SECTION 11.6 Systems of Nonlinear Equations 823 Figure 21 x y –6 6 (3, 2) (–3, 2) (2, –3) (–2, –3) x2 + y2 = 13 2 –8 x2 – y = 7 (y = x2 – 7) Figure 22 This quadratic equation in y can be solved by factoring. y y y y y y 6 0 3 2 0 3 or 2 2 ( )( ) + − = + − = = − = Use these values for y in equation (2) to find x. • If = y 2, then = + = x y 7 9, 2 so = x 3 or −3. • If = − y 3, then = + = x y 7 4, 2 so = x 2 or −2. There are four solutions: = = = − = = = − x y x y x y 3, 2; 3, 2; 2, 3; and = − = − x y 2, 3. You should verify that, in fact, these four solutions also satisfy equation (1), so all four are solutions of the system. The four points, ( ) ( ) ( ) − − 3,2, 3,2, 2, 3, and ( ) − − 2, 3 , are the points of intersection of the graphs. Look again at Figure 21. Select each point of intersection to find that the solutions to the system of equations are ( ) ( ) ( ) − − − 3,2 , 3,2 , 2, 3 ,and ( ) − 2, 3. Now Work PROBLEM 13 USING ELIMINATION Solving a System of Nonlinear Equations Solve: − = = ⎧ ⎨ ⎪ ⎩⎪⎪ x y y x 4 2 2 2 1 A hyperbola 2 A parabola ( ) ( ) Algebraic Solution Using Substitution Either substitution or elimination can be used here. To use substitution, replace x2 by y in equation (1). − = − = − + = x y y y y y 4 4 4 0 2 2 2 2 This is a quadratic equation whose discriminant is ( ) − −⋅⋅=−⋅=− < 1 414 144 15 0 2 The equation has no real solutions, so the system is inconsistent. The graphs of these two equations do not intersect. See Figure 23. Graphing Solution Graph = Y x 1 2 and − = x y 4, 2 2 as shown in Figure 24 using a TI-84 Plus CE. To graph − = x y 4, 2 2 use two functions: = − = − − Y x Y x 4 and 4 2 2 3 2 From Figure 24, observe that the graphs of these two equations do not intersect. The system is inconsistent. EXAMPLE 3 Equation (1) = y x2 Place in standard form. Figure 23 x y –5 5 y = x2 (2, 4) (–2, 4) x2 – y2 = 4 (3, ) 5 –5 (3, – ) (–3, ) (–3, – ) 5 5 5 5 Figure 24 Y1 5 x2 25 28 8 Y3 5 2 x2 2 4 Y2 5 x2 2 4 5
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