1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 10–6) Then/Now Example 1: Linear-Quadratic System Example 2: Quadratic-Quadratic System Example 3: Quadratic Inequalities Example 4: Quadratics with Absolute Value

3. Then/Now You solved systems of linear equations. (Lessons 3–1 and 3–2) Solve systems of linear and nonlinear equations algebraically and graphically. Solve systems of linear and nonlinear inequalities graphically.

4. Example 1 Linear-Quadratic System Solve the system of equations. 4x 2 – 16y 2 = 25 (1) 2y + x = 2 (2) Step 1 Solve the linear equation for y. 2y + x = 2Equation (2) Simplify. Solve for y.

5. Example 1 Linear-Quadratic System Step 2 Substitute into the quadratic equation and solve for x. Quadratic equation Substitute – x + 1 for y. __ 1 2 Simplify. 4x 2 – 4x 2 + 16x – 16 = 25Distribute.

6. Example 1 Linear-Quadratic System 16x = 41Add 16 to each side. Divide each side by 16. Step 3 Substitute x into the linear equation and solve for y. Linear equation Substitute for x.

7. Example 1 Linear-Quadratic System Answer: The solution is. Simplify.

8. A.A B.B C.C D.D Example 1 What is the solution to the system of equations? x 2 – y 2 = 4 2y + x = 2 A. B. C. D.

9. Example 2 Quadratic-Quadratic System Solve the system of equations. x 2 + y 2 = 16(1) 4x 2 + y 2 = 23(2) x 2 + y 2 = 16Equation (1) (–) 4x 2 + y 2 = 23Equation (2) –3x 2 = –7Subtract. Divide each side –3. Take the square root of each side.

10. Example 2 Quadratic-Quadratic System Multiply by a form of 1. Simplify. Substitute into one of the original equations and solve for y. Equation (1) Substitute for x.

11. Example 2 Quadratic-Quadratic System Simplify. Subtract from each side. Take the square root of each side.

12. Example 2 Quadratic-Quadratic System Answer: The solutions are

13. A.A B.B C.C D.D Example 2 Solve the system of equations. x 2 + y 2 = 36(1) x 2 + 3y 2 = 42(2) A. B. C. D.

14. Example 3 Quadratic Inequalities Solve the system of inequalities by graphing. y > x 2 + 1 x 2 + y 2 ≤ 9 The graph of y > x 2 + 1 is the parabola y = x 2 + 1 and the region inside and above it. The region is shaded blue. The graph of x 2 + y 2 ≤ 9 is the interior of the circle x 2 + y 2 = 9. This region is shaded yellow.

15. Example 3 Quadratic Inequalities The intersection of these regions, shaded green, represents the solution of the system of inequalities. Answer: Check (0, 2) is in the shaded area. Use this point to check your solution. y > x 2 + 1 x 2 + y 2 ≤ 9 2 > 1 4 ≤ 9 2 > 0 2 + 1 0 2 + 2 2 ≤ 9 ? ?

16. A.A B.B C.C D.D Example 3 Solve the system of inequalities by graphing. y < –x 2 + 1 x 2 + y 2 ≤ 4 A.B. C.D.

17. Example 4 Quadratics with Absolute Value Solve the system of inequalities by graphing. x 2 + y 2 > 16 y > │x + 4│ Graph the boundary equations. Then shade appropriately. The intersection of the graphs, shaded green, represents the solution to the system.

18. Example 4 Quadratics with Absolute Value Answer:

19. Example 4 x 2 + y 2 >16y > │x + 4│ Quadratics with Absolute Value Check(–4, 4) is in the shaded area. Use the points to check your solution. 32>164 > 0 ?? (–4) 2 + 4 2 >164 > │–4 + 4│

20. A.A B.B C.C D.D Example 4 Solve the system of inequalities by graphing. x 2 + y 2 > 9 y < │x + 1│+ 1 A.B. C.D.

21. End of the Lesson