182 CHAPTER 1 Equations and Inequalities LOOKING AHEAD TO CALCULUS The precise definition of a limit in calculus requires writing absolute value inequalities. A standard problem in calculus is to find the “interval of convergence” of a power series by solving the following inequality. x - a 6r This inequality says that x can be any number within r units of a on the number line, so its solution set is indeed an interval—namely the interval 1a - r, a + r2. (b) If the absolute values of two expressions are equal, then those expressions are either equal in value or opposite in value. 4x - 3 = x + 6 Special case a = b 4x - 3 = x + 6 or 4x - 3 = -1x + 62 Consider both possibilities. 3x = 9 or 4x - 3 = -x - 6 Solve each linear equation. x = 3 or 5x = -3 x = - 3 5 CHECK 4x - 3 = x + 6 Original equation Both solutions check. The solution set is E - 3 5 , 3F. S Now Try Exercises 11 and 19. P 4A - 3 5B - 3P ≟P - 3 5 + 6P Let x = - 3 5. P - 12 5 - 3P ≟P - 3 5 + 6P P - 27 5 P = P 27 5 P ✓ True 4132 - 3 ≟ 3 + 6 Let x = 3. 12 - 3 ≟ 3 + 6 9 = 9 ✓ True Absolute Value Inequalities EXAMPLE 2 Solving Absolute Value Inequalities (Cases 2 and 3) Solve each inequality. (a) 2x + 1 67 (b) 2x + 1 77 SOLUTION (a) This inequality fits Case 2. If the absolute value of an expression is less than 7, then the value of the expression is between -7 and 7. 2x + 1 67 -7 62x + 1 67 Case 2 -8 6 2x 66 Subtract 1 from each part. -4 6 x 63 Divide each part by 2. The final inequality gives the solution set 1-4, 32 in interval notation. (b) This inequality fits Case 3. If the absolute value of an expression is greater than 7, then the value of the expression is either less than -7 or greater than 7. 2x + 1 77 2x + 1 6 -7 or 2x + 1 77 Case 3 2x 6 -8 or 2x 76 Subtract 1 from each side. x 6 -4 or x 73 Divide each side by 2. The solution set written in interval notation is 1-∞, -42 ´13, ∞2. S Now Try Exercises 27 and 29.
RkJQdWJsaXNoZXIy NjM5ODQ=