College Algebra (Abridged)

Solution: We can solve this question by substituting the given values of and into the inequalities. The option that satisfies both the inequalities is the correct answer. Substitute = − 1 and = 2 : − 1– 2 > 1 − 3 > 1 Not true Now substitute (3, − 1) : 3– ( − 1) > 1 = 4 > 1 True − 1 < 2 × 3– ( − 1) = − 1 < 7 True Hence, option b is the answer. One may check for other options as well for double checking. Question 10: D Solution: The -intercept of the line is − 7 , therefore options (b) and (c) can be eliminated. The shaded area includes all points that will satisfy the equation. Therefore, select any point in the shaded area, plug it into each equation, and if it proves to be a true statement, that is the answer. For example, if the point (0,0) is chosen: ≤ 4 − 7 0 ≤ 4(0) − 7 0 ≤ − 7 This is false, so the answer cannot be (a). The answer must be (d). To check: ≥ 4 − 7 0 ≥ 4(0) − 7 0 ≥ − 7

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