Stitz-Zeager_College_Algebra_e-book

In general if p k is true ie ak a k 1d we set

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Unformatted text preview: 1 for x, we get x = y 2 − 4. Substituting this into E 2 gives y = y 2 − 4 − 2, or y 2 − y − 6 = 0. We find y = −2 and y = 3 and since x = y 2 − 4, we get that the graphs intersect at (0, −2) and (5, 3). Putting all of this together, we get our final answer below. y y y 3 x −5 4 − −3 y2 − 4 ≤ x 2 3 4 5 x −3 x<y+2 −5 4 − 2 3 4 5 x −3 y2 − 4 ≤ x < y + 2 2. To solve this system of inequalities, we need to find all of the points (x, y ) which satisfy both inequalities. To do this, we solve each inequality separately and take the set theoretic intersection of the solution sets. We begin with the inequality x2 + y 2 ≥ 4 which we rewrite as x2 + y 2 − 4 ≥ 0. The points which satisfy x2 + y 2 − 4 = 0 form our friendly circle x2 + y 2 = 4. Using test points (0, 0) and (0, 3) we find that our solution comprises the region outside the circle. As far as the circle itself, the point (0, 2) satisfies the inequality, so the circle itself is part of the solution set. Moving to the inequality x2 − 2x + y 2 − 2y ≤ 0, we start with x2 − 2x + y 2 − 2y = 0. Completing the squares, we obtain (x − 1)...
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