1 10.4 Graphing Nonlinear Sstems 10.4 OBJECTIVES 1. Graph a sstem of nonlinear equations 2. Find ordered pairs associated with the solution set of a nonlinear sstem 3. Graph a sstem of nonlinear inequalities 4. Use substitution to find the solution set for a nonlinear sstem. 5. Identif the solution set of a sstem of nonlinear inequalities In Section 5.1, we solved a sstem of linear equations b graphing the lines corresponding to those equations, and then recording the point of intersection. That point represented the solution to the sstem of equations. We will use a similar method to find the solution set for a nonlinear sstem. A sstem with two or more conic curves can have zero, one, two, three, or four solutions. The following graphs represent each of those possibilities. Zero Solutions One Solution Two Solutions Three Solutions Four Solutions For the remainder of this section, we will restrict our discussion to a sstem that has as its graph a line and a parabola. Such a sstem has either zero, one, or two solutions. Eample 1 Solving a Sstem of Nonlinear Equations Solve the following sstem of equations
2 796 CHAPTER 10 GRAPHS OF CONIC SECTIONS First, we will graph the sstem. From this graph we will be able to see the number of solutions. The graph will also give us a wa to check the reasonableness of our algebraic results. NOTE Use our calculator to approimate the solutions for the sstem. Let s use the method of substitution to solve the sstem. Substituting 6, from the second equation, for in the first equation, we get ( 4)( 1) The values for the solutions are 1 and 4. We can substitute these values for in either equation to solve for, but we know from the second equation that 6. The solution set is ( 1, 6), (4, 6). Looking at the graph, we see that this is a reasonable solution set for the sstem. CHECK YOURSELF 1 Solve the following sstem of equations Of course, not ever quadratic epression is factorable. In Eample 2, we must use the quadratic formula. Eample 2 Solving a Nonlinear Sstem Solve the following sstem of equations
3 GRAPHING NONLINEAR SYSTEMS SECTION Let s look at the graph of the sstem. We see two points of intersection, but neither seems to be an integer value for. Let s solve the sstem algebraicall. Using the method of substitution, we find The result is not factorable, so we use the quadratic formula to find the solutions The two points of intersection are, 7 It is difficult to 2 and 2. check these points against the graph, so we will approimate them. The approimate solutions (to the nearest tenth) are ( 2.6, 7) and (1.6, 7). The graph indicates that these are reasonable answers. CHECK YOURSELF 2 Solve the following sstem of equations As was stated earlier, not ever sstem has two solutions. In Eample 3, we will see a sstem with no real solution. Eample 3 Solving a Sstem of Nonlinear Equations Solve the following sstem of equations
4 798 CHAPTER 10 GRAPHS OF CONIC SECTIONS As we did with the previous sstems, we will first look at the graph of the sstem. Using the method of substitution, we get Using the quadratic formula, we can confirm that there are no real solutions to this sstem. ( 2) 2( 2) 2 4(1)(3) 2(1) CHECK YOURSELF Solve the following sstem of equations Consider the sstem consisting of the following two equations: The graph of the sstem indicates there are four solutions.
5 GRAPHING NONLINEAR SYSTEMS SECTION We could approimate the solutions, then check those approimations b substitution. But how could we find the solutions algebraicall? Eample 4 illustrates the elimination method. Eample 4 Solving a Nonlinear Sstem b Elimination Solve the following sstem algebraicall As was the case with linear sstems, we can eliminate one of the variables. In this case, adding the equations eliminates the variable Dividing b 4, we have 2 9, so 3 Substituting the value 3 into the first equation (3) Two of the ordered pairs in the solution set are (3, 4) and (3, 4). Substituting the value 3 into the first equation ( 3) The other two pairs in the solution set are ( 3, 4) and ( 3, 4).
6 800 CHAPTER 10 GRAPHS OF CONIC SECTIONS The solutions set is ( 3, 4),( 3, 4), (3, 4), (3, 4) ( 3, 4) (3, 4) ( 3, 4) (3, 4) CHECK YOURSELF 4 Solve b the elimination method Recall that a sstem of inequalities has as its solutions the set of all ordered pairs that make ever inequalit in the sstem a true statement. We almost alwas epress the solutions to a sstem of inequalities graphicall. We will do the same thing with nonlinear sstems. Eample 5 Solving a Sstem of Nonlinear Inequalities Solve the following sstem From Eample 1, we have the graph of the related sstem of equations.
7 GRAPHING NONLINEAR SYSTEMS SECTION The first inequalit has as its solution set ever ordered pair with a value that is greater than (above) the graph of the parabola. The second statement has as its solution set ever ordered pair with a value that is less than (below) the graph of the line. The solution set to the sstem is the set of ordered pairs that meet both of those criteria. Here is the graph of the solution set. NOTE The solution set is the shaded area above the parabola and below the line. CHECK YOURSELF 5 Solve the following sstem Eample 6 demonstrates that, even if the related sstem of equations has no solution, the sstem of inequalities could have a solution. Eample 6 Solving a Sstem of Nonlinear Inequalities Solve the following sstem As we did with the previous sstems, we will first look at the graph of the related sstem of equations (from Eample 3.)
8 802 CHAPTER 10 GRAPHS OF CONIC SECTIONS The solution set is now the set of all ordered pairs below the parabola ( 2 2 1) and above the line ( 2). Here is the graph of the solution set. NOTE The solution continues beond the borders of the grid. 2 CHECK YOURSELF 6 Solve the following sstem CHECK YOURSELF ANSWERS ( 1, 10), (6, 10) 2. (1.3, 8), ( 2.3, 8) 2 3. No real solution 4. ( 1, 3), ( 1, 3), (1, 3), (1, 3) 5. 6.,
9 Name 10.4 Eercises Section Date In eercises 1 to 8, the graph of a sstem of equations is given. Determine how man real solutions each sstem has ANSWERS
10 ANSWERS In eercises 9 to 12, draw the graph of a sstem that has the indicated number of solutions. Use the conic sections indicated solutions: (a) use a circle and an ellipse, and (b) use a parabola and a line. (a) (b) solution: (a) use a parabola and a circle, and (b) use a line and an ellipse. (a) (b) solutions: (a) use a parabola and a circle, and (b) use an ellipse and a parabola. (a) (b) 804
11 ANSWERS solutions: (a) use a circle and an ellipse, and (b) use a parabola and a circle. (a) (b) In eercises 13 to 24, graph each sstem and estimate the solutions
13 ANSWERS In eercises 25 to 32, solve using algebraic methods. (Note: These eercises have been solved graphicall in eercises 13 to 24.) (See eercise 13.) (See eercise 14.) (See eercise 15.) (See eercise 16.) (See eercise 19.) (See eercise 20.) (See eercise 23.) (See eercise 24.) In eercises 33 to 40, solve the sstems of inequalities graphicall. (Note: These have alread been graphed as sstems of equations in eercises 13 to 24.) (See eercise 13.) (See eercise 14.) 807
14 ANSWERS (See eercise 15.) (See eercise 16.) (See eercise 17.) (See eercise 18.) (See eercise 23.) (See eercise 24.) 808
15 ANSWERS In eercises 41 to 44, (a) graph each sstem and estimate the solution, and (b) use algebraic methods to solve each sstem Solve the following applications. 45. The manager of a large apartment comple has found that the profit, in dollars, is given b the equation P in which is the number of apartments rented. How man apartments must be rented to produce a profit of $3600? 46. The manager of a biccle shop has found that the revenue (in dollars) from the sale of biccles is given b the following equation. R How man biccles must be sold to produce a revenue of $12,500? 809
16 ANSWERS Find the equation of the line passing through the points of intersection of the graphs 2 and Write a sstem of inequalities to describe the following set of points: The points are in the interior of a circle whose center is the origin with a radius of 4, and above the line We are asked to solve the following sstem of equations Eplain how we can determine, before doing an work, that this sstem cannot have more than two solutions. 50. Without graphing, how can ou tell that the following sstem of inequalities has no solution? Solve the following sstems algebraicall Answers (a) (b) 810
17 11. (a) (b) 13. (3, 4) and ( 2, 4) 15. (4, 3) and (1, 3) 17. ( 3, 4) and ( 1, 4) 19. (0.6, 6) and ( 1.6, 6) 21. (6.2, 6) and (0.8, 6) 23. No solution 811
18 25. (3, 4) and ( 2, 4) 27. (4, 3) and (1, 3) , 6 or (0.618, 6) and ( 1.62, 6) 2 and, No solution (1, 1) and ( 2, 4) 43. ( 2, 1) and 38 25, ( 4, 1), ( 4, 1), (4, 1), (4, 1) 53. ( 2, 2), ( 2, 2), (2, 2), (2, 2) 812