We can check by substituting a value from the intersection of the sets and will find that this value makes both inequalities true. So, the solution set is a-∞, 4 3 b . 3c. 4 - 2x > 0 and 5x ≥ 15 5x 5 4- 2x - 4 > 0 - 4 ≥ 15 5 -2x -2 < -4 -2 x ≥ 3 x < 2 Now we find the intersection of the solution sets of the two inequalities. Inequality 1 x < 2 (-∞, 2) –5 –4 –3 –2 –1 0 1 2 3 4 5 Inequality 2 x ≥ 3 3, ∞) –5 –4 –3 –2 –1 0 1 2 3 4 5 Intersection of the inequalities ∅ –5 –4 –3 –2 –1 0 1 2 3 4 5 We can check by substituting values into the compound inequality. We will find that there are no values that make the inequality true. So, the solution set is the empty set, or ∅. 3d. 2x - 1 ≤ 5 and 2x - 1 ≥ -5 2x - 1 + 1 ≤ 5 + 1 2x - 1 + 1≥-5 + 1 2x ≤ 6 2x≥-4 x ≤3 x≥-2 Now we find the intersection of the solution sets of the two inequalities. Inequality 1 x ≤ 3 (-∞, 3 –4 –3 –2 –1 0 1 2 3 4 5 6 Inequality 2 x≥-2 -2, ∞) –4 –3 –2 –1 0 1 2 3 4 5 6 Intersection of the inequalities -2, 3 –4 –3 –2 –1 0 1 2 3 4 5 6 We can check by substituting a value from the intersection into the compound inequality. We will find that it makes the inequality true. So, the solution set is -2, 3. Note: This problem can also be worked using a compact form. The inequality 2x - 1≥-5 is the same as -5 ≤ 2x - 1. So, we have that -5 ≤ 2x - 1 and 2x - 1 ≤ 5 This is equivalent to -5 ≤ 2x - 1 ≤ 5 Subtract 4 from each side. Simplify. Divide each side by -2 and reverse the inequality sign. -2x>-4 Divide each side by 5. Simplify. Add 1 to each side. Divide each side by 2. Add 1 to each side. Divide each side by 2. 106 Chapter 2 Linear Equations and Inequalities in One Variable
hendricks_intermediate_algebra_1e_ch1_3
To see the actual publication please follow the link above