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Splash Screen. Lesson Menu Five-Minute Check (over Lesson 1–4) CCSS Then/Now New Vocabulary Key Concept: Addition / Subtraction Property of Inequality.

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Presentation on theme: "Splash Screen. Lesson Menu Five-Minute Check (over Lesson 1–4) CCSS Then/Now New Vocabulary Key Concept: Addition / Subtraction Property of Inequality."— Presentation transcript:

1 Splash Screen

2 Lesson Menu 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

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

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

5 Over Lesson 1–4 5-Minute Check 2 A.5 B.7 C.10 D.20 Evaluate the expression |2x + y| if x = 1.5 and y = 4.

6 Over Lesson 1–4 5-Minute Check 2 A.5 B.7 C.10 D.20 Evaluate the expression |2x + y| if x = 1.5 and y = 4.

7 Over Lesson 1–4 5-Minute Check 3 A.5 B.10 C.20 D.40 Evaluate the expression 5|xy – w| if w = –2, x = 1.5, and y = 4.

8 Over Lesson 1–4 5-Minute Check 3 A.5 B.10 C.20 D.40 Evaluate the expression 5|xy – w| if w = –2, x = 1.5, and y = 4.

9 Over Lesson 1–4 5-Minute Check 4 A.{–41, 1} B.{–41, –1} C.{–1, 1} D.{1, 2} Solve the equation |b + 20| = 21.

10 Over Lesson 1–4 5-Minute Check 4 A.{–41, 1} B.{–41, –1} C.{–1, 1} D.{1, 2} Solve the equation |b + 20| = 21.

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

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

13 Over Lesson 1–4 5-Minute Check 6 What is the solution to the equation 2|3x – 1| – 1 = –5? A. B. C. D.

14 Over Lesson 1–4 5-Minute Check 6 What is the solution to the equation 2|3x – 1| – 1 = –5? A. B. C. D.

15 CCSS 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.

16 Then/Now You solved equations involving absolute values. Solve one-step inequalities. Solve multi-step inequalities.

17 Vocabulary set-builder notation

18 Concept

19 Example 1 Solve an Inequality Using Addition or Subtraction Solve 4y – 3 < 5y + 2. Graph the solution set on a number line. 4y – 3<5y + 2Original inequality 4y – 3 – 4y<5y + 2 – 4ySubtract 4y from each side. –3<y + 2Simplify. –3 – 2<y + 2 – 2Subtract 2 from each side. –5<ySimplify. y>–5Rewrite with y first.

20 Example 1 Answer: Solve an Inequality Using Addition or Subtraction

21 Example 1 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. Solve an Inequality Using Addition or Subtraction

22 Example 1 Which graph represents the solution to 6x – 2 < 5x + 7? A. B. C. D.

23 Example 1 Which graph represents the solution to 6x – 2 < 5x + 7? A. B. C. D.

24 Concept

25 Example 2 Solve an Inequality Using Multiplication or Division Solve 12  –0.3p. 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.

26 Example 2 Solve an Inequality Using Multiplication or Division Answer:

27 Example 2 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.

28 Example 2 What is the solution to –3x  21? A.{x | x  –7} B.{x | x  –7} C.{x | x  7} D.{x | x  7}

29 Example 2 What is the solution to –3x  21? A.{x | x  –7} B.{x | x  –7} C.{x | x  7} D.{x | x  7}

30 Example 3 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.

31 Example 3 Solve Multi-Step Inequalities

32 Example 3 Solve Multi-Step Inequalities

33 Example 3 A. B. C. D.

34 Example 3 A. B. C. D.

35 Example 4 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? UnderstandLet g = the number of gallons of gasoline that Javier buys. The total cost of the gasoline is 2.89g. The cost of the pretzels and juice plus the total cost of the gasoline must be less than or equal to $15.00.

36 Example 4 Write and Solve an Inequality Original inequality Subtract 1.59 from each side. Simplify. The cost of pretzels & juice plus the cost of gasoline is less than or equal to $15.00. 1.59+2.89g  15.00 PlanWrite an inequality. Solve

37 Example 4 Write and Solve an Inequality Divide each side by 2.89. Simplify. Answer:

38 Example 4 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. CheckSince is actually greater than 4.6, Javier will have enough money if he gets no more than 4.6 gallons of gasoline.

39 Example 4 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 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?

40 Example 4 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 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?

41 End of the Lesson


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