930 CHAPTER 9 Systems and Matrices Sometimes a combination of the elimination method and the substitution method is effective in solving a system, as illustrated in Example 3. NOTE The elimination method works with the system in Example 2 because the system can be thought of as a system of linear equations where the variables are x2 and y2. To see this, substitute u for x2 and v for y2. The resulting system is linear in u and v. EXAMPLE 3 Solving a Nonlinear System (Combination of Methods) Solve the system. x2 + 3xy + y2 = 22 (1) x2 - xy + y2 = 6 (2) SOLUTION Begin as with the elimination method. x2 + 3xy + y2 = 22 (1) -x2 + xy - y2 = -6 Multiply (2) by -1. 4xy = 16 Add. (3) y = 4 x Solve for y 1x ≠02. (4) Now substitute 4 x for y in either equation (1) or equation (2). We use equation (2). x2 - xy + y2 = 6 (2) x2 - x a 4 xb + a 4 xb 2 = 6 Let y = 4 x . x2 - 4 + 16 x2 = 6 Multiply and square. x4 - 4x2 + 16 = 6x2 Multiply by x2 to clear fractions. x4 - 10x2 + 16 = 0 Subtract 6x2. 1x2 - 221x2 - 82 = 0 Factor. x2 - 2 = 0 or x2 - 8 = 0 Zero-factor property x2 = 2 or x2 = 8 Solve each equation. x = {22 or x = {222 Square root property; {28 = {24 # 22 = {222 Substitute these x-values into equation (4) y = 4 x to find corresponding values of y. This equation is quadratic in form. For each equation, include both square roots. Let x = 22 in (4). Let x = -22 in (4). Let x = 222 in (4). Let x = -222 in (4). y = 42 2 = 222 y = 4 -22 = -222 y = 4 222 = 22 y = 4 -222 = - 22 The solution set of the system is E A 2 2, 222 B, A -22, -222 B, A222, 22 B, A -222, -22 B F. Verify these solutions by substitution in the original system. S Now Try Exercise 41.
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