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Section 2.5 Compound Inequalities 105 Solutions 3a. x + 5 ≥ 8 and -2x≥-10 Subtract 5 from x + 5 - 5 ≥ 8 - 5 -2x -2 ≤ -10 -2 x ≥3 x ≤ 5 Now we find the intersection of the solution sets of the two inequalities. Inequality 1 x ≥ 3 3, ∞) –1 0 1 2 3 4 5 6 7 8 9 Inequality 2 x ≤ 5 (-∞, 5 –1 0 1 2 3 4 5 6 7 8 9 Intersection of the inequalities 3, 5 –1 0 1 2 3 4 5 6 7 8 9 Check: The numbers in the intersection of the two inequalities should make the compound inequality true. Numbers not in the intersection of the two inequalities will make the compound inequality false. x = 4: x + 5 ≥ 8 and -2x≥-10 4 + 5 ≥ 8 and -2(4)≥-10 9 ≥ 8 and -8≥-10 True x = 0: x + 5 ≥ 8 and -2x≥-10 0 + 5 ≥ 8 and -2(0)≥-10 5 ≥ 8 and 0≥-10 False Note that 4 makes both inequalities true and so is a solution of the compound inequality. The value 0 makes one of the inequalities false and is, therefore, not a solution of the compound inequality. So, the solution set is 3, 5. 3b. 3y + 1 < 5 and 2(y - 3) < 2 3y + 1 - 1 < 5 - 1 2y - 6 < 2 3y < 4 2y - 6 + 6 < 2 + 6 y < 4 3 2y < 8 y < 4 Now we find the intersection of the solution sets of the two inequalities. Inequality 1 y < 4 3 ¢-∞, 4 3 ≤ 4 3 –5 –4 –3 –2 –1 0 1 2 3 4 5 Inequality 2 y < 4 (-∞, 4) –5 –4 –3 –2 –1 0 1 2 3 4 5 Intersection of the inequalities ¢-∞, 4 3 ≤ 4 3 –5 –4 –3 –2 –1 0 1 2 3 4 5 each side. Simplify. Divide each side by -2 and reverse the inequality symbol. Subtract 1 from each side. Simplify. Divide each side by 3. Distribute. Add 6 to each side. Simplify. Divide each side by 2.


hendricks_intermediate_algebra_1e_ch1_3
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