1. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall

2. Chapter 11 Graphing Quadratic Functions, Rational Functions, and Conic Sections

3. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 11.7 Solving Nonlinear Systems of Equations

4. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall A nonlinear system of equations is a system of equations at least one of which is not linear. The substitution method or the elimination method may be used to solve the system.

5. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example Solve the system Solution y + 2 = x Second equation x 2 + y = 4 First equation (y + 2) 2 + y = 4 Replace x with y + 2. y 2 + 4y + 4 + y = 4 y 2 + 5y = 0 y(y + 5) = 0 continued

6. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall y(y + 5) = 0 y = 0 or y = –5 Let y = 0 and then let y = – 5 to find the corresponding x values. 0 + 2 = x x = 2 –5 + 2 = x x = –3 Let y = 0Let y = –5 y + 2 = x continued

7. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The solutions are (2, 0) and (–3, –5). Check both solutions in both equations. y + 2 = xx 2 + y = 4 2 2 + 0 = 4 4 = 4 0 + 2 = 2 2 = 2 y + 2 = xx 2 + y = 4 (–3) 2 + (–5) = 4 9 – 5 = 4 –5 + 2 = –3 –3 = –3 A graph of the system also verifies the solution. y + 2 = x x 2 + y = 4 (2, 0) (–3, –5). x y

8. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example Solve the system. Solution (–1) x 2 + (–1) 2y 2 = (–1) 4 Multiply the first equation by – 1. The elimination method will work best to solve this system. x 2 – y 2 = 4 – 3y 2 = 0 Add the two equations. – x 2 + – 2y 2 = – 4 continued

9. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall y = 0 Let y = 0 to find the corresponding x values. x 2 + 2y 2 = 4 – 3y 2 = 0 x 2 + 2(0) 2 = 4 x 2 = 4 x = 2 or x = –2 The solutions are (–2, 0) and (2, 0).