2. Lesson Menu Five-Minute Check (over Lesson 5–2) CCSS Then/Now Example 1:Real-World Example: Solve a Multi-Step Inequality Example 2:Inequality Involving a Negative Coefficient Example 3:Write and Solve an Inequality Example 4:Distributive Property Example 5:Empty Set and All Reals

3. Over Lesson 5–2 5-Minute Check 1 A.{a | a > 64} B.{a | a < 64} C.{a | a > 4} D.{a | a < 4}

4. Over Lesson 5–2 5-Minute Check 2 A.{p | p > 28 } B.{p | p < 28 } C. D.{p | p > –28 }

5. Over Lesson 5–2 5-Minute Check 3 A.{v | v ≥ 99} B.{v | v ≤ 12} C.{v | v ≥ 12} D.{v | v ≥ –12} Solve –9v ≥ –108.

6. Over Lesson 5–2 A.{c | c ≤ –5} B.{c | c ≤ –2} C. D. 5-Minute Check 4

7. Over Lesson 5–2 5-Minute Check 5 Which inequality represents one half of Dan’s savings is less than $60.00? A. B. C. D.

8. Over Lesson 5–2 5-Minute Check 6 A.1.59 – c > 20; 22 B.c + 1.59 < 20; 18 C.1.59c ≥ 20; 12 D.1.59c ≤ 20; 12 Marta wants to purchase charms for her necklace. Each charm costs $1.59. She wants to spend no more than $20 for the charms. Which inequality represents this situation? How many charms can Marta purchase?

9. CCSS Content Standards A.CED.1 Create equations and inequalities in one variable and use them to solve problems. A.REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. Mathematical Practices 7 Look for and make use of structure. Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

10. Then/Now You solved multi-step equations. Solve linear inequalities involving more than one operation. Solve linear inequalities involving the Distributive Property.

11. Multi-step inequalities can be solved by undoing the operations in the same way you would solve a multi-step equation. Example: 11y – 13 = 42 + 13 + 13 11y = 55 11 11 y = 5 However, instead of = sign you will have an inequality sign

12. Write and solve an inequality to find the sales Mrs. Jones needs if she earns a monthly salary of $2000 plus a 10% commission on her sales. Her goal is to make at least $4000 per month. What sales does she need to meet the goal? Base salary + (commission x sales) ≥ 4000 2000 + 0.10x ≥ 4000 -2000 -2000 0.10x ≥ 2000 0.10 0.10 x ≥ 20000

13. Example 1 Solve a Multi-Step Inequality FAXES Adriana has a budget of $115 for faxes. The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can Adriana fax and stay within her budget? Use the inequality 25 + 0.08p ≤ 115. Original inequality Subtract 25 from each side. Divide each side by 0.08. Simplify. Answer: Adriana can send at most 1125 faxes.

14. The Print Shop advertises a special to print 400 flyers for less than the competition. The price includes a $3.50 set-up fee. If the competition charges $35.50, what does the Print Shop charge for each flyer? 400n + 3.50 < 35.50 - 3.50 - 3.50 400n < 32 400 400 n < $0.08

15. Example 1 A.50 pictures B.55 pictures C.60 pictures D.70 pictures Rob has a budget of $425 for senior pictures. The cost for a basic package and sitting fee is $200. He wants to buy extra wallet-size pictures for his friends that cost $4.50 each. How many wallet-size pictures can he order and stay within his budget? Use the inequality 200 + 4.5p ≤ 425.

16. When multiplying or dividing by a negative the direction of the inequality changes. This holds true for multi-step inequalities. -11y – 13 > 42 + 13 +13 -11y > 55 -11 -11 y < -5 The solution set is { y│y < -5}

17. Example 2 Inequality Involving a Negative Coefficient Solve 13 – 11d ≥ 79. Answer: The solution set is {d | d ≤ –6}. 13 – 11d≥79Original inequality 13 – 11d – 13≥79 – 13Subtract 13 from each side. –11d≥66Simplify. Divide each side by –11 and change ≥ to ≤. d≤–6Simplify.

18. 23 ≥ 10 – 2w -10 -10 13 ≥ -2w -2 -2 -13/2 ≤ w The solution set is {w│w ≥ -13/2}

19. 43 > -4y + 11 -11 - 11 32 > -4y -4 -4 -8 < y The solution set is {y│y > -8}

20. Example 2 A.{y | y < –1} B.{y | y > 1} C.{y | y > –1} D.{y | y < 1} Solve –8y + 3 > –5.

21. You can translate sentences into multi-step inequalities and then solve them using the Properties of Inequalities

22. Example 3 Write and Solve an Inequality Define a variable, write an inequality, and solve the problem below. Four times a number plus twelve is less than the number minus three. a number minus three. is less thantwelveplus Four times a number n – 3<12+4n4n

23. Example 3 Write and Solve an Inequality 4n + 12<n – 3Original inequality Answer: The solution set is {n | n < –5}. n<–5Simplify. Divide each side by 3. 4n + 12 – n <n – 3 – nSubtract n from each side. 3n + 12<–3Simplify. 3n + 12 – 12<–3 – 12 Subtract 12 from each side. 3n<–15Simplify.

24. Five minus 6 times a number is more than four times the number plus 45 Five minus six times is more four times plus forty-five a number a number 5 – 6n > 4n + 45 - 4n -4n 5 – 10n > 45 -5 -5 -10n > 40 -10 -10 n < -4

25. Two more than half a number is greater than twenty-seven ½ n + 2 > 27 -2 -2 ½ n > 25 (2)(1/2 n) > 25(2) n > 50 {n│n > 50}

26. Example 3 Write an inequality for the sentence below. Then solve the inequality. 6 times a number is greater than 4 times the number minus 2. A.6n > 4n – 2; {n | n > –1} B.6n < 4n – 2; {n | n < –1} C.6n > 4n + 2; {n | n > 1} D.6n > 2 – 4n;

27. When solving inequalities that contain grouping symbols, first use the Distributive Property to remove the grouping symbols. Next use the order of operations to simplify the resulting inequality.

28. Example 4 Distributive Property Solve 6c + 3(2 – c) ≥ –2c + 1. Answer: The solution set is {c | c ≥ –1}. 6c + 3(2 – c)≥ –2c + 1 Original inequality 6c + 6 – 3c ≥–2c + 1 Distributive Property 3c + 6≥–2c + 1 Combine like terms. 3c + 6 + 2c ≥–2c + 1 + 2cAdd 2c to each side. 5c + 6≥1 Simplify. 5c + 6 – 6≥1 – 6 Subtract 6 from each side. 5c ≥–5 Simplify. c≥–1 Divide each side by 5.

29. 4(3t – 5) + 7 ≥ 8t + 3 12t – 20 + 7 ≥ 8t + 3 12t – 13 ≥ 8t + 3 -8t -8t 4t – 13 ≥ 3 +13 +13 4t ≥ 16 4 4 t ≥ 4

30. Solve each inequality. Graph the solution on a number line. 1)6(5z – 3) ≤ 36z 30z – 18 ≤ 36z -30z -30z -18 ≤ 6z 6 6 -3 ≤ z The solution set is {z│z ≥ -3}

31. 2) 2(h + 6) > -3(8 – h) 2h + 12 > -24 + 3h -3h - 3h -h + 12 > -24 - 12 - 12 -h > -36 Remember the variable can not be negative -1h > -36 h < 36 {h│h < 36}

32. Example 4 Solve 3p – 2(p – 4) < p – (2 – 3p). A. B. C. D. p | p

33. If solving an inequality results in a statement that is always true, the solution set is the set of all real numbers. This solution set is written as {x│x is a real number} If solving an inequality results in a statement that is never true, the solution set is the empty set, which is written as the symbol Ø. The empty set has no members.

34. Example 5 Empty Set and All Reals A. Solve –7(s + 4) + 11s ≥ 8s – 2(2s + 1). –7(s + 4) + 11s≥8s – 2(2s + 1)Original inequality –7s – 28 + 11s ≥8s – 4s – 2Distributive Property 4s – 28 ≥4s – 2Combine like terms. 4s – 28 – 4s≥4s – 2 – 4s Subtract 4s from each side. – 28≥– 2 Simplify. Answer: Since the inequality results in a false statement, the solution set is the empty set, Ø.

35. Example 5 Empty Set and All Reals B. Solve 2(4r + 3)  22 + 8(r – 2). Answer: All values of r make the inequality true. All real numbers are the solution. {r | r is a real number.} 2(4r + 3) ≤22 + 8(r – 2)Original inequality 8r + 6 ≤22 + 8r – 16Distributive Property 8r + 6 ≤6 + 8rSimplify. 8r + 6 – 8r ≤6 + 8r – 8rSubtract 8r from each side. 6 ≤6Simplify.

36. 9t – 5(t – 5) ≤ 4(t – 3) 9t – 5t + 25 ≤ 4t – 12 4t + 25 ≤ 4t – 12 -4t -4t 25 ≤ -12 Since the inequality results in a false statement, the solution set is the empty set Ø

37. 3(4m + 6) ≤ 42 + 6(2m – 4) 12m + 18 ≤ 42 + 12m – 24 12m + 18 ≤ 12m + 18 -12m -12m 18 ≤ 18 All values of m makes the inequality true. All real numbers is the solution. {m│m is a real numbers}

38. Solve each inequality 1)18 – 3(8c + 4) ≥ -6(4c – 1) {c│c is a real number} 2) 46 ≤ 8m -4(2m + 5) Ø 3) 3 – 8x ≥ 9 + 2(1 – 4x) Ø 4) 3(2 – b) < 10 – 3(b – 6) {b│b is a real number}

39. Example 5 A. Solve 8a + 5 ≤ 6a + 3(a + 4) – (a + 7). A.{a | a ≤ 3} B.{a | a ≤ 0} C.{a | a is a real number.} D.

40. Example 5 B. Solve 4r – 2(3 + r) < 7r – (8 + 5r). A.{r | r > 0} B.{r | r < –1} C.{r | r is a real number.} D.

41. End of the Lesson