Do Now 10418 Take out HW from last

Do Now 10/4/18 Ø Take out HW from last night. l Ø Chapter 2 - Pre. Assessment Copy HW in your planner. Text p. 58, #6 -18 evens, 41 -44 all (11 problems) Text p. 65, #8 -32 multiples of 4 (7 problems) Text p. 71, #4 -28 multiples of 4 (7 problems) Ø Go to Schoology and answer the discussion question. Outside of the math classroom, where have you heard phrases such as “at least” or “no more than”? Give examples. How would you write the phrases mathematically?

Chapter 2 Preview “Solving and Graphing Linear Inequalities” (2. 1) Write & Graph Inequalities (2. 2) Solve Inequalities Using Addition and Subtraction (2. 3) Solve Inequalities Using Multiplication and Division (2. 4) Solve Multi-Step Inequalities Mid-Chapter 2 Quiz (2. 5) Solve Compound Inequalities (2. 6) Solve Absolute Value Inequalities

Learning Goal SWBAT write, graph, and solve linear, compound, and absolute value inequatilies Learning Target Ø SWBAT write and graph and solve inequalities using addition, subtraction, multiplication, and division

Section 2. 1 “Writing and Graphing Inequalities” INEQUALITIES – mathematical sentence formed by placing a <, ≤, >, or ≥ between two expressions. 1 3 * 5 ÷ x > 3 2 1 0 < 8 – y² 11 - a ≤ 121 22 ≥ 14 - z

Writing Equations with Inequalities Symbol Meaning Key phrases = Is equal to The same as < Is less than Fewer than ≤ Is less than or equal At most, no more to than > Is greater than More than ≥ Is greater than or equal to At least, no less than

Ø On a number line, the GRAPH OF AN INEQUALITY is the set of points that represent ALL SOLUTIONS of the inequality. “Less than” and “greater than” are represented with an open circle. Graph x < 8 5 6 “Less than or equal to” and “greater than or equal to” are represented with a closed circle. 8 9 7 8 9 10 Graph x ≥ 11 10 11 12 13 14 11

Graph each inequality 7 ≥ y “Less than or equal to” and “greater than or equal to” are represented with a closed circle. 4 5 6 7 8 9 -1 < h -4 -3 -2 -1 10 “Less than” and “greater than” are represented with an open circle. 0 1 2

Write an inequality represented by the graph. SOLUTION The closed circle means that 8 is not a solution of the inequality. Because the arrow points to the left, all numbers less than 8 are solutions. ANSWER An inequality represented by the graph is x < 8.

Write an inequality represented by the graph. SOLUTION The closed circle means that – 2. 5 is a solution of the inequality. Because the arrow points to the right, all numbers greater than – 2. 5 are solutions. ANSWER An inequality represented by the graph is x > – 2. 5.

Section 2. 2 “Solve Inequalities Using Addition and Subtraction”

Solving an Inequality… Isolate the variable! Get ‘m’ by itself. To get the ‘m’ by itself get rid of “adding 4. ” Do the opposite. “Subtract 4. ” m + 4 < 12 - 4 -4 m < 8 Whatever you do to one side of the Inequality you must do the other side. 5 6 7 8 9 10 11

Solving an Inequality… Isolate the variable! Get ‘n’ by itself. To get the ‘n’ by itself get rid of “subtracting 5. ” Do the opposite. “Add 5. ” n - 5 ≥ 6 + 5 +5 n ≥ 11 Whatever you do to one side of the inequality you must do the other side. 8 9 10 11 12 13 14

Solve x – 5 > -3. 5 Graph your solution x – 5 > – 3. 5 +5 +5 x > 1. 5 Write original inequality. Add 5 to each side. Simplify. ANSWER The solutions are all real numbers greater than 1. 5. Check by substituting a number greater than 1. 5 for x in the original inequality.

Solve a real-world problem LUGGAGE WEIGHTS You are checking a bag at an airport. Bags can weigh no more than 50 pounds. Your bag weighs 16. 8 pounds. Find the possible weights w (in pounds) that you can add to the bag. SOLUTION Write a verbal model. Then write and solve an inequality. 16. 8 + w ≤ 50

Solve a real-world problem 16. 8+ w < 50 16. 8 + w – 16. 8 < 50 – 16. 8 w ≤ 33. 2 ANSWER You can add no more than 33. 2 pounds. Write inequality. Subtract 16. 8 from each side. Simplify.

Section 2. 3 “Solve Inequalities Using Multiplication and Division”

Solving a Multiplication Inequality 12 y < 36 12 12 y< 3 0 1 2 “Undo” multiplication by dividing both sides of the inequality 3 4 5 6

Solving a Division Inequality 5· j ≥ -7 · 5 5 j ≥ -35 -38 -37 -36 -35 -34 “Undo” division by multiplying both sides of the inequality -33 -32

Solve . Graph your solution 7 x > 91 7 7 x > 13 Write original inequality. Divide each side by 7. Simplify. Graph x > 13 10 11 12 13 14 15 16

Solve Inequalities When Multiplying and Dividing by a NEGATIVE” You can multiply or divide both sides of an inequality by the same negative number, but you must REVERSE the direction of the inequality symbol for the statement to be true. -2 y < 16 -2 -2 y > -8 Reverse inequality because of division by a negative number.

Solve . m – 7 < 1. 6 m – 7 > – 7 1. 6 m > – 11. 2 Write original inequality. Multiply each side by – 7. Reverse inequality symbol. Simplify. ANSWER The solutions are all real numbers greater than – 11. 2. Check by substituting a number greater than – 11. 2 in the original inequality.

Solve . Graph your solution. 8 m ≤– 2 8 8 m ≤ – 16 Write original inequality. Multiply each side by 8. Simplify. The solutions are all real numbers less than are equal to – 16. Check by substituting a number less than – 16 in the original inequality.

Solve . – 3 x > 24. – 3 x < 24 – 3 x<– 8 Write original inequality. Divide each side by – 3. Reverse inequality symbol. Simplify.

Solve a real-world problem A student pilot plans to spend 80 hours on flight training to earn a private license. The student has saved $6000 for training. Which inequality can you use to find the possible hourly rates r that the student can afford to pay for training? A 80 r < 6000 B 80 r < – 6000 C 80 r > – 6000 D 80 r > 6000 SOLUTION The total cost of training can be at most the amount of money that the student has saved. Write a verbal model for the situation. Then write an inequality. 80 r The correct answer is B. A < – B 6000 C D

Solve a real-world problem What are the possible hourly rates that the student can afford to pay for training? 80 r ≤ 6000 80 80 r ≤ 75 Write inequality. Divide each side by 80. Simplify. The student can afford to pay at most $75 per hour for training.

24 Homework Text p. 58, #6 -18 evens, 41 -44 all (11 problems) Text p. 65, #8 -32 multiples of 4 (7 problems) Text p. 71, #4 -28 multiples of 4 (7 problems)

Homework Text p. 48, #1 -9 all

Homework Text p. 48, #1 -9 all