10 7 Solving Nonlinear Systems Objective Solve systems

10 -7 Solving Nonlinear Systems Objective Solve systems of equations in 2 variables that contain at least 1 second-degree equation. Holt Algebra 2

10 -7 Solving Nonlinear Systems Notes - Identify the graph and solve systems 1. Solve 2. Solve 3. Solve 4 x 2 – 9 y 2 = 108 x 2 + y 2 = 40 2 x 2 + y 2 = 54 x 2 – 3 y 2 = 13 4 x 2 + 4 y 2 = 52 9 x 2 – 4 y 2 = 65 x + y = – 1 x 2 + y 2 = 25 Holt Algebra 2

10 -7 Solving Nonlinear Systems Example 1: Solving a Nonlinear System by Substitution x 2 + y 2 = 100 Solve y= 1 2 2 x – 26 by substitution. The graph of the first equation is a circle, and the graph of the second equation is a parabola, so there may be as many as 4 intersection points. Step 1 It is simplest to solve for x 2 because both equations have x 2 terms. 2 x = 2 y + 52 Holt Algebra 2

10 -7 Solving Nonlinear Systems Example 1 Continued Step 2 Use substitution. (2 y + 52) + y 2 = 100 y 2 + 2 y – 48 = 0 (y + 8) (y – 6) = 0 Substitute this value into the first equation. y = – 8 or y = 6 Step 3 Complete ordered pairs. The solution set of the system is {(6, – 8) (– 6, – 8), (8, 6), (– 8, 6)}. Holt Algebra 2

10 -7 Solving Nonlinear Systems Example 2: Solving a Nonlinear System by Addition Solve 4 x 2 + 25 y 2 = 41 36 x 2 + 25 y 2 = 169 by using addition. The graph of both are ellipses, so there may be as many as four intersection points Step 1 Make opposites to eliminate y 2. 36 x 2 + 25 y 2 = 169 – 4 x 2 – 25 y 2 = – 41 32 x 2 = 128 x 2 = 4, so x = ± 2 Step 2 Find the values for y. Holt Algebra 2 The solutions are (– 2, – 1), (– 2, 1), (2, – 1), (2, 1)

10 -7 Solving Nonlinear Systems Notes - Identify the graph and solve systems 4 x 2 – 9 y 2 = 108 1. Solve 2. Solve 3. Solve 4. Solve 2 2 x + y = 40 2 x 2 + y 2 = 54 x 2 – 3 y 2 = 13 4 x 2 + 4 y 2 = 52 2 2 9 x – 4 y = 65 x + y = – 1 x 2 + y 2 = 25 Holt Algebra 2 (± 6, ± 2) (± 5, ± 2) (± 3, ± 2) (3, – 4), (– 4, 3)

10 -7 Solving Nonlinear Systems Solving Systems: Extra Info The following power-point slides contain extra examples and information. Reminder: Lesson Objectives Solve systems of nonlinear systems of equations (by graphing, substitution, and the addition methods). Holt Algebra 2

10 -7 Solving Nonlinear Systems Check It Out! Example 3 Solve 25 x 2 + 9 y 2 = 225 25 x 2 – 16 y 2 = 400 by using the elimination method. The graph of the first equation is an ellipse, and the graph of the second equation is a hyperbola, There may be as many as four points of intersection. Holt Algebra 2

10 -7 Solving Nonlinear Systems Check It Out! Example 3 Continued Step 1 Eliminate x 2. 25 x 2 – 16 y 2 = 400 – 25 x 2 – 9 y 2 = – 225 – 25 y 2 = 175 y 2 = – 7 Subtract the first equation from the second. Solve for y. There is no real solution of the system. Holt Algebra 2

10 -7 Solving Nonlinear Systems Example 1: Solving a Nonlinear System by Graphing Solve x 2 + y 2 = 25 4 x 2 + 9 y 2 = 145 by graphing. The graph of the first equation is a circle, and the graph of the second equation is an ellipse, so there may be as many as four points of intersection. The points of intersection are ( – 4, – 3), (– 4, 3), (4, – 3), (4, 3). Holt Algebra 2