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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 2-5 Linear Equations in Two Variables Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Algebra 2

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2-5 Linear Equations in Two Variables Warm Up Find the intercepts of each line. 1. 3x + 2y = 18 2. 4x – y = 8 3. 5x + 10 = 2y Write the function in slope-intercept form. Then graph. 4. 2x + 3y = –3 (0, 9), (6, 0) (0, –8), (2, 0) (0, 5), (–2, 0)

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Graph linear inequalities on the coordinate plane. Solve problems using linear inequalities. Objectives

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables linear inequality boundary line Vocabulary

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Holt McDougal Algebra 2 2-5 Linear Equations 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.

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Holt McDougal Algebra 2 2-5 Linear Equations 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 is above the boundary line where y = 2x – 4.

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Holt McDougal Algebra 2 2-5 Linear Equations 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.

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Think of the underlines in the symbols and as representing solid lines on the graph. Helpful Hint

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Example 1A: Graphing Linear Inequalities Graph the inequality. The boundary line is which has a 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.

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Example 1A Continued Check Choose a point in the solution region, such as (3, 2) and test it in the inequality. The test point satisfies the inequality, so the solution region appears to be correct. ? 2 > 1 ?

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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.. Example 1B: Graphing Linear Inequalities

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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. Example 1B Continued

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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. Check It Out! Example 1a Graph the inequality y 3x –2.

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Check Choose a point in the solution region, such as (–3, 2) and test it in the inequality. The test point satisfies the inequality, so the solution region appears to be correct. y 3x –2 Check It Out! Example 1a Continued 2 3(–3) –2 ? 2 (–9) –2 ? 2 > –11 ?

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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.. Check It Out! Example 1b

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Holt McDougal Algebra 2 2-5 Linear Equations 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.

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Holt McDougal Algebra 2 2-5 Linear Equations 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. The point (0, 0) is the easiest point to test if it is not on the boundary line. Helpful Hint

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Graph 3x + 4y 12 using intercepts. Example 2: Graphing Linear Inequalities 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-interceptx-intercept 3x + 4y = 12 3(0) + 4y = 12 3x + 4(0) = 12 4y = 12 3x + 4y = 12 y = 3 3x = 12 x = 4

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Holt McDougal Algebra 2 2-5 Linear Equations 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). (0, 3) (4, 0)

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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-interceptx-intercept 3x – 4y = 12 3(0) – 4y = 123x – 4(0) = 12 – 4y = 12 3x – 4y = 12 y = – 3 3x = 12 x = 4 Check It Out! Example 2

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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). (4, 0) Check It Out! Example 2 (0, –3)

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Holt McDougal Algebra 2 2-5 Linear Equations 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. Dont forget which variable represents which quantity. Caution

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Holt McDougal Algebra 2 2-5 Linear Equations 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 number of adult tickets and y as the number of child tickets, 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?

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 1 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.

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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. An inequality that models the problem is 4.5x + 3y 135. 2 Make a Plan 135y3.00+x4.50 total. is at least number of child tickets times child price plus number of adult tickets times Adult price

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Find the intercepts of the boundary line. 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. Solve 3 4.5(0) + 3y = 135 4.5x + 3(0) = 135 y = 45 x = 30

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables If 25 child tickets are sold, 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. 14($4.50) + 25($3.00) = $138.00, so the answer is reasonable. Look Back 4 4.5x + 3(25) 135 4.5x + 75 135 4.5x 60, so x 13.3 _

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Holt McDougal Algebra 2 2-5 Linear Equations 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?

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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. 1 Understand the Problem

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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. An inequality that models the problem is 40x + 125y 135. 2 Make a Plan 1500y125+x40 total. is less than number awarded times large prize plus number awarded times Small prize

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Find the intercepts of the boundary line. 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. Solve 3 40(0) + 125y = 150040x + 125(0) = 1500 y = 12x = 37.5

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables If 4 large prizes are awarded, 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. $40(25) + $125(4) = $1500, so the answer is reasonable. Look Back 4 40x + 125(4) 1500 40x + 500 1500 40x 1000, so x 25

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Holt McDougal Algebra 2 2-5 Linear Equations 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.

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Solve for y. Graph the solution. Subtract 8x from both sides. Divide by –2, and reverse the inequality symbol. 8x – 2y > 8 –2y > –8x + 8 Example 4: Solving and Graphing Linear Inequalities y < 4x – 4 Multiply both sides by

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Note that the graph is shown in the STANDARD SQUARE window. ( 6:ZStandard followed by 5:ZSquare). Use the calculator option to shade below the line y < 4x – 4. Example 4 Continued

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Solve 2(3x – 4y) > 24 for y. Graph the solution. Subtract 3x from both sides. Divide by –4, and reverse the inequality symbol. –4y > –3x + 12 Check It Out! Example 4 Divide both sides by 2. 3x – 4y > 12

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables Note that the graph is shown in the STANDARD SQUARE window. ( 6:ZStandard followed by 5:ZSquare). Use the calculator option to shade below the line. Check It Out! Example 4 Continued

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Holt McDougal Algebra 2 2-5 Linear Equations 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

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Holt McDougal Algebra 2 2-5 Linear Equations in Two Variables 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. b. How many boxes of potatoes can the chef order if she orders 4 boxes of carrot? 18x + 12y 144 no more than 5 Lesson Quiz: Part II

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