4e. 1 2 b - 1ab + 3 4 b > 1 4 b + 1 2 1 2 b - b - 3 4 > 1 4 b + 1 2 4a1 2 b - b - 3 4 b > 4a1 4 b + 1 2 b - 2b - 4b - 3 > b + 2 -2b - 3 > b + 2 -2b - 3 - b > b + 2 - b 3b - 3 > 2 -3b - 3 + 3 > 2 + 3 -3b > 5 -3b -3 < 5 -3 b<- 5 3 Distribute -1 to ab + The graph consists of all numbers to the left of - 5 3 3 4 but not including - 5 3 . –5 –4 –3 –2 –1 0 1 2 3 4 5 Interval notation: a-∞, - 5 3 b Set-builder notation: e bP b <- 5 3 f 4f. 2(4x - 3) > 2x + 3(2x + 1) 8x - 6 > 2x + 6x + 3 8x - 6 > 8x + 3 8x - 6 - 8x > 8x + 3 - 8x -6 > 3 The resulting inequality, -6 > 3, is always false. Therefore, there is no solution of this inequality. The solution set is the empty set, or ∅. The graph is a blank number line as shown. –5 –4 –3 –2 –1 0 1 2 3 4 5 4g. x + 5(x - 2) < 3(2x + 1) + 7 x + 5x - 10 < 6x + 3 + 7 6x - 10 < 6x + 10 6x - 10 - 6x < 6x + 10 - 6x -10 < 10 The resulting inequality, -10 < 10, is always true. Therefore, any real number is a solution of this inequality. So, the solution set is all real numbers, or . –5 –4 –3 –2 –1 0 1 2 3 4 5 Interval notation: (-∞,��∞) Set-builder notation: Exux is a real numberF b . Multiply each side by the LCD, 4. Simplify. Combine like terms. Subtract b from each side. Simplify. Add 3 to each side. Simplify. Divide each side by -3. Remember to reverse the inequality symbol. Simplify. Apply the distributive property. Combine like terms. Subtract 8x from each side. False. Apply the distributive property. Combine like terms. Subtract 6x from each side. True. 172 Chapter 2 Linear Equations and Inequalities in One Variable
hendricks_beginning_algebra_1e_ch1_3
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