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A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 783 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 784 784 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 785 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 786 786 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 787 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 788 788 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 789 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 A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 790 790 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 A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 791 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 A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 792 792 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. A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 793 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.