Stitz-Zeager_College_Algebra_e-book

# 2a n 1d 2a n 1d the right hand side of

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Unformatted text preview: . Labeling the equations as before, we have 2 The calculator has trouble conﬁrming the solution (− ln(2), 0) due to its issues in graphing square root functions. If we mentally connect the two branches of the thicker curve, we see the intersection. 8.7 Systems of Non-Linear Equations and Inequalities 537 z (x − 2) = x E1 E2 yz = y E 3 (x − 2)2 + y 2 = 1 The easiest equation to start with appears to be E 2. While it may be tempting to divide both sides of E 2 by y , we caution against this practice because it presupposes y = 0. Instead, we take E 2 and rewrite it as yz − y = 0 so y (z − 1) = 0. From this, we get two cases: y = 0 or z = 1. We take each case in turn. Case 1: y = 0. Substituting y = 0 into E 1 and E 3, we get E 1 z (x − 2) = x E 3 (x − 2)2 = 1 Solving E 3 for x gives x = 1 or x = 3. Substituting these values into E 1 gives z = −1 when x = 1 and z = 3 when x = 3. We obtain two solutions, (1, 0, −1) and (3, 0, 3). Case 2: z = 1. Substituting z = 1 into E 1 and E 3 gives us E1 (1)(x −...
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