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Published byBritton Cobb Modified over 8 years ago

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Recall that the absolute value of a number x, written |x|, is the distance from x to zero on the number line. Because absolute value represents distance without regard to direction, the absolute value of any real number is nonnegative. Absolute-value equations and inequalities can be represented by compound statements. Consider the equation |x| = 3. The solutions of |x| = 3 are the two points that are 3 units from zero. The solution is a disjunction: x = –3 or x = 3.

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The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3. The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction: x < –3 or x > 3. Think: Greator inequalities involving > or ≥ symbols are disjunctions. Think: Less thand inequalities involving < or ≤ symbols are conjunctions. Helpful Hint

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**Note: The symbol ≤ can replace <, and the rules still apply**

Note: The symbol ≤ can replace <, and the rules still apply. The symbol ≥ can replace >, and the rules still apply. Ex 2A: Solve the equation. This can be read as “the distance from k to –3 is 10.” |–3 + k| = 10 Rewrite the absolute value as a disjunction. –3 + k = 10 or –3 + k = –10 k = 13 or k = –7 Add 3 to both sides of each equation.

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Ex 2B: Solve the equation. Isolate the absolute-value expression. Rewrite the absolute value as a disjunction. Multiply both sides of each equation by 4. x = 16 or x = –16 You can solve absolute-value inequalities using the same methods that are used to solve an absolute-value equation.

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**Example 3A: Solving Absolute-Value Inequalities with Disjunctions**

Solve the inequality. Then graph the solution. |–4q + 2| ≥ 10 Rewrite the absolute value as a disjunction. –4q + 2 ≥ 10 or –4q + 2 ≤ –10 Subtract 2 from both sides of each inequality. –4q ≥ 8 or –4q ≤ –12 Divide both sides of each inequality by –4 and reverse the inequality symbols. q ≤ –2 or q ≥ 3

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**To check, you can test a point in each of the three region.**

Example 3A Continued {q|q ≤ –2 or q ≥ 3} –3 –2 – (–∞, –2] U [3, ∞) To check, you can test a point in each of the three region. |–4(–3) + 2| ≥ 10 |14| ≥ 10 |–4(0) + 2| ≥ 10 |2| ≥ 10 x |–4(4) + 2| ≥ 10 |–14| ≥ 10

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**Example 3B: Solving Absolute-Value Inequalities with Disjunctions**

Solve the inequality. Then graph the solution. |0.5r| – 3 ≥ –3 Isolate the absolute value as a disjunction. |0.5r| ≥ 0 Rewrite the absolute value as a disjunction. 0.5r ≥ 0 or 0.5r ≤ 0 Divide both sides of each inequality by 0.5. r ≤ 0 or r ≥ 0 The solution is all real numbers, R. –3 –2 – (–∞, ∞)

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**Example 4A: Solving Absolute-Value Inequalities with Conjunctions**

Solve the compound inequality. Then graph the solution set. |2x +7| ≤ 3 Multiply both sides by 3. Rewrite the absolute value as a conjunction. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 Subtract 7 from both sides of each inequality. 2x ≤ –4 and 2x ≥ –10 Divide both sides of each inequality by 2. x ≤ –2 and x ≥ –5

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**The solution set is {x|–5 ≤ x ≤ 2}.**

Example 4A Continued The solution set is {x|–5 ≤ x ≤ 2}. –6 –5 –3 –2 –

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**Example 4B: Solving Absolute-Value Inequalities with Conjunctions**

Solve the compound inequality. Then graph the solution set. Multiply both sides by –2, and reverse the inequality symbol. |p – 2| ≤ –6 Rewrite the absolute value as a conjunction. |p – 2| ≤ –6 and p – 2 ≥ 6 Add 2 to both sides of each inequality. p ≤ –4 and p ≥ 8 Because no real number satisfies both p ≤ –4 and p ≥ 8, there is no solution. The solution set is ø.

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