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

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

2. Notes 4x2 – 9y2 = 108 1. Solve by graphing. x2 + y2 = 40 2x2 + y2 = 54 2. Solve by substitution. x2 – 3y2 = 13 4x2 + 4y2 = 52 3. Solve by elimination. 9x2 – 4y2 = 65

3. 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).

4. y = x2–26 1 2 Example 2: Solving a Nonlinear System by Substitution x2 + y2 = 100 Solve 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 four points of intersection.

5. Example 2 Continued Step 1 It is simplest to solve for x2 because both equations have x2 terms. Solve for x2 in the second equation. x2 = 2y + 52 Step 2 Use substitution. Substitute this value into the first equation. (2y + 52) + y2 = 100 y2 + 2y – 48= 0 Simplify, and set equal to 0. (y + 8) (y – 6)= 0 Factor. y = –8 or y = 6

6. Example 2 Continued Step 3 Substitute –8 and 6 into x2 = 2y + 52 to find valuesof x. The solution set of the system is {(6, –8) (–6, –8), (8, 6), (–8, 6)}.

7. Example 3: Solving a Nonlinear System by Elimination 4x2 + 25y2 = 41 Solve by using the elimination method. 36x2 + 25y2 = 169 The graph of the first equation is an ellipse, and the graph of the second equation is an ellipse, There may be as many as four points of intersection.

8. Example 3 Continued Step 1 Eliminate y2. 36x2+ 25y2 = 169 Add the two equations. –4x2– 25y2 = –41 32x2 = 128 Solve for x. x2 = 4, so x = ±2 Step 2 Find the values for y. The solution set of the system is {(–2, –1), (–2, 1), (2, –1), (2, 1)}.

9. Notes 4x2 – 9y2 = 108 1. Solve by graphing. (±6, ±2) x2 + y2 = 40 2x2 + y2 = 54 2. Solve by substitution. (±5, ±2) x2 – 3y2 = 13 4x2 + 4y2 = 52 (±3, ±2) 3. Solve by elimination. 9x2 – 4y2 = 65

10. 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, substitutions, and the addition methods).

11. Check It Out! Example 2a Solve the system of equations by using the substitution method. x + y = –1 x2 + y2 = 25 The graph of the first equation is a line, and the graph of the second equation is a circle, so there may be as many as two points of intersection.

12. Check It Out! Example 2a Continued Step 1 Solve for x. x = –y – 1 Solve for x in the first equation. Step 2 Use substitution. Substitute this value into the second equation. (–y –1)2+ y2 = 25 y2 + 2y + 1 + y2 – 25= 0 Simplify and set equal to 0. 2y2 +2y – 24= 0 2[(y2 + y – 12)]= 0 Factor. 2(y + 4)(y – 3)= 0 y = –4 or y = 3

13. Check It Out! Example 2a Continued Step 3 Substitute –4 and 3 into x+ y = –1 to find valuesof x. x+ (–4) = –1 x = 3 (3, –4) is a solution. x+ (3) = –1 x = –4 (–4, 3) is a solution. The solution set of the system is {(3, –4), (–4, 3)}.

14. Check It Out! Example 2a Continued Check Use a graphing calculator.The graph supports that there are two points of intersection.

15. 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.

16. 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 Solve for y. y2 = –7 There is no real solution of the system.