# Objective Solve systems of equations in 2 variables that contain at least 1 second-degree equation.

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Objective Solve systems of equations in 2 variables that contain at least 1 second-degree equation.

Notes - Identify the graph and solve systems
4x2 – 9y2 = 108 1. Solve x2 + y2 = 40 2x2 + y2 = 54 2. Solve x2 – 3y2 = 13 4x2 + 4y2 = 52 3. Solve 9x2 – 4y2 = 65 x + y = –1 4. Solve x2 + y2 = 25

Example 1: Solving a Nonlinear System by Substitution
x2 + y2 = 100 Solve by substitution. y = x2 – 26 1 2 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 x2 because both equations have x2 terms. x2 = 2y + 52

Substitute this value into the first equation.
Example 1 Continued Step 2 Use substitution. (2y + 52) + y2 = 100 Substitute this value into the first equation. y2 + 2y – 48 = 0 (y + 8) (y – 6) = 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)}.

Example 2: Solving a Nonlinear System by Addition
4x2 + 25y2 = 41 Solve by using addition. 36x2 + 25y2 = 169 The graph of both are ellipses, so there may be as many as four intersection points Step 1 Make opposites to eliminate y2. 36x2 + 25y2 = 169 –4x2 – 25y2 = –41 32x = 128 The solutions are (–2, –1),(–2, 1), (2, –1), (2, 1) x2 = 4, so x = ±2 Step 2 Find the values for y.

Notes - Identify the graph and solve systems
4x2 – 9y2 = 108 (±6, ±2) 1. Solve x2 + y2 = 40 2x2 + y2 = 54 2. Solve (±5, ±2) x2 – 3y2 = 13 4x2 + 4y2 = 52 3. Solve (±3, ±2) 9x2 – 4y2 = 65 x + y = –1 (3, –4), (–4, 3) 4. Solve x2 + y2 = 25

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). 7

Check It Out! Example 3 25x2 + 9y2 = 225 Solve by using the elimination method. 25x2 – 16y2 = 400 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.

Check It Out! Example 3 Continued
Step 1 Eliminate x2. 25x2 – 16y2 = 400 Subtract the first equation from the second. –25x2 – 9y2 = –225 –25y2 = 175 y2 = –7 Solve for y. There is no real solution of the system.

Example 1: Solving a Nonlinear System by Graphing
x2 + y2 = 25 Solve by graphing. 4x2 + 9y2 = 145 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).

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