SECTION A.9 Interval Notation; Solving Inequalities A83 Examples 10 and 11 illustrate the following results: Now Work problem 97 Now Work problem 103 THEOREM If a is any positive number, then • u a a u a is equivalent to < − < < (5) • u a a u a is equivalent to ≤ − ≤ ≤ (6) • u a u a u a is equivalent to or > < − > (7) • u a u a u a is equivalent to or ≥ ≤ − ≥ (8) Figure 37 x2 4 3 + ≤ 4 2 0 22 25 –1 22 –7 22 Solving an Inequality Involving Absolute Value Solve the inequality x2 4 3, + ≤ and graph the solution set. Solution EXAMPLE 12 x x x x x x 2 4 3 3 2 4 3 3 4 2 4 4 3 4 7 2 1 7 2 2 2 1 2 7 2 1 2 + ≤ − ≤ + ≤ − − ≤ + − ≤ − − ≤ ≤− − ≤ ≤ − − ≤ ≤− This follows the form of statement (6) where u x2 4 = + . Use statement (6). Subtract 4 from each part. Simplify. Divide each part by 2. Simplify. The solution set is x x 7 2 1 2 , { } − ≤ ≤ − that is, all x in the interval 7 2 , 1 2 . − − See Figure 37 for a graph of the solution set. Figure 38 x2 5 3 − > 0 1 2 3 4 5 6 21 22 7 Solving an Inequality Involving Absolute Value Solve the inequality x2 5 3, − > and graph the solution set. Solution EXAMPLE 13 x2 5 3 − > This follows the form of statement (7) where u x2 5 = − . x x x x x x x x x x 2 5 3 2 5 5 3 5 2 2 2 2 2 2 1 or or or or or 2 5 3 2 5 5 3 5 2 8 2 2 8 2 4 − < − − + < − + < < < − > − + > + > > > The solution set is x x x 1 or 4 , { } < > that is, all x in , 1 4, . ( ) ( ) −∞ ∪ ∞ See Figure 38 for a graph of the solution set. Use statement (7). Add 5 to each part. Simplify. Divide each part by 2. Simplify. CAUTION A common error to be avoided is to attempt to write the solution x 1 < or x 4 > as the combined inequality x 1 4, > > which is incorrect, since there are no numbers x for which x x 1 and 4. < > Another common error is to “mix” the symbols and write x 1 4, < > which makes no sense. j
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