824 CHAPTER 11 Systems of Equations and Inequalities Graphing Solution Figure 25 shows the solution using GeoGebra. The point of intersection of the graphs of the equations is ( ) −1, 1 . Note that the other apparent point of intersection is at ( ) 0, 1 . However, because x (does not equal) 0, there is a hole in the graph of equation (2), so there is no intersection of the two graphs there. Note: When using a tool like GeoGebra or Desmos it is not necessary to solve for the variable y. EXAMPLE 4 Figure 25 Algebraic Solution Using Elimination Because it is not straightforward how to graph the equations in the system, we proceed directly to use the method of elimination. First, multiply equation (2) by x to eliminate the fraction. The result is an equivalent system because x cannot be 0. [Look at the original equation (2) to see why.] + + − + = + + − = ⎧ ⎨ ⎪ ⎩⎪⎪ x x y y x x y y 3 2 0 0 2 2 2 2 ( ) ( ) ≠ x 1 2 0 Now subtract equation (2) from equation (1) to eliminate x. The result is − + = = y y 2 2 0 1 Solve for y. To find x, back-substitute = y 1 in equation (1). ( ) + + − + = + + − ⋅ + = + = + = x x y y x x x x x x 3 2 0 1 3 1 2 0 0 1 0 2 2 2 2 2 = = − x x 0 or 1 Because x cannot be 0, the value = x 0 is extraneous, so discard it. Check: Check = − = x y 1, 1: ( ) ( ) − +−+ −⋅+=−+−+= − + + − − = + − = ⎧ ⎨ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪ 1 1 1 312111320 1 1 1 1 1 0 0 1 0 2 2 2 (1) (2) The solution is = − = x y 1, 1. The point of intersection of the graphs of the equations is ( ) −1, 1 . Solving a System of Nonlinear Equations Solve: + + − + = + + − = ⎧ ⎨ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪ x x y y x y y x 3 2 0 1 0 2 2 2 (1) (2) Equation (1) Substitute 1 for y. Simplify. Factor. Use the Zero-Product Property. In Problem 55 you are asked to graph the equations given in Example 4 by hand. Be sure to show holes in the graph of equation (2) at = x 0. Now Work PROBLEMS 29 AND 49
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