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446 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? 448 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 449 (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. 450 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 6-3 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. 452 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. 454 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 ★★ 455 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)