929 9.5 Nonlinear Systems of Equations Visualizing the types of graphs involved in a nonlinear system helps predict the possible numbers of ordered pairs of real numbers that may be in the solution set of the system. For example, a line and a parabola may have 0, 1, or 2 points of intersection, as shown in Figure 12. LOOKING AHEAD TO CALCULUS In calculus, finding the maximum and minimum points for a function of several variables usually requires solving a nonlinear system of equations. CAUTION If we had solved for x in equation (2) to begin the algebraic solution in Example 1, we would have found y = 0 or y = -3. Substituting y = 0 into equation (1) gives x2 = 4, so x = 2 or x = -2, leading to the ordered pairs 12, 02 and 1-2, 02. The ordered pair 12, 02 does not satisfy equation (2), however. This illustrates the necessity of checking by substituting all proposed solutions into each equation of the system. Nonlinear systems where both variables are squared in both equations are best solved by elimination, as shown in the next example. The proposed solutions are 12, 02 and 1-2, 02. These satisfy both equations, confirming that the solution set is 512, 02, 1-2, 026. S Now Try Exercise 23. x2 + y2 = 4 (1) 22 + y2 = 4 Let x = 2. y2 = 0 y = 0 x2 + y2 = 4 (1) 1-222 + y2 = 4 Let x = -2. y2 = 0 y = 0 EXAMPLE 2 Solving a Nonlinear System (Elimination Method) Solve the system. x2 + y2 = 4 (1) 2x2 - y2 = 8 (2) SOLUTION The graph of equation (1) is a circle, and, as we will see in later work, the graph of equation (2) is a hyperbola. These graphs may intersect in 0, 1, 2, 3, or 4 points. We add to eliminate y2. x2 + y2 = 4 (1) 2x2 - y2 = 8 (2) 3x2 = 12 Add. x2 = 4 Divide by 3. x = {2 Square root property Find y by substituting the values of x in either equation (1) or equation (2). Remember to find both square roots. −6.6 −4.1 4.1 6.6 To solve the system in Example 2 graphically, solve equation (1) for y to obtain y = {24 - x2. Similarly, equation (2) yields y = {2-8 + 2x2. Graph these four functions to find the two solutions. The solution 1-2, 02 is indicated in the screen above. x y One point of intersection 0 x y Two points of intersection 0 x y No points of intersection 0 Figure 12
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