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Five-Minute Check (over Lesson 1– 4) CCSS Then/Now New Vocabulary Key Concept: Addition / Five-Minute Check (over Lesson 1– 4) CCSS Then/Now New Vocabulary Key Concept: Addition / Subtraction Property of Inequality Example 1: Solve an Inequality Using Addition or Subtraction Key Concept: Multiplication / Division Property of Inequality Example 2: Solve an Inequality Using Multiplication or Division Example 3: Solve Multi-Step Inequalities Example 4: Write and Solve an Inequality

Over Lesson 1– 4 Evaluate the expression |4 w + 3| if w = Over Lesson 1– 4 Evaluate the expression |4 w + 3| if w = – 2. A. 5 B. 7 C. 11 D. 12

Over Lesson 1– 4 Evaluate the expression |4 w + 3| if w = Over Lesson 1– 4 Evaluate the expression |4 w + 3| if w = – 2. A. 5 B. 7 C. 11 D. 12

Over Lesson 1– 4 Evaluate the expression |2 x + y| if x = Over Lesson 1– 4 Evaluate the expression |2 x + y| if x = 1. 5 and y = 4. A. 5 B. 7 C. 10 D. 20

Over Lesson 1– 4 Evaluate the expression |2 x + y| if x = Over Lesson 1– 4 Evaluate the expression |2 x + y| if x = 1. 5 and y = 4. A. 5 B. 7 C. 10 D. 20

Over Lesson 1– 4 Evaluate the expression 5|xy – w| if w = – Over Lesson 1– 4 Evaluate the expression 5|xy – w| if w = – 2, x = 1. 5, and y = 4. A. 5 B. 10 C. 20 D. 40

Over Lesson 1– 4 Evaluate the expression 5|xy – w| if w = – Over Lesson 1– 4 Evaluate the expression 5|xy – w| if w = – 2, x = 1. 5, and y = 4. A. 5 B. 10 C. 20 D. 40

Over Lesson 1– 4 Solve the equation |b + 20| = 21. A. {– Over Lesson 1– 4 Solve the equation |b + 20| = 21. A. {– 41, 1} B. {– 41, – 1} C. {– 1, 1} D. {1, 2}

Over Lesson 1– 4 Solve the equation |b + 20| = 21. A. {– Over Lesson 1– 4 Solve the equation |b + 20| = 21. A. {– 41, 1} B. {– 41, – 1} C. {– 1, 1} D. {1, 2}

Over Lesson 1– 4 Solve the equation – 4|a + 5| = – 8. Over Lesson 1– 4 Solve the equation – 4|a + 5| = – 8. A. {7, 3} B. {2, 3} C. {– 7, – 3} D. {– 2, 3}

Over Lesson 1– 4 Solve the equation – 4|a + 5| = – 8. Over Lesson 1– 4 Solve the equation – 4|a + 5| = – 8. A. {7, 3} B. {2, 3} C. {– 7, – 3} D. {– 2, 3}

Over Lesson 1– 4 What is the solution to the equation 2|3 x – Over Lesson 1– 4 What is the solution to the equation 2|3 x – 1| – 1 = – 5? A. B. C. D.

Over Lesson 1– 4 What is the solution to the equation 2|3 x – Over Lesson 1– 4 What is the solution to the equation 2|3 x – 1| – 1 = – 5? A. B. C. D.

Content Standards A. CED. 1 Create equations and inequalities in one variable and use Content Standards A. CED. 1 Create equations and inequalities in one variable and use them to solve problems. A. CED. 3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. Mathematical Practices 4 Model with mathematics.

You solved equations involving absolute values. • Solve one-step inequalities. • Solve multi-step inequalities. You solved equations involving absolute values. • Solve one-step inequalities. • Solve multi-step inequalities.

 • set-builder notation • set-builder notation

Solve an Inequality Using Addition or Subtraction Solve 4 y – 3 < 5 Solve an Inequality Using Addition or Subtraction Solve 4 y – 3 < 5 y + 2. Graph the solution set on a number line. 4 y – 3 < 5 y + 2 4 y – 3 – 4 y < 5 y + 2 – 4 y – 3 < y + 2 – 3 – 2 < y + 2 – 5 < y y > – 5 Original inequality Subtract 4 y from each side. Simplify. Subtract 2 from each side. Simplify. Rewrite with y first.

Solve an Inequality Using Addition or Subtraction Answer: Solve an Inequality Using Addition or Subtraction Answer:

Solve an Inequality Using Addition or Subtraction Answer: Any real number greater than – Solve an Inequality Using Addition or Subtraction Answer: Any real number greater than – 5 is a solution of this inequality. A circle means that this point is not included in the solution set.

Which graph represents the solution to 6 x – 2 < 5 x + Which graph represents the solution to 6 x – 2 < 5 x + 7? A. B. C. D.

Which graph represents the solution to 6 x – 2 < 5 x + Which graph represents the solution to 6 x – 2 < 5 x + 7? A. B. C. D.

Solve an Inequality Using Multiplication or Division Solve 12 – 0. 3 p. Graph Solve an Inequality Using Multiplication or Division Solve 12 – 0. 3 p. Graph the solution set on a number line. Original inequality Divide each side by – 0. 3, reversing the inequality symbol. Simplify. Rewrite with p first.

Solve an Inequality Using Multiplication or Division Answer: Solve an Inequality Using Multiplication or Division Answer:

Solve an Inequality Using Multiplication or Division Answer: The solution set is p | Solve an Inequality Using Multiplication or Division Answer: The solution set is p | p – 40. A dot means that this point is included in the solution set.

What is the solution to – 3 x 21? A. {x | x – What is the solution to – 3 x 21? A. {x | x – 7} B. {x | x – 7} C. {x | x 7} D. {x | x 7}

What is the solution to – 3 x 21? A. {x | x – What is the solution to – 3 x 21? A. {x | x – 7} B. {x | x – 7} C. {x | x 7} D. {x | x 7}

Solve Multi-Step Inequalities Original inequality Multiply each side by 2. Add –x to each Solve Multi-Step Inequalities Original inequality Multiply each side by 2. Add –x to each side. Divide each side by – 3, reversing the inequality symbol.

Solve Multi-Step Inequalities Solve Multi-Step Inequalities

Solve Multi-Step Inequalities Solve Multi-Step Inequalities

A. B. C. D. A. B. C. D.

A. B. C. D. A. B. C. D.

Write and Solve an Inequality CONSUMER COSTS Javier has at most $15. 00 to Write and Solve an Inequality CONSUMER COSTS Javier has at most $15. 00 to spend today. He buys a bag of pretzels and a bottle of juice for $1. 59. If gasoline at this store costs $2. 89 per gallon, how many gallons of gasoline, to the nearest tenth of a gallon, can Javier buy for his car? Understand Let g = the number of gallons of gasoline that Javier buys. The total cost of the gasoline is 2. 89 g. The cost of the pretzels and juice plus the total cost of the gasoline must be less than or equal to $15. 00.

Write and Solve an Inequality Plan Write an inequality. The cost of pretzels & Write and Solve an Inequality Plan Write an inequality. The cost of pretzels & juice plus 1. 59 + the cost of gasoline 2. 89 g is less than or equal to $15. 00 Solve Original inequality Subtract 1. 59 from each side. Simplify.

Write and Solve an Inequality Divide each side by 2. 89. Simplify. Answer: Write and Solve an Inequality Divide each side by 2. 89. Simplify. Answer:

Write and Solve an Inequality Divide each side by 2. 89. Simplify. Answer: Javier Write and Solve an Inequality Divide each side by 2. 89. Simplify. Answer: Javier can buy up to 4. 6 gallons of gasoline for his car. Check Since is actually greater than 4. 6, Javier will have enough money if he gets no more than 4. 6 gallons of gasoline.

RENTAL COSTS Jeb wants to rent a car for his vacation. Value Cars rents RENTAL COSTS Jeb wants to rent a car for his vacation. Value Cars rents cars for $25 per day plus $0. 25 per mile. How far can he drive for one day if he wants to spend no more that $200 on car rental? A. up to 700 miles B. up to 800 miles C. more than 700 miles D. more than 800 miles

RENTAL COSTS Jeb wants to rent a car for his vacation. Value Cars rents RENTAL COSTS Jeb wants to rent a car for his vacation. Value Cars rents cars for $25 per day plus $0. 25 per mile. How far can he drive for one day if he wants to spend no more that $200 on car rental? A. up to 700 miles B. up to 800 miles C. more than 700 miles D. more than 800 miles