Success Criteria:  I can interpret complicated expressions by viewing one or more of their parts as a single entity  Be able to create equations and inequalities in one variable and use them to solve problems Today 1.Do Now 2.Check HW #6 3.Lesson HW #7 5.Complete iReady Do Now (Turn on laptop to my calendar) I can write and solve absolute-value equations and inequalities. Write the inequality used to solve.

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.

Solve the equation. Example 2A: Solving Absolute-Value Equations |–3 + k| = 10 –3 + k = 10 or –3 + k = –10 k = 13 or k = –7 |6x| – 8 = 22 Solve the equation. |6x| = 30 6x = 30 or 6x = –30 x = 5 or x = –5

Solve the equation. Example 2B: Solving Absolute-Value Equations x = 16 or x = –16 Isolate the absolute-value expression. Rewrite the absolute value as a disjunction. Multiply both sides of each equation by 4.

Example 3A: Solving Absolute-Value Inequalities Solve the inequality. Then graph the solution. Rewrite the absolute value as or. |–4q + 2| ≥ 10 –4q + 2 ≥ 10 or –4q + 2 ≤ –10 –4q ≥ 8 or –4q ≤ –12 Divide both sides of each inequality by –4 and reverse the inequality symbols. Subtract 2 from both sides of each inequality. q ≤ –2 or q ≥ 3 –3 –2 –

Solve the compound inequality. Check for Extraneous Solutions |3x +2| = 4x + 5 x = –3 or x =-1 Check It Out! Example 4b

Assignment #7 Pg 46 #12-24 x 3, x3

HW#7 Pg 46 #12-24 x 3, x3

Solve the inequality. Then graph the solution. |0.5r| – 3 ≥ –3 0.5r ≥ 0 or 0.5r ≤ 0 Divide both sides of each inequality by 0.5. Isolate the absolute value as or. r ≤ 0 or r ≥ 0 |0.5r| ≥ 0 Rewrite the absolute value as or. Example 3B: Solving Absolute-Value Inequalities –3 –2 – (–∞, ∞) The solution is all real numbers, R.

The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is an “and” statement: –3 < x < 3. The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is an “or” statement: x 3.

Think: Greator inequalities involving > or ≥ symbols are disjunctions. Think: Less thand inequalities involving < or ≤ symbols are conjunctions. Helpful Hint Note: The symbol ≤ can replace, and the rules still apply.

Solve the compound inequality. –2|x +5| > 10 Divide both sides by –2, and reverse the inequality symbol. Subtract 5 from both sides of each inequality. Rewrite the absolute value as a conjunction. x x 0 Check It Out! Example 4b |x + 5| < –5

Lesson Quiz: Part I Solve. Then graph the solution y – 4 ≤ –6 or 2y >8 –7x < 21 and x + 7 ≤ 6 {y|y ≤ –2 ≤ or y > 4} {x|–3 < x ≤ –1} –4 –3 –2 – Solve each equation. 3. |2v + 5| = 94. |5b| – 7 = 13 2 or –7 + 4

Lesson Quiz: Part II Solve. Then graph the solution |1 – 2x| > 7{x|x 4} |3k| + 11 > 8 R –2|u + 7| ≥ 16ø –4 –3 –2 –

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