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

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

Vocabulary nonlinear system of equations

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.

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

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.

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).

Check It Out! Example 1 3x + y = 4.5 Solve by graphing. y = (x – 3)2 The graph of the first equation is a straight line, and the graph of the second equation is a parabola, so there may be as many as two points of intersection.

Check It Out! Example 1 Continued Step 1 Solve each equation for y. y = –3x + 4.5 Solve the first equation for y. y = (x – 3)2 1 2 Step 2 Graph the system on your calculator, and use the intersect feature to find the solution set. The point of intersection is (0, 4.5).

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

Example 2: 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 four points of intersection.

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

Example 2 Continued Step 3 Substitute –8 and 6 into x2 = 2y + 52 to find values of 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)}.

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

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.

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

Check It Out! Example 2a Continued Step 3 Substitute –4 and 3 into x + y = –1 to find values of 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)}.

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

Check It Out! Example 2b Continued Step 1 It is simplest to solve for x2 because both equations have x2 terms. Solve for x2 in the second equation. x2 = –y + 5 Step 2 Use substitution. Substitute this value into the first equation. (–y + 5) + y2 = 25 y2 – y – 20 = 0 Simplify, and set equal to 0. (y + 4) (y – 5) = 0 Factor. y = –4 or y = 5

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

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.

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

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

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.

What if …? Suppose the paths of the boats can be modeled by the system Check It Out! Example 4 What if …? Suppose the paths of the boats can be modeled by the system . Is there any danger of collision? 36x2 + 25y2 = 900 y + 2 = – x2 1 10

Understand the Problem Check It Out! Example 4 Continued 1 Understand the Problem There is a potential danger of a collision if the two paths cross. The paths will cross if the graphs of the equations intersect. List the important information: 36x2 + 25y2 = 900 represents the path of the first boat. represents the path of the second boat. y + 2 = – x2 1 10

Check It Out! Example 4 Continued 2 Make a Plan To see if the graphs intersect, solve the system 36x2 + 25y2 = 900 . y + 2 = – x2 1 10

Check It Out! Example 4 Continued Solve 3 The graph of the first equation is an ellipse, and the graph of the second equation is a parabola. There may be as many as four points of intersection. Solve the second equation for x2. x2 = –10y – 20 36(–10y – 20) + 25y2 = 900 Substitute this value into the first equation. 25y2 – 360y – 1620 = 0 Use the quadratic formula.

Check It Out! Example 4 Continued Substitute y = 18 and y = –3.6 into x2 = –10y – 20 to find the values for x. x2 = –10(18) – 20 x2 = –10(–3.6) – 20 x2 = –200 There are no real values. x = ±4 There are two real solutions to the system, (4, –3.6) and (–4, –3.6), so there are 2 possible collision points.

Check It Out! Example 4 Continued Look Back 4 The graph supports that there are two points of intersection. Because the paths intersect, the boats are in danger of colliding if they arrive at the intersections (4, –3.6) or (–4, –3.6) at the same time.