ALGEBRA 1 Lesson 6-6 Warm-Up

ALGEBRA 1 “Systems of Linear Inequalities” (6-6) What is a “system of linear inequalities”? What is a “solution of system of linear inequalities”? system of linear inequalities : two or more linear inequalities on a graph Solution of a System of Linear Inequalities : a solution that makes all of the inequalities in a system true In the graph below, the lavendar-shaded area include all of the solutions of the system represented by x ≥ 3 and y  -2 (in other words, it include all of the solutions that make both inequalities true).

ALGEBRA 1 “Systems of Linear Inequalities” (6-6) What is a “solution set”? Solution Set: the graph that shows the “set” of all solutions to the system Example: What is the solution set of y  2x – 5 and 3x + 4y  12. The points where the two inequalities overlap shaded in lavender is the solution set to the system.

ALGEBRA 1 “Systems of Linear Inequalities” (6-6) How can you describe the four quadrants of a coordinate grid with systems of linear inequalities”?

ALGEBRA 1 Solve by graphing.y < –x + 3 –2x + 4y 0 Check:The point (–1, 1) is in the region graphed by both inequalities. See if (–1, 1) satisfies both inequalities. > – Graph y = –x + 3 and –2x + 4y = 0. Systems of Linear Inequalities LESSON 6-6 Additional Examples –2x + 4y = 0 +2x 4y = 2x __  1y = x

ALGEBRA 1 (continued) The coordinates of the points in the [light blue] region where the graphs of the two inequalities overlap are solutions of the system. y < –x + 3 –2x + 4y 0 1 < –(–1) + 3 Substitute (–1, 1) for (x, y). –2(–1) + 4(1) 0 1 < > – > – > – Systems of Linear Inequalities LESSON 6-6 Additional Examples

ALGEBRA 1 Write a system of inequalities for each shaded region below. System for the [light blue] region: y < – x + 2 y < [red] region boundary: y = – x + 2 and y = The region lies between the boundary lines, so the inequality is y < – x + 2 and y  [blue] region Boundary lines: y = 4 and y = – x + 2 The region lies below the boundary line, so the inequality is y  – x + 2 and y < 4. Systems of Linear Inequalities LESSON 6-6 Additional Examples

ALGEBRA 1 You need to make a fence for a dog run. The length of the run can be no more than 60 ft, and you have 136 feet of fencing that you can use. What are the possible dimensions of the dog run? Define:Let = length of the dog run. Let w = width of the dog run. Words :Theis no 60 ft.Theis no136 ft. lengthmore thanperimetermore than Equation: w 136 < – < – Solve by graphing w 136 < – < – Systems of Linear Inequalities LESSON 6-6 Additional Examples

ALGEBRA 1 (continued) The solutions are the coordinates of the points that lie in the region shaded light-blue and on the lines = 60 and 2 + 2w = m = 0: b = 60 Shade below = 60. < – Test (0, 0). 2(0) + 2(0) So shade below 2 + 2w = 136 < – < – 2 + 2w 136 Graph the intercepts (68, 0) and (0, 68). < – Systems of Linear Inequalities LESSON 6-6 Additional Examples

ALGEBRA 1 Suppose you have two jobs, babysitting, which pays $5 per hour and sacking groceries, which pays $6 per hour. You can work no more than 20 hours each week, but you need to earn at least $90 per week. How many hours can you work at each job? Define: Let b = hours of babysitting. Let s = hours of sacking groceries. Words:The numberis less20.The amount is at90. of hoursthan orearnedleast workedequal to Equation: b + s 205 b + 6 s 90 > – < – Systems of Linear Inequalities LESSON 6-6 Additional Examples

ALGEBRA 1 (continued) The solutions are all the coordinates of the points that are nonnegative integers that lie in the region shaded light-blue and on the lines b + s = 20 and 5b + 6s = 90. Solve by graphing.b + s 20 5b + 6s 90 > – < – Systems of Linear Inequalities LESSON 6-6 Additional Examples

ALGEBRA 1 Solve each system by graphing. 1. x 02. 2x + 3y > 123. y x – 3 y < 3 2x + 2y < 12 2x – 3y –9 > – 2323 > – > – Systems of Linear Inequalities LESSON 6-6 Lesson Quiz

ALGEBRA 1 4. Write a system of inequalities for the following graph. y < x + 3 y > – x – Systems of Linear Inequalities LESSON 6-6 Lesson Quiz