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§ 4.3 Equations and Inequalities Involving Absolute Value Absolute Value Equations Rewriting an Absolute Value Equation Without Absolute Value Bars If c is a positive real number and X represents any algebraic expression, then |X| = c is equivalent to X = c or X = -c. Rewriting an Absolute Value Equation with Two Absolute Values, Without Absolute Value Bars If |X| = |Y|, then X = Y or X = -Y. Blitzer, Intermediate Algebra, 4e – Slide #42 Absolute Value Equations EXAMPLE Solve: 3|y + 5| = 12. SOLUTION We must first isolate the absolute value expression, |y + 5|. 3|y + 5| = 12 |y + 5| = 4 y+5=4 or y + 5 = -4 y = -1 y = -9 Divide both sides by 3 Rewrite equation without absolute value bars Subtract 5 from both sides Now we will check the two solutions using the original equations. Blitzer, Intermediate Algebra, 4e – Slide #43 Absolute Value Equations CONTINUED Check -1: Check -9: 3|y + 5| = 12 3|y + 5| = 12 3|(-1) + 5| =? 12 3|(-9) + 5| =? 12 Substitute the proposed solutions Simplify ? 3|-9 + 5| = 12 3|4| = 12 ? 3|-4| = 12 Add 3(4) =? 12 3(4) =? 12 Simplify 3|-1 + 5| = 12 12 = 12 true ? Original equation ? 12 = 12 true Multiply The solutions are -1 and -9. We can also say that the solution set is {-1,-9}. Blitzer, Intermediate Algebra, 4e – Slide #44 Absolute Value Equations EXAMPLE Solve: |3x - 5| = |3x + 5|. SOLUTION We rewrite the equation without absolute value bars. 3x - 5 = 3x + 5 or 3x - 5 = -(3x + 5) We now solve the two equations that do not contain absolute value bars. 3x - 5 = 3x + 5 -5 = 5 or 3x - 5 = -(3x + 5) 3x - 5 = -3x - 5 6x - 5 = - 5 6x = 0 x=0 Blitzer, Intermediate Algebra, 4e – Slide #45 Absolute Value Equations CONTINUED Since the first equation yielded -5 = 5, which is clearly a false statement, there is no solution for the first equation. However, the second equation yielded a potentially legitimate solution, x = 0. We now check it. Check 0: |3x - 5| = |3x + 5| Original equation ? Replace x with 0 |3(0) – 5| = |3(0) + 5| Multiply |0 – 5| =? |0 + 5| |-5| =? |5| Simplify 5 = 5 true Simplify The solution is 0. We can also say that the solution set is {0}. Blitzer, Intermediate Algebra, 4e – Slide #46 Absolute Value Inequalities Solving an Absolute Value Inequality If X is an algebraic expression and c is a positive number, 1) The solutions of |X| < c are the numbers that satisfy –c < X < c. 2) The solutions of |X| > c are the numbers that satisfy X < -c or X > c. These rules are valid if < is replaced by and > is replaced by . Absolute Value Inequalities with Unusual Solution Sets If X is an algebraic expression and c is a negative number, 1) The inequality |X| < c has no solution. 2) The inequality |X| > c is true for all real numbers for which X is defined. Blitzer, Intermediate Algebra, 4e – Slide #47 Absolute Value Inequalities EXAMPLE Solve and graph the solution set on a number line: |x - 2| > 5. SOLUTION We rewrite the inequality without absolute value bars. x - 2 < -5 or x-2>5 We now solve the compound inequality. x - 2 < -5 or x-2>5 x < -3 x>7 Add 2 to both sides Blitzer, Intermediate Algebra, 4e – Slide #48 Absolute Value Inequalities CONTINUED The solution set consists of all numbers that are less than -3 or greater than 7. The solution set is {x|x < -3 or x > 7}, or, in interval notation, ,3 7, . The graph of the solution set is shown as follows: x < -3 or x>7 ) -6 -5 -4 -3 ( -2 -1 0 1 2 3 4 5 Blitzer, Intermediate Algebra, 4e – Slide #49 6 7 Absolute Value Inequalities EXAMPLE Solve and graph the solution set on a number line: 3x 1 2 20 . SOLUTION We rewrite the inequality without absolute value bars. 20 3x 1 2 20 We now solve the compound inequality. 20 3x 3 2 20 Multiply 20 3x 1 20 Simplify 19 3x 21 Add 1 to all three sides 19 x7 3 Divide all three sides by 3 Blitzer, Intermediate Algebra, 4e – Slide #50 Absolute Value Inequalities CONTINUED 19 3 The solution set is all real numbers greater than or equal to 19 and less than or equal to 7, denoted by x | 319 x 7 or [ 3 ,7]. The graph of the solution set is as follows. 19 x7 3 [ -7 -6 ] -5 -4 -3 -2 -1 0 1 2 3 4 5 19 3 Blitzer, Intermediate Algebra, 4e – Slide #51 6 7 Absolute Value Inequalities EXAMPLE The specifications for machine parts are given with tolerance limits that describe a range of measurements for which the part is acceptable. In this example, x represents the length of a machine part, in centimeters. The tolerance limit is 0.01 centimeters. Solve: x 9.4 0.01. If the length of the machine part is supposed to be 9.4 centimeters, interpret the solution. SOLUTION First we solve the given inequality. x 9.4 0.01 0.01 x 9.4 0.01 9.39 x 9.41 Original inequality Rewrite inequality Add 9.4 to all three sides Blitzer, Intermediate Algebra, 4e – Slide #52 Absolute Value Inequalities CONTINUED Therefore, the solution set is x | 9.39 x 9.41 or 9.39,9.41. So, a machine part that is supposed to be 9.4 centimeters is acceptable between a low of 9.39 centimeters and a high of 9.41 centimeters. Blitzer, Intermediate Algebra, 4e – Slide #53