Download presentation

Presentation is loading. Please wait.

Published byTyrone Dominic Tyler Modified over 4 years ago

1
A linear inequality is similar to a linear equation, but the equal sign is replaced with an inequality symbol. A solution of a linear inequality is any ordered pair that makes the inequality true.

2
**Additional Example 1A: Identifying Solutions of Inequalities**

Tell whether the ordered pair is a solution of the inequality. (–2, 4); y < 2x + 1 y < 2x + 1 4 2(–2) + 1 4 –4 + 1 4 –3 < Substitute (–2, 4) for (x, y). (–2, 4) is not a solution.

3
**Additional Example 1B: Identifying Solutions of Inequalities**

Tell whether the ordered pair is a solution of the inequality. (3, 1); y > x – 4 y > x − 4 – 4 1 – 1 > Substitute (3, 1) for (x, y). (3, 1) is a solution.

4
**A linear inequality describes a region of a coordinate plane called a half-plane.**

All points in the half-plane region are solutions of the linear inequality. The boundary line of the region is the graph of the related equation.

6
**Graphing Linear Inequalities**

Step 1 Solve the inequality for y (slope-intercept form). Step 2 Graph the boundary line. Use a solid line for ≤ or ≥. Use a dashed line for < or >. Step 3 Shade the half-plane above the line for y > or ≥. Shade the half-plane below the line for y < or y ≤. Check your answer.

7
**Example 2A: Graphing Linear Inequalities in Two Variables**

Graph the solutions of the linear inequality. Check your answer. y 2x – 3 Step 1 The inequality is already solved for y. Step 2 Graph the boundary line y = 2x – 3. Use a solid line for . Step 3 The inequality is , so shade below the line.

8
**Additional Example 2A Continued**

Graph the solutions of the linear inequality. Check your answer. y 2x – 3 Substitute (0, 0) for (x, y) because it is not on the boundary line. Check y 2x – 3 (0) – 3 0 –3 A false statement means that the half-plane containing (0, 0) should NOT be shaded. (0, 0) is not one of the solutions, so the graph is shaded correctly.

9
Use the “test point” method shown in Example 2 to check your answers to linear inequalities. The point (0, 0) is a good test point to use if it does not lie on the boundary line. However, this method will check only that your shading is correct. It will not check the boundary line. Helpful Hint

10
**Example 2B: Graphing Linear Inequalities in Two Variables**

Graph the solutions of the linear inequality. Check your answer. 5x + 2y > –8 Step 1 Solve the inequality for y. 5x + 2y > –8 –5x –5x 2y > –5x – 8 y > x – 4 Step 2 Graph the boundary line Use a dashed line for >. y = x – 4.

11
**Additional Example 2B Continued**

Graph the solutions of the linear inequality. Check your answer. 5x + 2y > –8 Step 3 The inequality is >, so shade above the line.

12
**Additional Example 2B Continued**

Graph the solutions of the linear inequality. Check your answer. 5x + 2y > –8 Substitute ( 0, 0) for (x, y) because it is not on the boundary line. Check y > x – 4 (0) – 4 0 –4 > The point (0, 0) satisfies the inequality, so the graph is correctly shaded.

13
**Example 2C: Graphing Linear Inequalities in two Variables**

Graph the solutions of the linear inequality. Check your answer. 4x – y + 2 ≤ 0 Step 1 Solve the inequality for y. 4x – y + 2 ≤ 0 –y ≤ –4x – 2 –1 –1 y ≥ 4x + 2 Step 2 Graph the boundary line y = 4x + 2. Use a solid line for ≥.

14
**Additional Example 2C Continued**

Graph the solutions of the linear inequality. Check your answer. Step 3 The inequality is ≥, so shade above the line.

15
**Additional Example 2C Continued**

Check (–3)+ 2 3 –12 + 2 3 ≥ –10 y ≥ 4x + 2 Substitute ( –3, 3) for (x, y) because it is not on the boundary line. The point (–3, 3) satisfies the inequality, so the graph is correctly shaded.

16
**Additional Example 4A: Writing an Inequality from a Graph**

Write an inequality to represent the graph. y-intercept: 1; slope: Write an equation in slope-intercept form. The graph is shaded above a dashed boundary line. Replace = with > to write the inequality

17
**Additional Example 3a: Application**

Ada has at most 285 beads to make jewelry. A necklace requires 40 beads, and a bracelet requires 15 beads. a. Write a linear inequality to describe the situation. Let x represent the number of necklaces and y the number of bracelets. Write an inequality. Use ≤ for “at most.”

18
**Additional Example 3a Continued**

Necklace beads bracelet plus is at most 285 beads. 40x + 15y ≤ Solve the inequality for y. 40x + 15y ≤ 285 –40x –40x 15y ≤ –40x + 285 Subtract 40x from both sides. Divide both sides by 15.

19
Additional Example 3b b. Graph the solutions. = Step 1 Since Ada cannot make a negative amount of jewelry, the system is graphed only in Quadrant I. Graph the boundary line Use a solid line for ≤.

20
**Additional Example 3b Continued**

b. Graph the solutions. Step 2 Shade below the line. Ada can only make whole numbers of jewelry. All points on or below the line with whole number coordinates are the different combinations of bracelets and necklaces that Ada can make.

21
Additional Example 3c c. Give two combinations of necklaces and bracelets that Ada could make. Two different combinations of jewelry that Ada could make with 285 beads could be 2 necklaces and 8 bracelets or 5 necklaces and 3 bracelets. (2, 8) (5, 3)

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google