931 9.5 Nonlinear Systems of Equations To solve for the corresponding values of x, use either equation (1) or (2). x2 + y2 = 16 (1) x2 + 02 = 16 Let y = 0. x2 = 16 x = {4 x2 + y2 = 16 (1) x2 + 42 = 16 Let y = 4. x2 = 0 x = 0 S Now Try Exercise 45. The solution set, 514, 02, 1-4, 02, 10, 426, includes the points of intersection shown in Figure 13. Check the solutions in the original system. x x2 + y2 = 16 (–4, 0) (4, 0) (0, 4) y zxz + y = 4 0 Figure 13 x2 + y2 = 16 x + y = 4 −6.6 −4.1 4.1 6.6 EXAMPLE 4 Solving a Nonlinear System (Absolute Value Equation) Solve the system. x2 + y2 = 16 (1) x + y = 4 (2) SOLUTION Use the substitution method. Begin by solving equation (2) for x . x = 4 - y (3) In equation (1), the first term is x2, which is the same as x 2. Therefore, we substitute 4 - y for x in equation (1). x2 + y2 = 16 (1) 14 - y22 + y2 = 16 Let x = 4 - y. 116 - 8y + y22 + y2 = 16 Square the binomial. 2y2 - 8y = 0 Combine like terms. y2 - 4y = 0 Divide by 2. y1y - 42 = 0 Factor. y = 0 or y - 4 = 0 Zero-factor property y = 4 Add 4. Remember the middle term. x + y = 4 (2) x + 0 = 4 Let y = 0. x = 4 x = {4 x + y = 4 (2) x + 4 = 4 Let y = 4. x = 0 x = 0 NOTE After solving for y in Example 4, the corresponding values of x can be found using equation (2) instead of equation (1).
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