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10-7

Solving Nonlinear Systems. 10-7. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. Warm Up Solve by substitution. Solve by elimination. 3 x + 4 y = 15. 1. x = 6 y – 6. 3 x + 4 y = 57. 2. 5 x – 4 y = – 1. Objective.

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10-7

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  1. Solving Nonlinear Systems 10-7 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

  2. Warm Up Solve by substitution. Solve by elimination. 3x + 4y = 15 1. x = 6y – 6 3x + 4y = 57 2. 5x – 4y = – 1

  3. Objective Solve systems of equations in two variables that contain at least one second-degree equation.

  4. Vocabulary nonlinear system of equations

  5. A nonlinear system of equations is a system in which at least one of the equations is not linear. You have been studying one class of nonlinear equations, the conic sections. The solution set of a system of equations is the set of points that make all of the equations in the system true, or where the graphs intersect. For systems of nonlinear equations, you must be aware of the number of possible solutions.

  6. You can use your graphing calculator to find solutions to systems of nonlinear equations and to check algebraic solutions.

  7. 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.

  8. Example 1 Continued Step 1 Solve each equation for y. Solve the first equation for y. Solve the second equation for y. Step 2 Graph the system on your calculator, and use the intersect feature to find the solution set. The points of intersection are (–4, –3), (–4, 3), (4, –3), (4, 3).

  9. 1 y = (x –3)2 2 Check It Out! Example 1 3x + y = 4.5 Solve by graphing.

  10. The substitution method for solving linear systems can also be used to solve nonlinear systems algebraically.

  11. 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.

  12. 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

  13. Example 2 Continued Step 3 Substitute –8 and 6 into x2 = 2y + 52 to find valuesof x. x2 = 2(–8) + 52 = 36 x = ±6 (6, –8) and (–6, –8) are solutions. x2 = 2(6) + 52 = 64 x = ±8 (8, 6) and (–8, 6) are solutions. The solution set of the system is {(6, –8) (–6, –8), (8, 6), (–8, 6)}.

  14. Example 2 Continued Check Use a graphing calculator.The graph supports that there are four points of intersection.

  15. Check It Out! Example 2a Solve the system of equations by using the substitution method. x + y = –1 x2 + y2 = 25

  16. Check It Out! Example 2b Solve the system of equations by using the substitution method. x2 + y2 = 25 y – 5 = –x2

  17. Remember! In Example 3, you can check your work on a graphing calculator. The elimination method can also be used to solve systems of nonlinear equations.

  18. 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.

  19. Example 3 Continued Step 1 Eliminate y2. 36x2+ 25y2 = 169 Subtract the first equation from the second. –4x2– 25y2 = –41 32x2 = 128 Solve for x. x2 = 4, so x = ±2

  20. Example 3 Continued Step 2 Find the values for y. Substitute 4 for x2. 4(4) + 25y2 = 41 16 + 25y2 = 41 Simplify. 25y2 = 25 y = ±1 The solution set of the system is {(–2, –1), (–2, 1), (2, –1), (2, 1)}.

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

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