1. Chapter 2 Linear Systems in Two Variables and Inequalities with Applications Section 2.3

2. Section 2.3 Linear Inequalities in Two Variables Solution of a Linear Inequality in Two Variables Graphical Solutions and Applications

3. Linear Inequalities A linear inequality in two variables can be written in one of the following forms: a x + b y > c a x + b y < c a x + b y ≥ c a x + b y ≤ c where a, b, and c are real numbers, and a and b are not both zero. An ordered pair (x, y) will be a solution of a linear inequality in two variables if it satisfies the inequality.

4. Determine if the point (3, –5) is a solution of –3x + 2y < 0. –3(3) + 2(–5) < 0 ? –9 – 10 < 0 ? –19 < 0 ? True The ordered pair (3, –5) is a solution of the given linear inequality in two variables because it satisfies the inequality.

5. Graphical Solution (1) Replace the inequality symbol with an equal sign and graph the line, which we call the “boundary line.”  If the inequality symbol is ≥ or ≤, the line will be part of the solution set. Draw a solid line to include the line in the solution.  If the inequality symbol is > or <, the line will not be part of the solution set. Draw a dashed line to exclude the line. ( 2) Select any point (“test point”) that does not lie on the line to determine the region whose points will satisfy the inequality.  If the point (0, 0) is not on the line, using it as a test point is convenient for calculations. ( 3) If the test point satisfies the inequality, shade the region that contains the point. If it does not satisfy the inequality, then shade the opposite region.  The shaded region will contain all the points that satisfy the inequality.

6. Find the solution set of the inequality 2x + y ≥ 6. Step 1: Graph the boundary line. We can use the x- and y-intercepts to sketch the graph. x-intercept: (3, 0) y-intercept: (0, 6) Since the inequality symbol is ≥, we will use a solid line. (continued on the next slide)

7. (Contd.) Find the solution set of the inequality 2x + y ≥ 6. Step 2: Since (0, 0) is not on the line, we will use it as our test point. 2(0) + (0) ≥ 6 0 ≥ 6 False (continued on the next slide)

8. (Contd.) Find the solution set of the inequality 2x + y ≥ 6. Step 3: Our test point (0, 0) does not satisfy the inequality, therefore we shade the opposite region. The solution set to 2x + y ≥ 6 will be all the points on the coordinate plane that are on or above the corresponding line.

9. Find the solution set of the inequality y < 4.5. (1) Graph the boundary line: y = 4.5 Since the inequality symbol is <, we will use a dashed line. ( 2) Since (0, 0) is not on the line, we will use it as our test point. 0 < 4.5 True (continued on the next slide)

10. (Contd.) Find the solution set of the inequality y < 4.5. ( 3) Our test point satisfies the inequality, therefore we shade the region containing the point (0, 0). The solution set to y < 4.5 will be all the points on the coordinate plane that are below the corresponding line.

11. A group of friends can spend no more than $30 for candy bars and sodas at a concert. Their favorite candy costs $4 and a medium soda is $6.25. a. Write a linear inequality that represents how many of each snack the group can buy for $30 or less. cost of candy bars + cost of sodas ≤ $30 Let x equal the number of candy bars and y represent the number of sodas. Our inequality is: 4x + 6.25y ≤ 30

12. (Contd.) A group of friends can spend no more than $30 for candy bars and sodas at a concert. Their favorite candy costs $4 and a medium soda is $6.25. b. Graph your inequality and shade the solution set. The inequality symbol in 4x + 6.25y ≤ 30 implies the use of a solid line. Using (0, 0) as the test point: 4(0) + 6.25(0) ≤ 30 True The price for the candy bars and the sodas cannot be negative, thus we can restrict our values to positive numbers.

13. Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 2.3.