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

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Presentation on theme: "Holt Algebra 2 10-7 Solving Nonlinear Systems Solve systems of equations in 2 variables that contain at least 1 second-degree equation. Objective."— Presentation transcript:

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

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

3 Holt Algebra Solving Nonlinear Systems Example 1: Solving a Nonlinear System 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. Solve by substitution. x 2 + y 2 = 100 y = x 2 – Step 1 It is simplest to solve for x 2 because both equations have x 2 terms. x 2 = 2y + 52

4 Holt Algebra Solving Nonlinear Systems Step 2 Use substitution. Example 1 Continued Substitute this value into the first equation. (2y + 52) + y 2 = 100 (y + 8) (y – 6) = 0 y 2 + 2y – 48 = 0 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)}.

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

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

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

8 Holt Algebra Solving Nonlinear Systems 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. Solve by using the elimination method. 25x 2 + 9y 2 = x 2 – 16y 2 = 400 Check It Out! Example 3

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

10 Holt Algebra Solving Nonlinear Systems Example 1: Solving a Nonlinear System 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. Solve by graphing. x 2 + y 2 = 25 4x 2 + 9y 2 = 145 The points of intersection are (–4, –3), (–4, 3), (4, –3), (4, 3).


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