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Chapter 7 Systems of Equations and Inequalities © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved 1 SECTION 7.4 Nonlinear Systems of Equations and Inequalities OBJECTIVES 1 2 Solve nonlinear systems of equations. Solve a nonlinear system of inequalities. You will not need to know the details of inequalities. © 2010 Pearson Education, Inc. All rights reserved 2 EXAMPLE 1 Using Substitution to Solve a Nonlinear System Solve the system of equations by the substitution method. 4 x y 3 (1) 2 x y 1 (2) Solution Step 1 Solve for one variable. Express y in terms of x in equation (2). 2 y x 1 Step 2 Substitute. Substitute x2 +1 for y in equation (1). © 2010 Pearson Education, Inc. All rights reserved 3 EXAMPLE 1 Using Substitution to Solve a Nonlinear System Solution continued 4 x y 3 4 x x 1 3 2 4 x x 1 3 2 x 4x 4 0 2 Step 3 Solve the equation resulting from step (2). x 2 x 2 0 x20 x 2 © 2010 Pearson Education, Inc. All rights reserved 4 EXAMPLE 1 Using Substitution to Solve a Nonlinear System Solution continued Step 4 Back substitution. Substitute x = –2 in equation (3) to obtain the corresponding y-value. y x2 1 y 2 1 2 y5 Since x = –2 and y = 5, the apparent solution set of the system is {(–2, 5)}. © 2010 Pearson Education, Inc. All rights reserved 5 EXAMPLE 1 Using Substitution to Solve a Nonlinear System Solution continued Step 5 Check. Replace x with –2 and y with 5 in both equations (1) and (2). 4 x y 3 x2 y 1 4 2 5 3 2 5 ? 1 ? ? 8 5 3 3 3 2 4 5 1 11 ? Confirm the solution with a graph. © 2010 Pearson Education, Inc. All rights reserved 6 EXAMPLE 1 Using Substitution to Solve a Nonlinear System Solution continued The graphs of the line 4x + y = –3 and the parabola y = x 2 + 1 confirm that the solution set is {(–2, 5)}. © 2010 Pearson Education, Inc. All rights reserved 7 © 2010 Pearson Education, Inc. All rights reserved 8 EXAMPLE 2 Using Elimination to Solve a Nonlinear System Solve the system of equations by the elimination method. x 2 y 2 25 (1) 2 x y 5 (2) Solution Step 1 Adjust the coefficients. Multiply equation (2) by –1 to eliminate x. 2 2 x y 25 (1) –x y –5 (3) 2 y y 20 2 Step 2 © 2010 Pearson Education, Inc. All rights reserved 9 EXAMPLE 2 Using Elimination to Solve a Nonlinear System Solution continued Step 3 Solve the equation obtained in Step 2. 2 y y 20 y 2 y 20 0 y 5 y 4 0 y 5 0 or y 4 0 y 5 or y 4 Step 4 Back-substitute the values in one of the original equations to solve for the other variable. © 2010 Pearson Education, Inc. All rights reserved 10 EXAMPLE 2 Using Elimination to Solve a Nonlinear System Solution continued (i) Substitute y = –5 in equation (2) & solve for x. x2 y 5 x 2 5 5 x2 5 5 x2 0 x0 Thus (0, –5) is a solution of the system. © 2010 Pearson Education, Inc. All rights reserved 11 EXAMPLE 2 Using Elimination to Solve a Nonlinear System Solution continued (ii) Substitute y = 4 in equation (2) & solve for x. 2 x y5 x 4 5 2 x 45 2 x 9 x 3 2 Thus (3, 4) and (–3, 4) are the solutions of the system. © 2010 Pearson Education, Inc. All rights reserved 12 Using Elimination to Solve a Nonlinear System EXAMPLE 2 Solution continued Step 5 Check (0, –5), (3, 4), and (–3, 4) in the equations x2 + y2 = 25 and x2 – y = 5. 0 5 25 3 4 25 ? 25 25 9 16 25 25 25 2 2 ? 0 5 5 2 ? 55 2 2 ? 3 45 ? 945 55 2 ? © 2010 Pearson Education, Inc. All rights reserved 3 4 25 2 2 ? ? 9 16 25 25 25 3 2 ? 45 ? 945 55 13 EXAMPLE 2 Using Elimination to Solve a Nonlinear System Solution continued The graphs of the circle x2 + y2 = 25 and the parabola y = x2 – 5 confirm that the solution set is {(0, –5), (3, 4), (–3, 4)}. © 2010 Pearson Education, Inc. All rights reserved 14 © 2010 Pearson Education, Inc. All rights reserved 15 Since the nature of such a problem requires a rather lengthy list of formulas, it is unlikely that this would be a FE problem. Business majors might wish to look at Example 3 and this related problem. It seems specialized. © 2010 Pearson Education, Inc. All rights reserved 16