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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

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.

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.

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 x – 16 = 25Distribute.

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.

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

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.

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.

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.

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

Example 2 Quadratic-Quadratic System Answer: The solutions are

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.

Example 3 Quadratic Inequalities Solve the system of inequalities by graphing. y > x x 2 + y 2 ≤ 9 The graph of y > x is the parabola y = x 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.

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 x 2 + y 2 ≤ 9 2 > 1 4 ≤ 9 2 > ≤ 9 ? ?

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

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.

Example 4 Quadratics with Absolute Value Answer:

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) >164 > │–4 + 4│

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.

End of the Lesson