The values of the variable that make at least one inequality true are solutions of a compound inequality joined by “or.” Thus, values that lie in the union of the solution sets of each inequality are solutions of the compound inequality. Procedure: Solving a Compound Inequality Involving “Or” Step 1: Find the solution set of inequality 1. Step 2: Find the solution set of inequality 2. Step 3: Find the union of the solution sets of inequalities 1 and 2. Step 4: Write the final solution set in interval notation and provide its graph. Objective 4 Examples Solve each compound inequality. Write each solution set in interval notation and graph the solution set. 4a. x + 3<-1 or 2x>-4 4b. 2 3 x - 4 ≥ 2 or -3(x + 1) ≥ 5 4c. 2x - 4>-8 or 7x - 5 < 2 Solutions 4a. x + 3<-1 or 2x>-4 x + 3 - 3<-1 - 3 2x 2 > -4 2 x<-4 x>-2 Now we find the union of the two solution sets. Inequality 1 x<-4 (-∞, -4) –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 Inequality 2 x>-2 (-2, ∞) –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 Union of two sets (-∞, -4) ∪ (-2, ∞) –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 We can check by substituting values from the union in the original compound inequality and will find that these values make at least one of the inequalities true. So, the solution set is (-∞, -4) ∪ (-2, ∞). 4b. 2 3 x - 4 ≥ 2 or -3(x + 1) ≥ 5 -3x - 3 ≥ 5 -3x - 3 + 3 ≥ 5 + 3 -3x ≥ 8 3x -≤ -3 8 -3 x≤- 8 3 3a2 3 x - 4b ≥ 3(2) Multiply each side by 3. Simplify. Add 12 to each side. Simplify. Divide each side by 2. 2x - 12 ≥ 6 2x - 12 + 12 ≥ 6 + 12 2x ≥ 18 x ≥ 9 Distribute. Add 3 to each side. Simplify. Divide each side by -3 and reverse the inequality symbol. Subtract 3 from each side. Simplify. Divide each side by 2. Simplify. 108 Chapter 2 Linear Equations and Inequalities in One Variable
hendricks_intermediate_algebra_1e_ch1_3
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