(u 4)(u 2) 0
u 4, 2
Thus,
x2 4
x 2
y
5
For x 2, y xy 2
or
2
1.
2
For x 2, y 5
or
2
1.
2
x 2 2
x 2 i
2
For x i
2, y 2
i
2.
i
2
For x i
2, y 2
i
2.
i
2
x
x 2 2y 2 2
Thus, the four solutions to this system are (2, 1), (2, 1), (i
2, i
2), and
(i
2, i
2). Notice that two of the solutions involve imaginary numbers. These
imaginary solutions cannot be illustrated graphically (see Fig. 2); however, they do
satisfy both equations in the system (verify this).
FIGURE 2
Matched Problem 2
EXPLORE-DISCUSS 2
Solve:
3x2 y2 6
xy 3
(A) Refer to the system in Example 2. Could a graphing utility be used to find
the real solutions of this system? The imaginary solutions?
6-3
Systems Involving Second-Degree Equations
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(B) In general, explain why graphic approximation techniques can be used to
approximate the real solutions of a system, but not the complex solutions.
EXAMPLE 3
Design
An engineer is to design a rectangular computer screen with a 19-inch diagonal and
a 175-square-inch area. Find the dimensions of the screen to the nearest tenth of an
inch.
Solution
es
ch
19
in
Sketch a rectangle letting x be the width and y the height (Fig. 3). We obtain the
following system using the Pythagorean theorem and the formula for the area of a
rectangle:
y
x 2 y2 192
xy 175
x
FIGURE 3
This system is solved using the procedures outlined in Example 2. However, in this
case, we are only interested in real solutions. We start by solving the second equation for y in terms of x and substituting the result into the first equation.
y
x2 175
x
1752
192
x2
x 4 30,625 361x 2
x4 361x 2 30,625 0
Multiply both sides by x2 and simplify
Quadratic in x2
Solve the last equation for x 2 using the quadratic formula, then solve for x:
x
361 3612 4(1)(30,625)
2
15.0 inches or 11.7 inches
Substitute each choice of x into y 175/x to find the corresponding y values:
For x 15.0 inches,
y
175
11.7 inches
15
For x 11.7 inches,
y
175
15.0 inches
11.7
Assuming the screen is wider than it is high, the dimensions are 15.0 by 11.7 inches.
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6 Systems of Equations and Inequalities
Matched Problem 3
An engineer is to design a rectangular television screen with a 21-inch diagonal and
a 209-square-inch area. Find the dimensions of the screen to the nearest tenth of an
inch.
Since Example 3 is only concerned with real solutions, graphic techniques can
also be used to approximate the solutions (see Fig. 4). As we saw in Section 3-1,
graphing a circle on a graphing utility requires two functions, one for the upper half
of the circle and another for the lower half. [Note: since x and y must be nonnegative
real numbers, we ignore the intersection points in the third quadrant—see Fig 4(a).]
FIGURE 4 Graphic
solution of x 2 y2 192,
xy 175.
40
16
60
60
8
8
18
40
10
(a) y1 361 x2
y2 361 x2
175
y3 x
• Other Solution
16
18
10
(b) Intersection point:
(11.7, 15.0)
(c) Intersection point:
(15.0, 11.7)
We now look at some other techniques for solving nonlinear systems of equations.
Methods
EXAMPLE 4
Solving a Nonlinear System by Elimination
Solve:
Solution
x2 y2 5
x2 2y2 17
This type of system can be solved using elimination by addition. Multiply the second
equation by 1 and add:
x 2 y2 5
x 2y2 17
3y2 12
y2 4
y 2
2
Now substitute y 2 and y 2 back into either original equation to find x.
For y 2,
For y 2,
x 2 (2)2 5
x 3
x 2 (2)2 5
x 3
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Systems Involving Second-Degree Equations
451
Thus, (3, 2), (3, 2), (3, 2), and (3, 2), are the four solutions to the system.
The check of the solutions is left to you.
Matched Problem 4
EXAMPLE 5
Solve:
Solving a Nonlinear System Using Factoring and Substitution
Solve:
Solution
2x 2 3y2 5
3x 2 4y2 16
x 2 3xy y2 20
x 2 xy y2 0
Factor the left side of the equation that has a 0 constant term:
xy y2 0
y(x y) 0
y0
yx
or
Thus, the original system is equivalent to the two systems:
y 0
or
y x
x 2 3xy y2 20
or
x 2 3xy y2 20
These systems are solved by substitution.
y 0
First System
x 3xy y2 20
2
Substitute y 0 in the second equation, and solve for x.
x 2 3x(0) (0)2 20
x 2 20
x 20 2
5
Second System
y x
x 2 3xy y2 20
Substitute y x in the second equation and solve for x.
x 2 3xx x 2 20
5x 2 20
x2 4
x 2
Substitute these values back into y x to find y.
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6 Systems of Equations and Inequalities
For x 2, y 2.
For x 2, y 2.
The solutions for the original system are (2
5, 0), (2
5, 0), (2, 2), and
(2, 2). The check of the solutions is left to you.
Matched Problem 5
Solve:
x 2 xy y2 9
2x 2 xy 0
Example 5 is somewhat specialized. However, it suggests a procedure that is
effective for some problems.
EXAMPLE 6
Graphic Approximations of Real Solutions
Use a graphing utility to approximate real solutions to two decimal places:
x 2 4xy y2 12
2x 2 2xy y2 6
Solution
Before we can enter these equations in our graphing utility, we must solve for y:
x 2 4xy y2 12
y2 4xy (x 2 12) 0
2x 2 2xy y2 6
y2 2xy (2x 2 6) 0
Applying the quadratic formula to each equation, we have
y
4x 16x 2 4(x 2 12)
2
y
4x 12x 2 48
2
2x 3x 2 12
2x 4x 2 4(2x 2 6)
2
2x 24 4x 2
2
x 6 x 2
Since each equation has two solutions, we must enter four functions in the graphing utility, as shown in Figure 5(a). Examining the graph in Figure 5(b), we see that
there are four intersection points. Using the built-in intersection routine repeatedly
(details omitted), we find that the solutions to two decimal places are (2.10, 0.83),
(0.37, 2.79), (0.37, 2.79), and (2.10, 0.83).
FIGURE 5
5
7.6
7.6
5
(a)
(b)
6-3
Matched Problem 6
Systems Involving Second-Degree Equations
453
Use a graphing utility to approximate real solutions to two decimal places:
x 2 8xy y2 70
2x 2 2xy y2 20
Answers to Matched Problems
1. (1, 3), (59 , 13
2. (
3, 3), (
3, 3), (i, 3i), (i, 3i)
3. 17.1 by 12.2 in
5)
4. (2, 1), (2, 1), (2, 1), (2, 1)
5. (0, 3), (0, 3), (
3, 2
3), (
3, 2
3)
6. (3.89, 1.68), (0.96, 5.32), (0.96, 5.32), (3.89, 1.68)
EXERCISE
6-3
A
Solve each system in Problems 1–12.
1. x 2 y2 169
x 12
2. x 2 y2 25
y 4
3. 8x 2 y2 16
y 2x
4. y2 2x
xy
5. 3x 2 2y2 25
xy0
6. x 2 4y2 32
x 2y 0
7.
y2 x
x 2y 2
9. 2x 2 y2 24
x 2 y2 12
11.
x 2 y2 10
16x 2 y2 25
1
2
8. x 2 2y
3x y 2
10. x 2 y2 3
x 2 y2 5
12. x 2 2y2 1
x 2 4y2 25
B
Solve each system in Problems 13–24.
13. xy 4 0
xy2
14. xy 6 0
xy4
15. x2 2y2 6
xy 2
16. 2x2 y2 18
xy 4
17. 2x2 3y2 4
4x2 2y2 8
18. 2x2 3y2 10
x2 4y2 17
19. x2 y2 2
x2 y2 x
20. x2 y2 20
y2 x2 y
21. x2 y2 9
y2 x2 9 2y
22. x2 y2 16
x2 y2 4 x
23. x2 y2 3
xy 2
24. y2 5x2 1
xy 2
An important type of calculus problem is to find the area
between the graphs of two functions. To solve some of these
problems it is necessary to find the coordinates of the points
of intersections of the two graphs. In Problems 25–32, find
the coordinates of the points of intersections of the two
given equations.
25. y 5 x2, y 2 2x
26. y 5x x2, y x 3
27. y x2 x, y 2x
28. y x2 2x, y 3x
29. y x2 6x 9, y 5 x
30. y x2 2x 3, y 2x 4
31. y 8 4x x2, y x2 2x
32. y x2 4x 10, y 14 2x x2
33. Consider the circle with equation x2 y2 5 and the
family of lines given by 2x y b, where b is any real
number.
(A) Illustrate graphically the lines in this family that intersect the circle in exactly one point, and describe the relationship between the circle and these lines.
(B) Find the values of b corresponding to the lines in part
A, and find the intersection points of the lines and the
circle.
(C) How is the line with equation x 2y 0 related to this
family of lines? How could this line be used to find the
intersection points in part B?
34. Consider the circle with equation x2 y2 25 and the
family of lines given by 3x 4y b, where b is any real
number.
(A) Illustrate graphically the lines in this family that intersect the circle in exactly one point, and describe the relationship between the circle and these lines.
(B) Find the values of b corresponding to the lines in part
A, and find the intersection points of the lines and the
circle.
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6 Systems of Equations and Inequalities
(C) How is the line with equation 4x 3y 0 related to
this family of lines? How could this line be used to find
the intersection points and the values of b in part B?
of the circle is 6.5 inches, and the area of the rectangle is 15
square inches. Find the dimensions of the rectangle.
C
Solve each system in Problems 35–42.
6.5 inches
35. 2x 5y 7xy 8
xy 3 0
36. 2x 3y xy 16
xy 5 0
37. x2 2xy y2 1
x 2y 2
38. x2 xy y2 5
yx3
39. 2x2 xy y2 8
x2 y2 0
40. x2 2xy y2 36
x2 xy 0
41. x2 xy 3y2 3
x2 4xy 3y2 0
42. x2 2xy 2y2 16
xy x2 y2 0
★
55. Construction. A rectangular swimming pool with a deck 5
feet wide is enclosed by a fence as shown in the figure. The
surface area of the pool is 572 square feet, and the total area
enclosed by the fence (including the pool and the deck) is
1,152 square feet. Find the dimensions of the pool.
In Problems 43–48, use a graphing utility to approximate
the real solutions of each system to two decimal places.
43. x2 2xy y2 1
3x2 4xy y2 2
44. x2 4xy y2 2
8x2 2xy y2 9
45. 3x 4xy y 2
2x2 2xy y2 9
46. 5x 4xy y 4
4x2 2xy y2 16
2
2
2
Fence
2
5 ft
5 ft
47. 2x 2xy y 9
4x2 4xy y2 x 3
2
2
Pool
48. 2x2 2xy y2 12
4x2 4xy y2 x 2y 9
5 ft
APPLICATIONS
49. Numbers. Find two numbers such that their sum is 3 and
their product is 1.
★
50. Numbers. Find two numbers such that their difference is 1
and their product is 1. (Let x be the larger number and y the
smaller number.)
51. Geometry. Find the lengths of the legs of a right triangle
with an area of 30 square inches if its hypotenuse is 13
inches long.
56. Construction. An open-topped rectangular box is formed
by cutting a 6-inch square from each corner of a rectangular piece of cardboard and bending up the ends and sides.
The area of the cardboard before the corners are removed is
768 square inches, and the volume of the box is 1,440 cubic
inches. Find the dimensions of the original piece of cardboard.
6 in.
52. Geometry. Find the dimensions of a rectangle with an area
of 32 square meters if its perimeter is 36 meters long.
6 in.
53. Design. An engineer is designing a small portable television set. According to the design specifications, the set
must have a rectangular screen with a 7.5-inch diagonal and
an area of 27 square inches. Find the dimensions of the
screen.
6 in.
54. Design. An artist is designing a logo for a business in the
shape of a circle with an inscribed rectangle. The diameter
6 in.
6 in.
6 in.
6 in.
★★
5 ft
6 in.
57. Transportation. Two boats leave Bournemouth, England,
6-4 Systems of Linear Inequalities in Two Variables
at the same time and follow the same route on the 75-mile
trip across the English Channel to Cherbourg, France. The
average speed of boat A is 5 miles per hour greater than the
average speed of boat B. Consequently, boat A arrives at
Cherbourg 30 minutes before boat B. Find the average
speed of each boat.
SECTION
6-4
★★
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58. Transportation. Bus A leaves Milwaukee at noon and
travels west on Interstate 94. Bus B leaves Milwaukee 30
minutes later, travels the same route, and overtakes bus A at
a point 210 miles west of Milwaukee. If the average speed
of bus B is 10 miles per hour greater than the average speed
of bus A, at what time did bus B overtake bus A?
Systems of Linear Inequalities in Two Variables
• Graphing Linear Inequalities in Two Variables
• Solving Systems of Linear Inequalities Graphically
• Application
Many applications of mathematics involve systems of inequalities rather than systems
of equations. A graph is often the most convenient way to represent the solutions of
a system of inequalities in two variables. In this section, we discuss techniques for
graphing both a single linear inequality in two variables and a system of linear
inequalities in two variables.
• Graphing Linear
We know how to graph first-degree equations such as
Inequalities in
Two Variables
y 2x 3
2x 3y 5
and
but how do we graph first-degree inequalities such as
y 2x 3
2x 3y 5
and
Actually, graphing these inequalities is almost as easy as graphing the equations. But
before we begin, we must discuss some important subsets of a plane in a rectangular
coordinate system.
A line divides a plane into two halves called half-planes. A vertical line divides
a plane into left and right half-planes [Fig. 1(a)]; a nonvertical line divides a plane
into upper and lower half-planes [Fig. 1(b)].
y
y
FIGURE 1 Half-planes.
Left
half-plane
Upper
half-plane
Right
half-plane
x
x
Lower
half-plane
(a)
(b)