Stitz-Zeager_College_Algebra_e-book

# 3 we get some more practice with induction in the

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Unformatted text preview: 2 + (y − 1)2 = 2, which is √ a circle centered at (1, 1) with a radius of 2. Choosing (1, 1) to represent the inside of the circle, (1, 3) as a point outside of the circle and (0, 0) as a point on the circle, we ﬁnd that the solution to the inequality is the inside of the circle, including the circle itself. Our ﬁnal answer, then, consists of the points on or outside of the circle x2 + y 2 = 4 which lie on or inside the circle (x − 1)2 + (y − 1)2 = 2. To produce the most accurate graph, we need to ﬁnd where these circles intersect. To that end, we solve the system (E 1) x2 + y 2 = 4 2 − 2x + y 2 − 2y = 0 (E 2) x 8.7 Systems of Non-Linear Equations and Inequalities 543 We can eliminate both the x2 and y 2 by replacing E 2 with −E 1 + E 2. Doing so produces −2x − 2y = −4. Solving this for y , we get y = 2 − x. Substituting this into E 1 gives x2 + (2 − x)2 = 4 which simpliﬁes to x2 + 4 − 4x + x2 = 4 or 2x2 − 4x = 0. Factoring yields 2x(x − 2) which gives x = 0 or x = 2. Substituting these values into y = 2 − x gives the points (0, 2) and (2, 0). The intermediate graphs and ﬁna...
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