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

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

2. Section 2.3Linear 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: • ax + by > c • ax + by < c • ax + by ≥ c • ax + by ≤ c • where a, b, and c are real numbers, and aandbare 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 • 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 lineto 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 lineto 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. • 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.