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Linear Inequalities in 2-5 2-5 Linear Inequalities in Two Variables Two Variables Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 22 2-5 Linear Inequalities in Two Variables Warm Up Find the intercepts of each line. 1. 3x + 2y = 18 (0, 9), (6, 0) 2. 4x – y = 8 (0, –8), (2, 0) 3. 5x + 10 = 2y (0, 5), (–2, 0) Write the function in slope-intercept form. Then graph. 4. 2x + 3y = –3 Holt Algebra 2 2-5 Linear Inequalities in Two Variables Objectives Graph linear inequalities on the coordinate plane. Solve problems using linear inequalities. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Vocabulary linear inequality boundary line Holt Algebra 2 2-5 Linear Inequalities in Two Variables Linear functions form the basis of linear inequalities. A linear inequality in two variables relates two variables using an inequality symbol, such as y > 2x – 4. Its graph is a region of the coordinate plane bounded by a line. The line is a boundary line, which divides the coordinate plane into two regions. Holt Algebra 2 2-5 Linear Inequalities in Two Variables For example, the line y = 2x – 4, shown at right, divides the coordinate plane into two parts: one where y > 2x – 4 and one where y < 2x – 4. In the coordinate plane higher points have larger y values, so the region where y > 2x – 4 is above the boundary line where y = 2x – 4. Holt Algebra 2 2-5 Linear Inequalities in Two Variables To graph y ≥ 2x – 4, make the boundary line solid, and shade the region above the line. To graph y > 2x – 4, make the boundary line dashed because y-values equal to 2x – 4 are not included. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Helpful Hint Think of the underlines in the symbols ≤ and ≥ as representing solid lines on the graph. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 1A: Graphing Linear Inequalities Graph the inequality The boundary line is y-intercept of 2 and a slope of Draw the boundary line dashed because it is not part of the solution. Then shade the region above the boundary line to show . Holt Algebra 2 . which has a . 2-5 Linear Inequalities in Two Variables Example 1A Continued Check Choose a point in the solution region, such as (3, 2) and test it in the inequality. ? ? 2>1 The test point satisfies the inequality, so the solution region appears to be correct. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 1B: Graphing Linear Inequalities Graph the inequality y ≤ –1. Recall that y= –1 is a horizontal line. Step 1 Draw a solid line for y=–1 because the boundary line is part of the graph. Step 2 Shade the region below the boundary line to show where y < –1. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 1B Continued Check The point (0, –2) is a solution because –2 ≤ –1. Note that any point on or below y = –1 is a solution, regardless of the value of x. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 1a Graph the inequality y ≥ 3x –2. The boundary line is y = 3x – 2 which has a y–intercept of –2 and a slope of 3. Draw a solid line because it is part of the solution. Then shade the region above the boundary line to show y > 3x – 2. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 1a Continued Check Choose a point in the solution region, such as (–3, 2) and test it in the inequality. y ≥ 3x –2 ? 2 ≥ 3(–3) –2 ? 2 ≥ (–9) –2 ? 2 > –11 The test point satisfies the inequality, so the solution region appears to be correct. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 1b Graph the inequality y < –3. Recall that y = –3 is a horizontal line. Step 1 Draw the boundary line dashed because it is not part of the solution. Step 2 Shade the region below the boundary line to show where y < –3. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 1b Continued Check The point (0, –4) is a solution because –4 < –3. Note that any point below y < –4 is a solution, regardless of the value of x. Holt Algebra 2 2-5 Linear Inequalities in Two Variables If the equation of the boundary line is not in slope-intercept form, you can choose a test point that is not on the line to determine which region to shade. If the point satisfies the inequality, then shade the region containing that point. Otherwise, shade the other region. Helpful Hint The point (0, 0) is the easiest point to test if it is not on the boundary line. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 2: Graphing Linear Inequalities Using Intercepts Graph 3x + 4y ≤ 12 using intercepts. Step 1 Find the intercepts. Substitute x = 0 and y = 0 into 3x + 4y = 12 to find the intercepts of the boundary line. y-intercept x-intercept 3x + 4y = 12 3(0) + 4y = 12 3x + 4y = 12 3x + 4(0) = 12 4y = 12 y=3 3x = 12 x=4 Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 2 Continued Step 2 Draw the boundary line. The line goes through (0, 3) and (4, 0). Draw a solid line for the boundary line because it is part of the graph. Step 3 Find the correct region to shade. Substitute (0, 0) into the inequality. Because 0 + 0 ≤ 12 is true, shade the region that contains (0, 0). Holt Algebra 2 (0, 3) (4, 0) 2-5 Linear Inequalities in Two Variables Check It Out! Example 2 Graph 3x – 4y > 12 using intercepts. Step 1 Find the intercepts. Substitute x = 0 and y = 0 into 3x – 4y = 12 to find the intercepts of the boundary line. y-intercept 3x – 4y = 12 3(0) – 4y = 12 x-intercept 3x – 4y = 12 3x – 4(0) = 12 – 4y = 12 y=–3 3x = 12 x=4 Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 2 Step 2 Draw the boundary line. The line goes through (0, –3) and (4, 0). Draw the boundary line dashed because it is not part of the solution. Step 3 Find the correct region to shade. Substitute (0, 0) into the inequality. Because 0 + 0 >12 is false, shade the region that does not contain (0, 0). Holt Algebra 2 (4, 0) (0, –3) 2-5 Linear Inequalities in Two Variables Many applications of inequalities in two variables use only nonnegative values for the variables. Graph only the part of the plane that includes realistic solutions. Caution Don’t forget which variable represents which quantity. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 3: Problem-Solving Application A school carnival charges $4.50 for adults and $3.00 for children. The school needs to make at least $135 to cover expenses. A. Using x as the adult ticket price and y as the child ticket price, write and graph an inequality for the amount the school makes on ticket sales. B. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses? Holt Algebra 2 2-5 1 Linear Inequalities in Two Variables Understand the Problem The answer will be in two parts: (1) an inequality graph showing the number of each type of ticket that must be sold to cover expenses (2) the number of adult tickets that must be sold to make at least $135 if 25 child tickets are sold. List the important information: • The school sells tickets at $4.50 for adults and $3.00 for children. • The school needs to make at least $135. Holt Algebra 2 2-5 2 Linear Inequalities in Two Variables Make a Plan Let x represent the number of adult tickets and y represent the number of child tickets that must be sold. Write an inequality to represent the situation. Adult price 4.50 times number of adult tickets • x plus child price + 3.00 times number of child tickets is at least total. • y 135 An inequality that models the problem is 4.5x + 3y ≥ 135. Holt Algebra 2 2-5 3 Linear Inequalities in Two Variables Solve Find the intercepts of the boundary line. 4.5(0) + 3y = 135 4.5x + 3(0) = 135 y = 45 Graph the boundary line through (0, 45) and (30, 0) as a solid line. Shade the region above the line that is in the first quadrant, as ticket sales cannot be negative. Holt Algebra 2 x = 30 2-5 Linear Inequalities in Two Variables If 25 child tickets are sold, 4.5x + 3(25) ≥ 135 4.5x + 75 ≥ 135 _ 4.5x ≥ 60, so x ≥ 13.3 Substitute 25 for y in 4.5x + 3y ≥ 135. Multiply 3 by 25. A whole number of tickets must be sold. At least 14 adult tickets must be sold. 4 Look Back 14($4.50) + 25($3.00) = $138.00, so the answer is reasonable. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 3 A café gives away prizes. A large prize costs the café $125, and the small prize costs $40. The café will not spend more than $1500. How many of each prize can be awarded? How many small prizes can be awarded if 4 large prizes are given away? Holt Algebra 2 2-5 1 Linear Inequalities in Two Variables Understand the Problem The answer will be in two parts: (1) an inequality graph showing the number of each type of prize awarded not too exceed a certain amount (2) the number of small prizes awarded if 4 large prizes are awarded. List the important information: • The café awarded large prizes valued at $125 and $40 for small prizes. • The café will not spend over $1500. Holt Algebra 2 2-5 2 Linear Inequalities in Two Variables Make a Plan Let x represent the number of small prizes and y represent the number of large prizes, the total not too exceed $1500. Write an inequality to represent the situation. Small prize times 40 number awarded plus large prize x + 125 times number awarded is less than total. y ≤ 1500 An inequality that models the problem is 40x + 125y ≤ 135. Holt Algebra 2 2-5 3 Linear Inequalities in Two Variables Solve Find the intercepts of the boundary line. 40(0) + 125y = 1500 y = 12 40x + 125(0) = 1500 x = 37.5 Graph the boundary line through (0, 12) and (37.5, 0) as a solid line. Shade the region below the line that is in the first quadrant, as prizes awarded cannot be negative. Holt Algebra 2 2-5 Linear Inequalities in Two Variables If 4 large prizes are awarded, 40x + 125(4) ≤ 1500 40x + 500 ≤ 1500 40x ≥ 1000, so x ≤ 25 Substitute 4 for y in 40x + 125y ≤ 135. Multiply 125 by 4. A whole number of small prizes must be awarded. No more than 25 small prizes can be awarded. 4 Look Back $40(25) + $125(4) = $1500, so the answer is reasonable. Holt Algebra 2 2-5 Linear Inequalities in Two Variables You can graph a linear inequality that is solved for y with a graphing calculator. Press and use the left arrow key to move to the left side. Each time you press you will see one of the graph styles shown here. You are already familiar with the line style. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Example 4: Solving and Graphing Linear Inequalities Solve for y. Graph the solution. Multiply both sides by 8x – 2y > 8 –2y > –8x + 8 Subtract 8x from both sides. y < 4x – 4 Holt Algebra 2 Divide by –2, and reverse the inequality symbol. 2-5 Linear Inequalities in Two Variables Example 4 Continued Use the calculator option to shade below the line y < 4x – 4. Note that the graph is shown in the STANDARD SQUARE window. ( by 6:ZStandard followed 5:ZSquare). Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 4 Solve 2(3x – 4y) > 24 for y. Graph the solution. 3x – 4y > 12 –4y > –3x + 12 Divide both sides by 2. Subtract 3x from both sides. Divide by –4, and reverse the inequality symbol. Holt Algebra 2 2-5 Linear Inequalities in Two Variables Check It Out! Example 4 Continued Use the calculator option to shade below the line . Note that the graph is shown in the STANDARD SQUARE window. ( by 6:ZStandard followed 5:ZSquare). Holt Algebra 2 2-5 Linear Inequalities in Two Variables Lesson Quiz: Part I 1. Graph 2x –5y 10 using intercepts. 2. Solve –6y < 18x – 12 for y. Graph the solution. y > –3x + 2 Holt Algebra 2 2-5 Linear Inequalities in Two Variables Lesson Quiz: Part II 3. Potatoes cost a chef $18 a box, and carrots cost $12 a box. The chef wants to spend no more than $144. Use x as the number of boxes of potatoes and y as the number of boxes of carrots. a. Write an inequality for the number of boxes the chef can buy. 18x + 12y ≤ 144 b. How many boxes of potatoes can the chef order if she orders 4 boxes of carrot? no more than 5 Holt Algebra 2