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10.8 Systems of Nonlinear Equations Copyright © Cengage Learning. All rights reserved.

Objectives Substitution and Elimination Methods Graphical Method

Systems of Nonlinear Equations In this section we solve systems of equations in which the equations are not all linear.

Substitution and Elimination Methods

Substitution and Elimination Methods To solve a system of nonlinear equations, we can use the substitution or elimination method, as illustrated in the next examples.

Example 1 – Substitution Method Find all solutions of the system. x2 + y2 = 100 3x – y = 10 Solution: Solve for one variable. We start by solving for y in the second equation. y = 3x – 10 Equation 1 Equation 2 Solve for y in Equation 2

Example 1 – Solution cont’d Substitute. Next we substitute for y in the first equation and solve for x. x2 + (3x – 10)2 = 100 x2 + (9x2 – 60x + 100) = 100 10x2 – 60x = 0 10x(x – 6) = 0 x = 0 or x = 6 Substitute y = 3x – 10 into Equation 1 Expand Simplify Factor Solve for x

Example 1 – Solution cont’d Back-substitute. Now we back-substitute these values of x into the equation y = 3x – 10. For x = 0: y = 3(0) – 10 = –10 For x = 6: y = 3(6) – 10 = 8 So we have two solutions: (0, –10) and (6, 8). Back-substitute Back-substitute

Example 1 – Solution cont’d The graph of the first equation is a circle, and the graph of the second equation is a line. Figure 1 shows that the graphs intersect at the two points (0, –10) and (6, 8). Figure 1

Example 1 – Solution Check Your Answers: x = 0, y = –10: cont’d Check Your Answers: x = 0, y = –10: (0)2 + (–10)2 = 100 3(0) – (–10) = 10 x = 6, y = 8: (6)2 + (8)2 = 36 + 64 = 100 3(6) – (8) = 18 – 8 = 10

Example 2 – Elimination Method Find all solutions of the system. 3x2 + 2y = 26 5x2 + 7y = 3 Solution: We choose to eliminate the x-term, so we multiply the first equation by 5 and the second equation by –3. Then we add the two equations and solve for y. 15x2 + 10y = 130 –15x2 – 21y = –9 –11y = 121 y = –11 Equation 1 Equation 2 5  Equation 1 (–3)  Equation 2 Add Solve for y

Example 2 – Solution cont’d Now we back-substitute y = –11 into one of the original equations, say 3x2 + 2y = 26, and solve for x. 3x2 + 2(–11) = 26 3x2 = 48 x2 = 16 x = –4 or x = 4 So we have two solutions: (–4, –11) and (4, –11). Back-substitute y = –11 into Equation 1 Add 22 Divide by 3 Solve for x

Example 2 – Solution cont’d The graphs of both equations are parabolas. Figure 2 shows that the graphs intersect at the two points (–4, –11) and (4, –11). Figure 2

Example 2 – Solution Check Your Answers: x = –4, y = –11: cont’d Check Your Answers: x = –4, y = –11: 3(–4)2 + 2(–11) = 26 5(–4)2 + 7(–11) = 3 x = 4, y = –11: 3(4)2 + 2(–11) = 26 5(4)2 + 7(–11) = 3

Graphical Method

Graphical Method The graphical method is particularly useful in solving systems of nonlinear equations.

Example 3 – Graphical Method Find all solutions of the system x2 – y = 2 2x – y = –1 Solution: Graph each equation. To graph, we solve for y in each equation. y = x2 – 2 y = 2x + 1

Example 3 – Solution cont’d Find intersection points. Figure 3 shows that the graphs of these equations intersect at two points. Zooming in, we see that the solutions are (–1, –1) and (3, 7) Figure 3

Example 3 – Solution Check Your Answers: x = –1, y = –1: cont’d Check Your Answers: x = –1, y = –1: (–1)2 – (–1) = 2 2(–1) – (–1) = –1 x = 3, y = 7: 32 – 7 = 2 2(3) – 7 = –1

Example 4 – Solving a System of Equations Graphically Find all solutions of the system, rounded to one decimal place. x2 + y2 = 12 y = 2x2 – 5x Solution: The graph of the first equation is a circle, and the graph of the second is a parabola. To graph the circle on a graphing calculator, we must first solve for y in terms of x. Equation 1 Equation 2

Example 4 – Solution x2 + y2 = 12 y2 = 12 – x2 cont’d x2 + y2 = 12 y2 = 12 – x2 To graph the circle, we must graph both functions. and Isolate y2 on LHS Take square roots

Example 4 – Solution cont’d In Figure 4 the graph of the circle is shown in red, and the parabola is shown in blue. The graphs intersect in Quadrants I and II. Zooming in, or using the Intersect command, we see that the intersection points are (–0.559, 3.419) and (2.847, 1.974). (a) (b) x2 + y2 = 12, y = 2x2 – 5x Figure 4

Example 4 – Solution cont’d There also appears to be an intersection point in Quadrant IV. However, when we zoom in, we see that the curves come close to each other but don’t intersect (see Figure 5). Thus the system has two solutions; rounded to the nearest tenth, they are (–0.6, 3.4) and (2.8, 2.0) Zooming in Figure 5