822 CHAPTER 11 Systems of Equations and Inequalities EXAMPLE 2 Substitute this expression for y in equation (2). The result is an equation containing just the variable x, which can then be solved. x y x x x x x x x x x x 2 0 2 3 2 0 2 3 2 0 2 1 2 0 2 1 0 or 2 0 1 2 or 2 2 2 2 ( ) ( )( ) − = − + = − − = + − = + = − = = − = Use these values for x in = + y x3 2 to find ( ) = − + = = ⋅ + = y y 3 1 2 2 1 2 or 3 2 2 8 The apparent solutions are = − = x y 1 2 , 1 2 and = = x y 2, 8. Check: For = − = x y 1 2 , 1 2 : ( ) ( ) − − =− − = − − − = ⋅ − = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪⎪ ⎪ 3 1 2 1 2 3 2 1 2 2 2 1 2 1 2 2 1 4 1 2 0 2 (1) (2) For = = x y 2, 8: ⋅ − = − =− ⋅ − = ⋅ − = ⎧ ⎨ ⎪⎪⎪ ⎩ ⎪⎪⎪ (1) (2) 3 2 8 6 8 2 2 2 8 2 4 8 0 2 Each solution checks. The graphs of the two equations intersect at the points ( ) − 1 2 , 1 2 and ( ) 2,8 , as shown in Figure 19. Now Work PROBLEM 15 USING SUBSTITUTION 2 Solve a System of Nonlinear Equations Using Elimination Equation (2) Substitute +x3 2 for y. Remove parentheses. Factor. Use the Zero-Product Property. Solving a System of Nonlinear Equations Solve: + = − = ⎧ ⎨ ⎪⎪⎪ ⎩ ⎪⎪⎪ x y x y 13 7 2 2 2 ( ) ( ) 1 Acircle 2 A parabola Algebraic Solution Using Elimination First graph each equation, as shown in Figure 21 on the next page. Based on the graph, four solutions are expected. Notice that subtracting equation (2) from equation (1) eliminates the variable x. + = − = ⎧ ⎨ ⎪⎪⎪ ⎩ ⎪⎪⎪ + = x y x y y y 13 7 6 2 2 2 2 Subtract. Graphing Solution Use a graphing utility to graph + = x y 13 2 2 and − = x y 7. 2 (Remember that to graph + = x y 13 2 2 on most graphing utilities requires two functions, = − Y x 13 1 2 and = − − Y x 13 , 2 2 and a square screen.) Figure 22 on the next page shows the graphs using Desmos. Observe that the system apparently has four solutions. (continued) (continued)
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