1. Slide 10- 1 Copyright © 2012 Pearson Education, Inc.

2. 10.4 Nonlinear Systems of Equations ■ Systems Involving One Nonlinear Equation ■ Systems of Two Nonlinear Equations ■ Problem Solving

3. Slide 10- 3 Copyright © 2012 Pearson Education, Inc. Systems Involving One Nonlinear Equation Suppose that a system consists of an equation of a circle and an equation of a line. In what ways can the circle and the line intersect? The figures on the next slide represent three ways in which the situation can occur.

4. Slide 10- 4 Copyright © 2012 Pearson Education, Inc. 0 real solutions x y 1 real solution x y 2 real solutions y

5. Slide 10- 5 Copyright © 2012 Pearson Education, Inc. Recall that graphing, elimination, and substitution were all used to solve systems of linear equations. To solve systems in which one equation is of first degree and one is of second degree, it is preferable to use the substitution method.

6. Slide 10- 6 Copyright © 2012 Pearson Education, Inc. Example Solution Solve the system x 2 + y 2 = 20, (1) (The graph is a circle.) y – 2x = 0. (2) (The graph is a line.) First solve the linear equation, (2), for y: y = 2x. (3) We could have solved for x instead. Then substitute 2x for y in equation (1) and solve for x: x 2 + (2x) 2 = 20 x 2 + 4x 2 = 20 5x 2 = 20

7. Slide 10- 7 Copyright © 2012 Pearson Education, Inc. Solution continued x 2 = 4 Now substitute these numbers for x in equation (3) and solve for y: For x = 2, y = 2(2) = 4; For x = –2, y = 2(–2) = –4. It is left to the student to confirm that the pairs (2, 4) and (–2, –4) check in both equations. The pairs (2, 4) and (–2, –4) are solutions to the system. Using the principle of square roots Check:

8. Slide 10- 8 Copyright © 2012 Pearson Education, Inc. Systems of Two Nonlinear Equations We now consider systems of two second- degree equations. Graphs of such systems can involve any two conic sections. On the next slide are some ways in which a circle and a hyperbola can intersect.

9. Slide 10- 9 Copyright © 2012 Pearson Education, Inc.

10. Slide 10- 10 Copyright © 2012 Pearson Education, Inc. To solve systems of two second-degree equations, we either substitute or eliminate. The elimination method is generally better when both equations are of the form Ax 2 + By 2 = C. Then we can eliminate an x 2 - or y 2 -term in a manner similar to the procedure used in Chapter 3.

11. Slide 10- 11 Copyright © 2012 Pearson Education, Inc. Example Solution Solve the system 3x 2 + 2y 2 = 66, (1) x 2 – y 2 = 7. (2) Multiply equation (2) by 2 and add: 5x 2 = 80 x 2 = 16 3x 2 + 2y 2 = 66 2x 2 – 2y 2 = 14 x =

12. Slide 10- 12 Copyright © 2012 Pearson Education, Inc. There is no x-term, and x 2 = 16 whether x = 4 or –4, thus we can substitute 4 and –4 for x simultaneously in equation (2): ( ) 2 – y 2 = 7 16 – y 2 = 7 –y 2 = – 9 y 2 = 9 y =. Thus if x = 4, then y = 3 or y = –3; and if x = –4, then y = 3 or y = –3. The possible solutions are (4, 3), (4, –3), (–4, 3), and (–4, –3). Solution continued

13. Slide 10- 13 Copyright © 2012 Pearson Education, Inc. Check: The solutions are (4, 3), (4, –3), (–4, 3), and (–4, –3). 3x 2 + 2y 2 = 66 x 2 – y 2 = 7 ( ) 2 – ( ) 2 7 16 – 9 3( ) 2 + 2( ) 2 66 3(16) + 2(9) 7 = 748 + 18 66 = 66 True Solution continued

14. Slide 10- 14 Copyright © 2012 Pearson Education, Inc. Example Solution Solve the system x 2 + 4y 2 = 20, (1) (ellipse) xy = 4. (2) (hyperbola) Solve equation (2) for y. Substitute into equation (1) and solve for x.

15. Slide 10- 15 Copyright © 2012 Pearson Education, Inc. Solution continued x 2 + 4y 2 = 20, xy = 4 Since y = 4/x, for x = 2, we have y = 4/2, or 2. Thus (2, 2) is a solution. similarly, (–2, 2), (4, 1), and (–4, –1) are solutions. You can show that all four pairs check.

16. Slide 10- 16 Copyright © 2012 Pearson Education, Inc. Problem Solving We now consider applications that can be modeled by a system of equations in which at least one equation is nonlinear.

17. Slide 10- 17 Copyright © 2012 Pearson Education, Inc. Example Solution A picture frame with a picture takes up an area of 216 in 2. The perimeter of the frame is 60 in. Find the dimensions of the frame. 1. Familiarize. We draw and label a sketch, letting l = the length and w = width, both in inches. Area = lw = 216 l w Perimeter = 2l + 2w = 60

18. Slide 10- 18 Copyright © 2012 Pearson Education, Inc. 2. Translate. We have the following translation: Perimeter: 2l + 2w = 60; Area: lw = 216. 3. Carry out. We solve the system 2l + 2w = 60, lw = 216 to get (18, 12) and (12, 18). Since length is usually considered to be longer than width, we have the solution l = 18 and w = 12. Completing the solution is left to the student.

19. Slide 10- 19 Copyright © 2012 Pearson Education, Inc. 4. Check. If l = 18 and w = 12, the perimeter is 2(12) + 2(18), or 60. The area is (18)(12), or 196. The numbers check. 5. State. The length is 18 in. and the width is 12 in.