183 1.8 Absolute Value Equations and Inequalities Cases 1, 2, and 3 require that the absolute value expression be isolated on one side of the equation or inequality. EXAMPLE 3 Solving an Absolute Value Inequality (Case 3) Solve 2 - 7x - 1 74. SOLUTION 2 - 7x - 1 74 2 - 7x 75 Add 1 to each side. 2 - 7x 6 -5 or 2 - 7x 75 Case 3 -7x 6 -7 or -7x 73 Subtract 2 from each side. x 71 or x 6 - 3 7 Divide by -7. Reverse the direction of each inequality. The solution set written in interval notation is A -∞, - 3 7B ´11, ∞2. S Now Try Exercise 51. Special Cases The three cases given in this section require the constant k to be positive. When k "0, use the fact that the absolute value of any expression must be nonnegative, and consider the conditions necessary for the statement to be true. EXAMPLE 4 Solving Special Cases Solve each equation or inequality. (a) 2 - 5x Ú -4 (b) 4x - 7 6 -3 (c) 5x + 15 = 0 SOLUTION (a) Since the absolute value of a number is always nonnegative, the inequality 2 - 5x Ú -4 is always true. The solution set includes all real numbers, written 1-∞, ∞2. (b) There is no number whose absolute value is less than -3 (or less than any negative number). The solution set of 4x - 7 6 -3 is ∅. (c) The absolute value of a number will be 0 only if that number is 0. Therefore, 5x + 15 = 0 is equivalent to 5x + 15 = 0, which has solution set 5-36. CHECK Substitute -3 into the original equation. 5x + 15 = 0 Original equation 51-32 + 15 ≟0 Let x = -3. 0 = 0 ✓ True S Now Try Exercises 55, 57, and 59.
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