928 CHAPTER 9 Systems and Matrices 9.5 Nonlinear Systems of Equations ■ Nonlinear Systems with Real Solutions ■ Nonlinear Systems with Nonreal Complex Solutions ■ An Application of Nonlinear Systems Nonlinear Systems with Real Solutions A system of equations in which at least one equation is not linear is a nonlinear system. x2 - y = 4 (1) x + y = -2 (2) x2 + y2 = 16 (1) Nonlinear systems x + y = 4 (2) The substitution method works well for solving many such systems, particularly when one of the equations is linear, as in the next example. EXAMPLE 1 Solving a Nonlinear System (Substitution Method) Solve the system. x2 - y = 4 (1) x + y = -2 (2) ALGEBRAIC SOLUTION When one of the equations in a nonlinear system is linear, it is usually best to begin by solving the linear equation for one of the variables. y = -2 - x Solve equation (2) for y. Substitute this result for y in equation (1). x2 - y = 4 (1) x2 - 1-2 - x2 = 4 Let y = -2 - x. x2 + 2 + x = 4 Distributive property x2 + x - 2 = 0 Standard form 1x + 221x - 12 = 0 Factor. x + 2 = 0 or x - 1 = 0 Zero-factor property x = -2 or x = 1 Solve each equation. Substituting -2 for x in equation (2) gives y = 0. If x = 1, then y = -3. The solution set of the given system is 51-2, 02, 11, -326. A graph of the system is shown in Figure 10. GRAPHING CALCULATOR SOLUTION Solve each equation for y and graph them in the same viewing window. We obtain y1 = x 2 - 4 and y 2 = -x - 2. The screens in Figure 11, which indicate that the points of intersection are 1-2, 02 and 11, -32, support the solution found algebraically. 2 0 x y x + y = –2 x2 – y = 4 (1, –3) (–2, 0) 2 4 –2 –4 Figure 10 y1 = x 2 − 4 y2 = −x − 2 −10 −10 10 10 Figure 11 y1 = x 2 − 4 y2 = −x − 2 −10 −10 10 10 S Now Try Exercise 15.
RkJQdWJsaXNoZXIy NjM5ODQ=