90 Chapter 2 Linear Equations and Inequalities in One Variable Objective 2 Examples Solve each inequality. Graph the solution set and write the solution set in interval notation and set-builder notation. 2a. x + 4<-1 2b. -2.5y ≤ 50 2c. -2a - 4 ≥ 10 2d. 3(2y - 4) - (y + 3) ≥ 5(3y + 1) 2e. 3 2 (x - 5) > 1 4 x + 4 2f. 4x - 5(x - 2) < 6(x + 1) - 7x 2g. x + 5(x - 2) < 3(2x + 1) + 7 Solutions 2a. x + 4<-1 x + 4 - 4<-1 - 4 Subtract 4 from each side. x<-5 Simplify. We graph the solution set and write the solution set in interval notation and set-builder notation. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 Interval notation: (-∞, -5) Set-builder notation: 5xux<-56 Check: It is impossible to check every solution but we can check two values to make sure our answer is reasonable. A value in the shaded region should make the original inequality true, while other values will not satisfy the inequality. x=-6 (in solution set): x = 1 (not in solution set): x + 4<-1 x + 4<-1 -6 + 4<-1 1 + 4<-1 - 2<-1 True 5<-1 False 2b. -2.5y ≤ 50 -2.5y -2.5 ≥ 50 -2.5 y≥-20 We graph the solution set and write the solution set in interval notation and set-builder notation. –25 –24 –23 –22 –21 –20 –19 –18 –17 –16 –15 Interval notation: -20, ∞) Set-builder notation: 5yuy≥-206 Check: We will check two values to make sure our answer is reasonable. y=-10 (in solution set): y=-30 (not in solution set): -2.5y ≤ 50 -2.5y ≤ 50 -2.5(-10) ≤ 50 -2.5(-30) ≤ 50 25 ≤ 50 True 75 ≤ 50 False 2c. -2a - 4 ≥ 10 -2a - 4 + 4 ≥ 10 + 4 -2a ≥ 14 -2a ≤ -2 14 -2 a≤-7 Divide each side by -2.5 and reverse the inequality symbol. Simplify. Add 4 to each side. Simplify. Divide each side by -2 and reverse the inequality symbol. Simplify.
hendricks_intermediate_algebra_1e_ch1_3
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