10-7 10-7Solving SolvingNonlinear NonlinearSystems Systems Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 22 10-7 Solving Nonlinear Systems Warm Up Solve by substitution. 1. 3x + 4y = 15 x = 3; y = 1.5 x = 6y – 6 Solve by elimination. 2. Holt Algebra 2 3x + 4y = 57 5x – 4y = – 1 x = 7; y = 9 10-7 Solving Nonlinear Systems Objective Solve systems of equations in two variables that contain at least one second-degree equation. Holt Algebra 2 10-7 Solving Nonlinear Systems Vocabulary nonlinear system of equations Holt Algebra 2 10-7 Solving Nonlinear Systems 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. Holt Algebra 2 10-7 Solving Nonlinear Systems You can use your graphing calculator to find solutions to systems of nonlinear equations and to check algebraic solutions. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 1: Solving a Nonlinear System by Graphing Solve x2 + y2 = 25 2 2 4x + 9y = 145 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. Holt Algebra 2 10-7 Solving Nonlinear Systems 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). Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 1 Solve 3x + y = 4.5 y= 1 (x – 3)2 2 by graphing. 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 1 Continued Step 1 Solve each equation for y. y = –3x + 4.5 y= 1 2 Solve the first equation for y. (x – 3)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). Holt Algebra 2 10-7 Solving Nonlinear Systems The substitution method for solving linear systems can also be used to solve nonlinear systems algebraically. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 2: Solving a Nonlinear System by Substitution Solve x2 + y2 = 100 1 y= 2 2 x – 26 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 2 Continued Step 1 It is simplest to solve for x2 because both equations have x2 terms. 2 in the second 2 Solve for x x = 2y + 52 equation. Step 2 Use substitution. (2y + 52) + y2 = 100 y2 + 2y – 48 = 0 (y + 8) (y – 6) = 0 y = –8 or y = 6 Holt Algebra 2 Substitute this value into the first equation. Simplify, and set equal to 0. Factor. 10-7 Solving Nonlinear Systems 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)}. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 2 Continued Check Use a graphing calculator. The graph supports that there are four points of intersection. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 2a Solve the system of equations by using the substitution method. x + y = –1 2 2 x + y = 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. Holt Algebra 2 10-7 Solving Nonlinear Systems 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. 2 2 (–y – 1) + y = 25 y2 + 2y + 1 + y2 – 25 = 0 Substitute this value into the second equation. Simplify and set equal to 0. 2y2 +2y – 24 = 0 2[(y2 + y – 12)] = 0 2(y + 4)(y – 3) = 0 y = –4 or y = 3 Holt Algebra 2 Factor. 10-7 Solving Nonlinear Systems 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)}. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 2a Continued Check Use a graphing calculator. The graph supports that there are two points of intersection. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 2b Solve the system of equations by using the substitution method. x2 + y2 = 25 y – 5 = –x2 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 2b Continued 2 Step 1 It is simplest to solve for x because both equations have x2 terms. 2 in the second 2 Solve for x x = –y + 5 equation. Step 2 Use substitution. (–y + 5) + y2 = 25 y2 – y – 20 = 0 (y + 4) (y – 5) = 0 y = –4 or y = 5 Holt Algebra 2 Substitute this value into the first equation. Simplify, and set equal to 0. Factor. 10-7 Solving Nonlinear Systems Check It Out! Example 2b Continued Step 3 Substitute –4 and 5 into x2 + y2 = 25 to find values of x. x2 + (–4)2 = 25 x = ±3 (3, –4) and (–3, –4) are solutions. x2 +(5)2 = 25 x=0 (0, 5) is a solution. The solution set of the system is {(3, –4), (–3, –4), (0, 5)}. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 2b Continued Check Use a graphing calculator. The graph supports that there are three points of intersection. Holt Algebra 2 10-7 Solving Nonlinear Systems The elimination method can also be used to solve systems of nonlinear equations. Remember! In Example 3, you can check your work on a graphing calculator. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 3: Solving a Nonlinear System by Elimination Solve 4x2 + 25y2 = 41 36x2 + 25y2 = 169 by using the elimination method. 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 3 Continued Step 1 Eliminate y2. 36x2 + 25y2 = 169 –4x2 – 25y2 = –41 32x2 = 128 x2 = 4, so x = ±2 Holt Algebra 2 Subtract the first equation from the second. Solve for x. 10-7 Solving Nonlinear Systems Example 3 Continued Step 2 Find the values for y. 2 4(4) + 25y = 41 2 16 + 25y = 41 Substitute 4 for x2. Simplify. 25y2 = 25 y = ±1 The solution set of the system is {(–2, –1), (–2, 1), (2, –1), (2, 1)}. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 3 Solve 25x2 + 9y2 = 225 25x2 – 16y2 = 400 by using the elimination method. 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 3 Continued Step 1 Eliminate x2. 25x2 – 16y2 = 400 –25x2 – 9y2 = –225 –25y2 = 175 y2 = –7 Subtract the first equation from the second. Solve for y. There is no real solution of the system. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 4: Problem-Solving Application Suppose that the paths of two boats are 2 2 modeled by 36x + 25y = 900 and y = 0.25x2 – 6. How many possible collision points are there? 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 4 Continued 1 Understand the Problem List the important information: • 36x2 + 25y2 = 900 represents the path of the first boat. • y = 0.25x2 – 6 represents the path of the second boat. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 4 Continued 2 Make a Plan To see if the graphs intersect, solve the system 36x2 + 25y2 = 900 y = 0.25x2 – 6 Holt Algebra 2 . 10-7 Solving Nonlinear Systems Example 4 Continued 3 Solve 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. x2 = 4y + 24 Solve the second equation 2 for x . 36(4y + 24) + 25y2 = 900 Substitute this value into the first equation. 2 25y + 144y – 36 = 0 Simplify, and set equal to 0. Use the quadratic formula. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 4 Continued Substitute y = 0.24 and y = –6 into x2 = 4y + 24 to find the values for x. x2 = 4(0.24) + 24 x ±5 x2 = 4(–6) + 24 x=0 There are three real solutions to the system, (0, –6), (5, 0.24), and (–5, 0.24), so there are 3 possible collision points. Holt Algebra 2 10-7 Solving Nonlinear Systems Example 4 Continued 4 Look Back The graph supports that there are three points of intersection. Because the paths intersect, the boats are in danger of colliding if they arrive at one of the three points of intersection at the same time. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 4 What if …? Suppose the paths of the boats can be modeled by the system 36x2 + 25y2 = 900 1 y+2=– 10 x2 Is there any danger of collision? Holt Algebra 2 . 10-7 Solving Nonlinear Systems 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. 1 • y+2=– x2 represents the path of the 10 second boat. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 4 Continued 2 Make a Plan To see if the graphs intersect, solve the system 36x2 + 25y2 = 900 y+2=– Holt Algebra 2 1 x2 10 . 10-7 Solving Nonlinear Systems Check It Out! Example 4 Continued 3 Solve 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. x2 = –10y – 20 Solve the second equation for x2. 2 36(–10y – 20) + 25y = 900 Substitute this value into the first equation. 2 25y – 360y – 1620 = 0 Use the quadratic formula. Holt Algebra 2 10-7 Solving Nonlinear Systems 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Check It Out! Example 4 Continued 4 Look Back 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. Holt Algebra 2 10-7 Solving Nonlinear Systems Lesson Quiz 1. Solve 4x2 – 9y2 = 108 x2 + y2 = 40 2. Solve 2x2 + y2 = 54 by graphing. (±6, ±2) by substitution. (±5, ±2) x2 – 3y2 = 13 3. Solve 4x2 + 4y2 = 52 2 2 9x – 4y = 65 Holt Algebra 2 by elimination. (±3, ±2)