Lesson 5 MI/Vocab absolute value Solve absolute value equations. Solve absolute value inequalities.

Lesson 5 Ex1 A. WEATHER The average January temperature in a northern Canadian city is 1 degree Fahrenheit. The actual January temperature for that city may be about 5 degrees Fahrenheit warmer or colder. Solve |t – 1| = 5 to find the range of temperatures. Method 1 Graphing |t – 1| = 5 means that the distance between t and 1 is 5 units. To find t on the number line, start at 1 and move 5 units in either direction. The distance from 1 to 6 is 5 units. The distance from 1 to –4 is 5 units. The solution set is {–4, 6}. Solve an Absolute Value Equation

Lesson 5 Ex1 Method 2 Compound Sentence Write |t –1| = 5 as t – 1 = 5 or t – 1 = –5. Answer: The solution set is {–4, 6}. The range of temperatures is –4°F to 6°F. Solve an Absolute Value Equation Case 1Case 2 t – 1 = 5t – 1 = –5 t – = 5 + 1Add 1 to each side. t – = –5 + 1Add 1 to each side. t = 6Simplify. t = –4 Simplify.

A.A B.B C.C D.D Lesson 5 CYP1 A.{–60, 60} B.{0, 60} C.{–45, 45} D.{30, 60} WEATHER The average temperature for Columbus on Tuesday was 45ºF. The actual temperature for anytime during the day may have actually varied from the average temperature by 15ºF. Solve to find the range of temperatures.

Lesson 5 Ex1 B. Solve |x + 2| = –1. Answer: |x + 2| = –1 means that the distance between x and –2 is –1. Since distance cannot be negative, the solution is the empty set, Ø. Solve an Absolute Value Equation

Lesson 5 CYP1 Answer: Ø B. Solve |x – 3| = –5.

Lesson 5 Ex2 Write an Absolute Value Equation Write an open sentence involving the absolute value for the graph. Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1.

Lesson 5 Ex2 Write an Absolute Value Equation The distance from 1 to –4 is 5 units. The distance from 1 to 6 is 5 units. Answer: |x –1| = 5 So, an equation is |x –1| = 5.

Lesson 5 CYP2 1.A 2.B 3.C 4.D A.|x – 2| = 4 B.|x + 2| = 4 C.|x – 4| = 2 D.|x + 4| = 2 Write an equation involving the absolute value for the graph.

Lesson 5 Ex3 Solve an Absolute Value Equation (<) A. Solve |s – 3| ≤ 12. Then graph the solution set. Write |s – 3| ≤ 12 as s – 3 ≤ 12 and s – 3 ≥ –12. Answer: The solution set is {s | –9 ≤ s ≤ 15}. Case 1Case 2 s – 3 ≤ 12 Original inequality s –3 ≥ –12 s – ≤ Add 3 to each side. s – ≥ – s ≤ 15Simplify.s ≥ –9

Lesson 5 Ex3 Solve an Absolute Value Equation (<) B. Solve |x + 6| < –8. Since |x + 6| cannot be negative, |x + 6| cannot be less than –8. So, the solution is the empty set Ø. Answer: Ø

1.A 2.B 3.C 4.D Lesson 5 CYP3 A. Solve |p + 4| < 6. Then graph the solution set. A.{p | p < 2} B.{p | p > –10} C.{p | –10 < p < 2} D.{p | –2 < p < 10}

Lesson 5 CYP3 B. Solve |p – 5| < –2. Then graph the solution set. Answer: Ø

Lesson 5 Ex4 A. Solve |3y –3| > 9. Then graph the solution set. Solve an Absolute Value Equation (>) Original inequality Add 3 to each side. Simplify. Divide each side by 3. Simplify.

Lesson 5 Ex4 Answer: The solution set is {y | y 4}. Solve an Absolute Value Equation (>)

Lesson 5 Ex4 B. Solve |2x + 7| ≥ –11. Solve an Absolute Value Equation (>) Answer:Since |2x + 7| is always greater than or equal to 0, the solution set is {x | x is a real number}.

A.A B.B C.C D.D Lesson 5 CYP4 A. Solve |2m – 2| > 6. Then graph the solution set. A.{m | m > –2 or m < 4}. B.{m | m > –2 or m > 4}. C.{m | –2 < m < 4}. D.{m | m 4}.

Lesson 5 CYP4 Answer: {x | x is a real number} B. Solve |5x – 1| ≥ –2. Then graph the solution set.

Lesson 5 KC2 Animation: Absolute Value Equations and Inequalities