10.6 Systems of Nonlinear Equations

Transcription

1 SECTION 0.6 Systems of Nonlinear Equations Systems of Nonlinear Equations PREPARING FOR THIS SECTION Before getting started, review the following: Lines (Section., pp. 7 38) Ellipses (Section 9.3, pp ) Circles (Section.5, pp. 9) Hyperbolas (Section 9., pp ) Parabolas (Section 9., pp ) Now work the Are You Prepared? problems on page 800. OBJECTIVES Solve a System of Nonlinear Equations Using Substitution Solve a System of Nonlinear Equations Using Elimination Solve a System of Nonlinear Equations Using Substitution In Section 0. we observed that the solution to a system of linear equations could be found geometrically by determining the point(s) of intersection (if any) of the equations in the system. Similarly, when solving systems of nonlinear equations, the solution(s) also represents the point(s) of intersection (if any) of the graphs of the equations. There is no general methodology for solving a system of nonlinear equations. At times substitution is best; other times, elimination is best; and sometimes neither of these methods works. Experience and a certain degree of imagination are your allies here. Before we begin, two comments are in order.. If the system contains two variables and if the equations in the system are easy to graph, then graph them. By graphing each equation in the system, you can get an idea of how many solutions a system has and approximately where they are located.. Extraneous solutions can creep in when solving nonlinear systems, so it is imperative that all apparent solutions be checked.

2 79 CHAPTER 0 Analytic Geometry EXAMPLE Solve the following system of equations: 3x - y = - () A line b x - y = 0 () A parabola Algebraic Solution Using Substitution First, we notice that the system contains two variables and that we know how to graph each equation by hand. See Figure 8. The system apparently has two solutions. Figure 8 x y = 0 (y = x ) y 0 3x y = (y = 3x + ) (, 8) We use a graphing utility to graph Y and Y = x = 3x +. From Figure 9 we see that the system apparently has two solutions. Using INTERSECT, the solutions to the system of equations are -0.5, 0.5 and, 8. (, ) 6 6 We will use substitution to solve the system. Equation () is easily solved for y. 3x - y = - Equation () y = 3x + We substitute this expression for y in equation (). The result is an equation containing just the variable x, which we can then solve. x - y = 0 x - 3x + = 0 Equation () Substitute 3x + for y x - 3x - = 0 x + x - = 0 Remove parentheses. Factor. x + = 0 or x - = 0 Apply the Zero-Product Property. x = - or x = Using these values for x in y = 3x +, we find y = 3a - b + = or y = 3 + = 8 The apparent solutions are x = - and x =, y = 8., y = CHECK: For x = -, y = : 3a - b - = = - d a - b - = a b - = 0 For x =, y = 8: 3-8 = 6-8 = - b - 8 = - 8 = 0 Each solution checks. Now we know that the graphs in Figure 8 intersect at a - and at, 8., b x () () () () Figure 9 Y x Y 0 3x 6 6 NOW WORK PROBLEM 5 USING SUBSTITUTION.

3 SECTION 0.6 Systems of Nonlinear Equations 795 Solve a System of Nonlinear Equations Using Elimination Our next example illustrates how the method of elimination works for nonlinear systems. EXAMPLE Solve: b x + y = 3 x - y = 7 () A circle () A parabola Algebraic Solution Using Elimination First, we graph each equation, as shown in Figure 0. Based on the graph, we expect four solutions. By subtracting equation () from equation (), the variable x can be eliminated. b x + y = 3 x - y = 7 y + y = 6 This quadratic equation in y can be solved by factoring. y + y - 6 = 0 y + 3y - = 0 y = -3 or y = Subtract We use these values for y in equation () to find x. If y =, then x = y + 7 = 9, so x = 3 or -3. If y = -3, then x = y + 7 =, so x = or -. We have four solutions: x = 3, y = ; x = -3, y = ; x =, y = -3; and x = -, y = -3. You should verify that, in fact, these four solutions also satisfy equation (), so all four are solutions of the system. The four points, 3,, -3,,, -3, and -, -3, are the points of intersection of the graphs. Look again at Figure 0. We use a graphing utility to graph x + y = 3 and x - y = 7. (Remember that to graph x + y = 3 requires two functions, Y and Y =-33 - x = 33 - x, and a square screen.) From Figure we see that the system apparently has four solutions. Using INTERSECT, the solutions to the system of equations are -3,, 3,, -, -3, and, -3. Figure Y 3 x 5 7 Y 3 x Y 3 x Figure 0 y x y = 7 (y = x 7) ( 3, ) (3, ) x + y = x (, 3) (, 3) 8 NOW WORK PROBLEM 3 USING ELIMINATION.

4 796 CHAPTER 0 Analytic Geometry EXAMPLE 3 Solve: x + x + y - 3y + = 0 c x + + y - y = 0 x () () Algebraic Solution Using Elimination Since it is not straightforward how to graph the equations in the system, we proceed directly to use the method of elimination. First, we multiply equation () by x to eliminate the fraction. The result is an equivalent system because x cannot be 0. [Look at equation () to see why.] b x + x + y - 3y + = 0 x + x + y - y = 0 Now subtract equation () from equation () to eliminate x. The result is -y + = 0 y = Solve for y. To find x, we back-substitute y = in equation (): x + x + y - 3y + = 0 x + x = 0 x = 0 or x = - x + x = 0 xx + = 0 () () Equation () Substitute for y in (). Simplify. Factor. Apply the Zero-Product Property. Because x cannot be 0, the value x = 0 is extraneous, and we discard it. The solution is x = -, y =. CHECK: We now check x = -, y = : = = 0 c = = 0 () () First, we multiply equation () by x to eliminate the fraction. The result is an equivalent system because x cannot be 0 [look at equation () to see why]: b x + x + y - 3y + = 0 x + x + y - y = 0 We need to solve each equation for y. First, we solve equation () for y: x + x + y - 3y + = 0 y - 3y = -x - x - y - 3y + 9 = -x - x ay - 3 b = -x - x + y - 3 = ; A -x - x + y = 3 ; A -x - x + Now we solve equation () for y: x + x + y - y = 0 y - y = -x - x y - y + = -x - x + ay - b = -x - x + y - = ; A -x - x + () () Equation () Rearrange so that terms involving y are on left side. Complete the square involving y. Factor; simplify Square Root Method Solve for y. Equation () Rearrange so that terms involving y are on left side. Complete the square involving y. Factor Square Root Method y = ; A -x - x + Solve for y. Now graph each equation using a graphing utility. See Figure. Using INTERSECT, the points of intersection are -, and 0,. Since x Z 0 [look back at the original equation ()], the graph of Y 3 has a hole at the point 0, and Y has a hole at 0, 0. The value x = 0 is extraneous, and we discard it. The only solution is x = - and y =.

5 SECTION 0.6 Systems of Nonlinear Equations 797 Figure Y 3 x x Y 3 x x Y 3 x x Y x x NOW WORK PROBLEMS 9 AND 9. EXAMPLE Solve: b x - y = y = x () A hyperbola () A parabola Algebraic Solution Either substitution or elimination can be used here. We use substitution and replace x by y in equation (). The result is y - y = y - y + = 0 This is a quadratic equation whose discriminant is - - # # = - # = The equation has no real solutions, so the system is inconsistent. The graphs of these two equations do not intersect. See Figure 3. We graph Y and x - y = x = in Figure. You will need to graph x - y = as two functions: Y = 3x - and Y 3 = -3x - From Figure we see that the graphs of these two equations do not intersect. The system is inconsistent. Figure Y x Figure 3 Y x (, ) y 5 y = x (, ) 6 6 ( 3, 5) (3, 5 ) 5 5 ( 3, 5 ) x (3, 5 ) Y 3 x 5 x y =

6 798 CHAPTER 0 Analytic Geometry EXAMPLE 5 Algebraic Solution Solve: b 3xy - y = - 9x + y = 0 We multiply equation () by and add the result to equation () to eliminate the y terms. () b 6xy - y = - 9x + y = 0 () 9x + 6xy = 6 Add. 3x + xy = Divide each side by 3. Since x Z 0 (do you see why?), we can solve for y in this equation to get y = - 3x, x Z 0 (3) x Now substitute for y in equation () of the system. 9x + y = 0 Equation () 9x + - 3x = 0 Substitute y = - 3x in (). x x 9x + - x + 9x x = 0 9x + - x + 9x = 0x Multiply both sides by x. 8x - x + = 0 Subtract 0x from both sides. 9x - x + = 0 Divide both sides by. This quadratic equation (in x ) can be factored: 9x - x - = 0 9x - = 0 or x - = 0 x = 9 x = () () To graph 3xy - y = -, we need to solve for y.in this instance, it is easier to view the equation as a quadratic equation in the variable y. 3xy - y = - y - 3xy - = 0 Place in standard form. y = --3x ; -3x - - Use the quadratic formula with a =, b = -3x, c = -. y = 3x ; 39x + 6 Simplify. Using a graphing utility, we graph Y = 3x + 39x + 6 and From equation (), we graph Y 3 = Y = x. Figure 5 Y 3 See Figure x Y = 3x - 39x x Y 3x 9x 6 and x = ; A 9 = ; x = ; 3 To find y, we use equation (3): 3 3 If If If x = 3 : y = - 3x x x = - 3 : y = - 3x x x = : y = - 3x x = - 3 = - If x = -: y = - 3x = - 3- = x - The system has four solutions. Check them for yourself. = = - 3 a 3 b a- - 3 = 3 b = = - = - Y 0 9x Y 3x 9x 6 Using INTERSECT, the solutions to the system of equations are -, 0.5, 0.7,.,, -0.5, and -0.7, -., each rounded to two decimal places. NOW WORK PROBLEM 7.

7 SECTION 0.6 Systems of Nonlinear Equations 799 EXAMPLE 6 Running a Long Distance Race In a 50-mile race, the winner crosses the finish line mile ahead of the second-place runner and miles ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many miles does the second-place runner beat the third-place runner? 3 miles mile Solution Let v, v, and v 3 denote the speeds of the first-, second-, and third-place runners, respectively. Let t and t denote the times (in hours) required for the first-place runner and second-place runner to finish the race. Then we have the system of equations 50 = v t 9 = v d t 6 = v 3 t 50 = v t () First-place runner goes 50 miles in t. () Second-place runner goes 9 miles in t. (3) Third-place runner goes 6 miles in t. () Second-place runner goes 50 miles in t. We seek the distance d of the third-place runner from the finish at time t. At time t, the third-place runner has gone a distance of v 3 t miles, so the distance d remaining is 50 - v 3 t. Now d = 50 - v 3 t = 50 - v 3 t # t t = 50 - v 3 t # t t = 50-6 # = 50-6 # v v = 50-6 # 50 9 L 3.06 miles 50 v 50 v From (3), v 3 t = 6 From (), t = 50 e v From (), t = 50 v Form the quotient of () and ().

8 800 CHAPTER 0 Analytic Geometry HISTORICAL FEATURE In the beginning of this section, we said that imagination and experience are important in solving systems of nonlinear equations. Indeed, these kinds of problems lead into some of the deepest and most difficult parts of modern mathematics. Look again at the graphs in Examples and of this section (Figures 8 and 0). We see that Example has two solutions, and Example has four solutions. We might conjecture that the number of solutions is equal to the product of the degrees of the equations involved. Historical Problem A papyrus dating back to 950 BC contains the following problem: A given surface area of 00 units of area shall be represented as the sum of two squares whose sides are to each other as : 3. This conjecture was indeed made by Etienne Bezout ( ), but working out the details took about 50 years. It turns out that, to arrive at the correct number of intersections, we must count not only the complex number intersections, but also those intersections that, in a certain sense, lie at infinity. For example, a parabola and a line lying on the axis of the parabola intersect at the vertex and at infinity. This topic is part of the study of algebraic geometry. Solve for the sides by solving the system of equations x + y = 00 c x = 3 y