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###### Communicate Your Answer

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###### Exercises

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Exercises 1 If you have a linear inequality and want to know if a specific ordered pair is a solution, you can test it just like you would a linear equation. Substitute the point into the inequality for the appropriate variables and see if the statement remains true after you simplify it. Let's say we have the inequality y>x+1 and we want to know if the points (3,5) and (1,1) are solutions. We can substitute the points into the inequality for x and y and check the results. Let's test the first point. y>x+1x=3, y=55>?3+1Add terms5>4 Because 5 is greater than 4, we know that the point (3,5) is a solution to the inequality. Now let's test the second point. y>x+1x=1, y=11>?1+1Add terms1≯2 Because 1 is not greater than 2, we know that (1,1) is not a solution to the inequality. | |

Exercises 2 In order to compare the differences between graphs of linear equations in two variables and linear inequalities in two variables, let's compare the graphs for functions that are nearly identical, that differ only in their symbols. y=2x+1y>2x+1 Let's first graph each of these functions individually and then make our comparison.y=2x+1 We can graph this line on a coordinate plane.y>2x+1 To graph this inequality, we need to first graph our boundary line. This is actually going to be the same line as our previous function. However, because our inequality is strictly values where y is greater than >, our line will be dashed as points along this line are not included.Now, we can show the entire solution set by adding some shading. To decide which side of the line to shade, we can substitute in any point, such as (0,0). y>2x+1x=0, y=00>?2⋅0+1Multiply0>?0+1Add terms0≯1 Because (0,0) is not a solution to the inequality, we need to shade on the side of the line that does not contain the origin.Comparison Creating the graph of a linear inequality in two variables begins with graphing a linear equation in two variables. However, because the solutions to equations are only the points that lie directly on the line, the graph of an equation is only that line. Linear inequalities, on the other hand, must have graphs that show the entire range of all possible solutions. Where we shade and the look of the line depends on the inequality sign. | |

Exercises 3 We can determine whether (2,3) is a solution by substituting it into the inequality and simplifying using the order of operations. x+y<7x=2, y=32+3<?7Add terms5<7 Since 5 is less than 7, (2,3) is a solution. | |

Exercises 4 We can determine whether (5,2) is a solution by substituting it into the inequality and simplifying using the order of operations. x−y≤0x=5, y=25−2≤?0Subtract term3≤0 Since 3 is not less than or equal to 0, (5,2) is not a solution. | |

Exercises 5 We can determine whether (-9,2) is a solution by substituting it into the inequality and simplifying. x+3y≥-2x=-9, y=2-9+3(2)≥?-2Multiply-9+6≥?-2Add terms-3≱-2 × Since -3 is not greater than or equal to -2, the ordered pair (-9,2) is not a solution. | |

Exercises 6 We can determine whether (-1,2) is a solution by substituting it into the inequality and simplifying using the order of operations. 8x+y>-6x=-1, y=28(-1)+2>?-6a(-b)=-a⋅b-8+2>?-6Add terms-6>?-6 Since -6 is not greater than -6, (-1,2) is not a solution. | |

Exercises 7 We can determine whether (-3,-3) is a solution by substituting it into the inequality and simplifying using the order of operations. -6x+4y≤6x=-3, y=-3-6(-3)+4(-3)≤?6Multiply18−12≤?6Subtract term6≤6 Since 6 is less than or equal to 6, (-3,-3) is a solution. | |

Exercises 8 To determine whether (-1,-1) is a solution, we will substitute it into the inequality. If it produces a true statement, then the point is part of the solution set. If not, the point is not part of the solution set. 3x−5y≥2x=-1, y=-13(-1)−5(-1)≥?2a(-b)=-a⋅b-3−5(-1)≥?2-a(-b)=a⋅b-3+5≥?2Add terms2≥2 ✓ Since 2 is greater than or equal to 2, the point (-1,-1) is a solution to the inequality. | |

Exercises 9 We can determine whether (-8,2) is a solution by substituting it into the inequality and simplifying using the order of operations. -x−6y>12x=-8, y=2-(-8)−6(2)>?12-(-a)=a8−6(2)>?12(-a)b=-ab8−12>?12Subtract term-4≯12 Since -4 is not greater than 12, (-8,2) is not a solution. | |

Exercises 10 We can determine whether (-6,3) is a solution by substituting it into the inequality and simplifying using the order of operations. -4x−8y<15x=-6, y=3-4(-6)−8(3)<?15Multiply24−24<?15Subtract term0<15 Since 0 is less than 15, (-6,3) is a solution. | |

Exercises 11 To check whether the ordered pair (0,-1) is a solution of the inequality shown on the graph, we plot this point and check whether it lies within the shaded area.As we can see, the point is outside of the shaded area which means it is not a solution to this inequality. | |

Exercises 12 To check whether the ordered pair (-1,3) is a solution of the inequality shown on the graph, let's plot this point and check whether it lies in the shaded area.As we can see, the point lies within the shaded area, so it is a solution to this inequality. | |

Exercises 13 To check whether the ordered pair (1,4) is a solution of the inequality shown on the graph, let's plot this point and check whether it lies in the shaded area.As we can see, the point lies within the shaded area, so it is a solution to this inequality. | |

Exercises 14 To check whether the ordered pair (0,0) is a solution of the inequality shown on the graph, let's plot this point and check whether it lies in the solution set.At first glance one might think that this point is within the solution set. However, because it lies on the dashed line, and the dashed line is not a part of the inequality, it is not a solution. | |

Exercises 15 To check whether the ordered pair (3,3) is a solution of the inequality shown on the graph, let's plot this point and check whether it lies in the shaded area.At first glance one might think that this point is within the solution set. However, because it lies on the dashed line, and the dashed line is not included in the inequality, it is not a solution. | |

Exercises 16 To check whether the ordered pair (2,1) is a solution of the inequality shown on the graph, let's plot this point and check whether it lies in the shaded area.As we can see, the point does not lie within the shaded area, so it is not a solution to this inequality. | |

Exercises 17 To check whether the carpenter can buy twelve 2-by-8 boards and fourteen 4-by-4 boards, let's substitute x=12 and y=14 in the given inequality and check if it is satisfied. 8x+12y≤250x=12, y=148⋅12+12⋅14≤?250Multiply96+168≤?250Add terms264≰250 We got a contradiction! Therefore the ordered pair (12,14) is not a solution of the inequality and the carpenter can't buy this set of boards. | |

Exercises 18 We are given the following inequality. 3x+2y≥93 Since we know that x represents the number of multiple-choice questions and y represents the number of matching questions, coefficients of 3 and 2 has to represent the number of points you get for respective type of question. 3x+2y≥93 Meanwhile, the right-hand side of the inequality is equal to 93, which is the minimum number of points you need to obtain in total to get an A. 3x+2y≥93 In order to tell if you receive an A we will substitute 20 for x and 18 for y, then we will check if the inequality holds true. 3x+2y≥?93x=20, y=183(20)+2(18)≥?93Multiply60+36≥?93Add terms96≥93 Th score you get for 20 mutiple-choice and 18 matching exercises is equal to 96, which is greater than or equal to 93. Therefore you receive an A. | |

Exercises 19 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequalityy≤ 5 Boundary Line y = 5 Notice that the equation for this boundary line only has one variable. This equation is telling us that each and every point that lies on the line will have a y-coordinate equal to 5. This means that it is a horizontal line. Also notice that the inequality is not strict, so the points on the boundary line are included in the solution set. We have shown this by drawing a solid line.Shading the Plane The inequality y≤5 describes all values of y that are less than or equal to 5. This means that every possible (x,y) coordinate pair with a y-value that is less than or equal to 5 needs to be included in the shading. | |

Exercises 20 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequalityy> 6 Boundary Line y = 6 Notice that the equation for this boundary line only has one variable. This equation is telling us that each and every point that lies on the line will have a y-coordinate equal to 6. This means that it is a horizontal line. Also, the inequality is strict, so the points on the boundary line are not included in the solution set. We show this by drawing a dashed line.Shading the Plane The inequality y>6 describes all values of y that are greater than 6. This means that every possible (x,y) coordinate pair with a y-value that is greater than 6 needs to be included in the shading. | |

Exercises 21 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequalityx < 2 Boundary Line x = 2 Notice that the equation for this boundary line only has one variable. This equation is telling us that each and every point that lies on the line will have an x-coordinate equal to 2. This gives us a vertical line. Also, the inequality is strict, so the points on the boundary line are not included in the solution set. We show this by drawing a dashed line.Shading the Plane The inequality x<2 describes all values of x that are less than 2. This means that every possible (x,y) coordinate pair with an x-value that is less than 2 needs to be included in the shading. | |

Exercises 22 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequalityx ≥ -3 Boundary Line x = -3 Notice that the equation for this boundary line only has one variable. This equation is telling us that each and every point that lies on the line will have an x-coordinate equal to -3. This gives us a vertical line. Also, the inequality is not strict, so the points on the boundary line are included in the solution set. We have shown this by drawing a solid line.Shading the Plane The inequality x≥-3 describes all values of x that are greater than or equal to -3. This means that every possible (x,y) coordinate pair with an x-value that is greater than or equal to -3 needs to be included in the shading. | |

Exercises 23 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequalityy> -7 Boundary Line y = -7 Notice that the equation for this boundary line only has one variable. This equation is telling us that each and every point that lies on the line will have a y-coordinate equal to -7. This means that it is a horizontal line. Also, the inequality is strict, so the points on the boundary line are not included in the solution set. We show this by drawing a dashed line.Shading the Plane The inequality y>-7 describes all values of y that are greater than -7. This means that every possible (x,y) coordinate pair with a y-value that is greater than -7 needs to be included in the shading. | |

Exercises 24 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequalityx < 9 Boundary Line x = 9 Notice that the equation for this boundary line only has one variable. This equation is telling us that each and every point that lies on the line will have an x-coordinate equal to 9. This gives us a vertical line. Also, the inequality is strict, so the points on the boundary line are not included in the solution set. We show this by drawing a dashed line.Shading the Plane The inequality x<9 describes all values of x that are less than 9. This means that every possible (x,y) coordinate pair with an x-value that is less than 9 needs to be included in the shading. | |

Exercises 25 First, let's graph the function y=-2x−4. This is the boundary line for our inequality. Because it is already in slope-intercept form, we can immediately identify the slope to be -2 and the y-intercept to be -4.Notice that our boundary line is dashed. This is because the inequality symbol is > and points lying on the line are not solutions to the inequality. Next, let's select a test point to help us learn which side of the boundary line should be shaded. We can use (0,0). y>-2x−4x=0, y=00>?-2(0)−4Zero Property of Multiplication0>-4 Since 0 is greater than -4, this is a true statement. Therefore, we must shade the region which includes the point (0,0). | |

Exercises 26 First, let's graph the function y=3x−1. This is the boundary line for our inequality. Because it is already in slope-intercept form, we can immediately identify the y-intercept as (0,-1) and the slope as 3.Notice that our boundary line is solid. This is because the inequality symbol is ≤ and points lying on the line are solutions to the inequality. Next, we select a test point to help us learn which side of the boundary line should be shaded. Let's test (0,0). y≤3x−1x=0, y=00≤?3(0)−1Zero Property of Multiplication0≰-1 Since 0 is not less than -1, we can conclude that (0,0) is not a solution to the inequality. Thus, the solution set is the region that does not include (0,0). | |

Exercises 27 Graphing an inequality involves two main steps.Plotting the boundary line. Shading the corresponding half-plane.Boundary Line To write the equation of the boundary line, replace the inequality sign with an equals sign. Inequality-4x+y < -7Boundary Line -4x+y = -7 To draw this line, we will first rewrite the equation in slope-intercept form. -4x+y=-7LHS+4x=RHS+4xy=4x−7 Now that the equation is in slope-intercept form, we can identify the slope m and the y-intercept b. y=4x−7⇔y=4x+(-7) We will plot the y-intercept b=-7, and then use the slope m=4 to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since we have a strict inequality, the boundary line will be dashed.Shading the Corresponding Half-plane To decide which side of the boundary line to shade, we will substitute a test point that is not on this line. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. -4x+y<-7x=0, y=0-4(0)+0<?-7Zero Property of Multiplication0≮-7 × Since the substitution of the test point did not create a true statement, we will shade the region that does not contain the point. | |

Exercises 28 Graphing an inequality involves two main steps.Plotting the boundary line. Shading one half-plane to show the solution set.Boundary Line The equation of the boundary line is obtained by replacing the inequality symbol with an equals sign. Inequality 3x−y ≥ 5 Boundary Line 3x−y = 5 To draw this line, we will first rewrite the equation in slope-intercept form. 3x−y=5 Write in slope-intercept form LHS+y=RHS+y3x=5+yLHS−5=RHS−53x−5=yRearrange equation y=3x−5 Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept b. y=3x−5⇔y=3x+(-5) We will plot the y-intercept b=-5, and use the slope m=3 to plot another point on the line. We will then connect these two points with a straight line. This is the boundary line. Since the inequality is not strict, the line will be solid.Shading the Half-plane To determine the solution set, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. 3x−y≥5x=0, y=03(0)−0≥?5Zero Property of Multiplication0−0≥?5Subtract term0≱5 × Since the substitution did not create a true statement, we will shade the region that does not contain the point. | |

Exercises 29 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequality5x−2y ≤ 6Boundary Line5x−2y = 6 To draw this line, we will first rewrite the equation in slope-intercept form. 5x−2y=6LHS−5x=RHS−5x-2y=-5x+6LHS/-2=RHS/-2y=25x−3 Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept (0,b). y=25x+(-3) We will plot the y-intercept (0,-3), then use the slope m=25 to plot another point on the line. Connecting these points with a solid line will give us the boundary line of our inequality. Note that the boundary line is solid, not dashed, because the inequality is not strict.Shading the Plane To decide which side of the boundary line to shade, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. 5x−2y≤6x=0, y=05(0)−2(0)≤?6Zero Property of Multiplication0≤6 Since the substitution of the test point created a true statement, we will shade the region that contains the point. | |

Exercises 30 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequality-x+4y > -12 Boundary Line -x+4y = -12 To draw this line, we will first rewrite the equation in slope-intercept form. -x+4y=-12LHS+x=RHS+x4y=x−12LHS/4=RHS/4y=41x−3 Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept (0,b). y=41x+(-3) We will plot the y-intercept (0,-3), then use the slope m=41 to plot another point on the line. Connecting these points with a dashed line will give us the boundary line of our inequality. Note that the boundary line is dashed, not solid, because the inequality is strict.Shading the Plane To decide which side of the boundary line to shade, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. -x+4y>-12x=0, y=0-(0)+4(0)>?-12Zero Property of Multiplication0>-12 Since the substitution of the test point created a true statement, we will shade the region that contains the point. | |

Exercises 31 To graph the given inequality, we will begin by graphing the boundary line, y=-x+1. Note that, since the inequality is strict, the boundary line will be dashed.Let's now use (0,0) as a test point to check whether the shaded half-plane is correct. y<-x+1x=0, y=00<-0+1Add terms0<1 ✓ Since the test point produced a true statement, we will shade the region that contains (0,0).We can see that our graph is almost the same as the given graph. The difference is that our boundary line is dashed, and the one in the given graph is solid. That is the error. | |

Exercises 32 To graph the given inequality, we will begin by graphing the boundary line, y=3x−2. Note that, since the inequality is not strict, the boundary line will be solid.Let's now use (0,0) as a test point to check whether the shaded half-plane is correct. y≤3x−2x=0, y=00≤?3(0)−2Zero Property of Multiplication0≤?0−2Subtract term0≰-2 × Since the test point did not produce a true statement, we will shade the region that does not contain (0,0).The error was shading the incorrect half-plane. | |

Exercises 33 Before we can write an equation to represent the inequality, we need to identify our variables. Let x represent the number of arcade games played and let y represent the number of snacks purchased. This makes the total cost of games 0.75x and the total cost of snacks 2.25y. These totals, together, cannot exceed $20. 0.75x+2.25y≤20 Now that we have an inequality, we can graph it. To make it easier, we can solve the inequality for y and put it into slope-intercept form. 0.75x+2.25y≤20LHS−0.75x≤RHS−0.75x2.25y≤-0.75x+20LHS/2.25≤RHS/2.25y≤-31x+8.88 The inequality is now in slope-intercept form. To graph, we plot the y-intercept and use the slope to find a second point on the line before drawing a line through the two points. Since this inequality is equal to, the line will be solid.All that is left to do is to shade the appropriate region. To decide where to shade, we pick a test point above or below the line. We then test the point in the inequality. For simplicity, let's try the origin. y≤-31x+8.888888x=0, y=00≤-31⋅0+8.888888Use the Zero Product Property0≤8.88 Since we have a true statement, the origin is in the region that needs to be shaded, below the line.Note that we do not shade past the x- or y-axes in this case. That would represent negative numbers of games and snacks, respectively, which does not make sense in this scenario.Interpreting Solutions As we saw, the origin is a solution to the inequality. It represents not purchasing any snack and not playing any arcade games. This means no money is spent, which is less than $20. Any other point in the shaded region will also a be a solution to the inequality. Say (1,1), where we play one arcade game and buy one snack. | |

Exercises 34 To solve this exercise, first, we have to write the inequality representing the problem. Next, we will graph it. Finally, from the graph, we will be able to identify and interpret two solutions of the inequality.Writing the Inequality We are told that the drama club must sell at least $1500 worth of tickets to cover the expenses of producing the play. This means that the income from selling the tickets must be greater than or equal to $1500. Income from selling the tickets≥ $1500 Now, we must rewrite the left-hand side of the inequality using variables. Let x be the number of adult tickets and y the number of student tickets. We know that one adult ticket costs $10 and one student ticket costs $6. Using this information, we can write an expression to represent the income. Income from selling the tickets=x⋅$10+y⋅$6 Let's substitute the above expression into our original inequality. x⋅$10+y⋅$6≥$1500⇔10x+6y≥1500Graphing the Inequality Next, we will graph the obtained inequality. 10x+6y≥1500 To do so, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol with an equals sign. Inequality 10x+6y ≥ 1500 Boundary Line 10x+6y = 1500 To draw this line, we will first rewrite the equation in slope-intercept form. 10x+6y=1500 Write in slope-intercept form LHS−10x=RHS−10x6y=-10x+1500LHS/6=RHS/6y=6-10x+1500Write as a sum of fractionsy=6-10x+61500Calculate quotienty=6-10x+250Put minus sign in front of fractiony=-610x+250ca⋅b=ca⋅by=-610x+250ba=b/2a/2 y=-35x+250 Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept b. y=-35x+250 We will plot the y-intercept, and then use the slope to plot another point on the line. The boundary line will be solid because the inequality is not strict. Since x and y both represent the number of tickets, we need to restrict the graph to nonnegative values of x and y. Negative values do not make sense in this real-life context.To decide which side of the boundary line to shade, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. 10x+6y≥1500x=0, y=010(0)+6(0)≥?1500 Simplify left-hand side Zero Property of Multiplication0+0≥?1500Add terms 0≱1500 × Since the substitution of the test point did not create a true statement, we will shade the region that does not contain the point.Identifying and Interpreting the Solutions Finally, we can identify two solutions of the inequality. A solution of the graphed inequality is a point that lies in the shaded region. Note that, since the boundary line is solid, every point lying on it is a solution of the inequality as well. Let's choose our two points!Both (30,200) and (100,140) are solutions of the inequality. The solution (30,200) means that the drama club sells 30 adult tickets and 200 student tickets. Similarly, (100,140) means that the club sells 100 adult tickets and 140 student tickets. | |

Exercises 35 There are two major steps to writing an inequality when given its graph.Write an equation for the boundary line Determine the inequality symbol and complete the inequalityLet's get started by focusing on the boundary line.Writing the Boundary Line Equation It only takes two points to create a unique equation for any line, so let's identify two points on the boundary line.Here we've identified two points, (0,1) and (1,3), and indicated the horizontal and vertical changes between them. This gives us the "rise" and "run" of the graph, which will give us the slope m. runrise=12⇔m=2 One of the points we selected, (0,1), is also the y-intercept. This means that we can combine the slope m and the y-intercept at the point (0,b) to write an equation for the boundary line in slope-intercept form. y=mx+b⇒y=2x+1Forming the Inequality To finish forming the inequality, we need to determine the inequality symbol. This means replacing the equals sign with a blank space, since it is still unknown to us. y ? 2x+1 To figure out what the symbol should be, we need a test point that lies within the solution set.We'll substitute (-2,2) for this test, then make the inequality symbol fit the resulting statement. y ? 2x+1x=-2, y=22 ? 2(-2)+1a(-b)=-a⋅b2 ? -4+1Add terms2 ? -3 Two is greater than -3. We can infer that, of the four inequality symbols, only two would make this a true statement, > or ≥. Returning to the given graph one last time, we see that the boundary line is dashed, not solid. This implies that the inequality is strict. We can now form our final inequality. y>2x+1 | |

Exercises 36 There are two major steps to writing an inequality when given its graph.Write an equation for the boundary line Determine the inequality symbol and complete the inequalityLet's get started by focusing on the boundary line.Writing the Boundary Line Equation It only takes two points to create a unique equation for any line, so let's identify two points on the boundary line.Here we've identified two points, (0,2) and (2,3), and indicated the horizontal and vertical changes between them. This gives us the "rise" and "run" of the graph, which will give us the slope m. runrise=21⇔m=21 One of the points we selected, (0,2), is also the y-intercept. This means that we can combine the slope m and the y-intercept at the point (0,b) to write an equation for the boundary line in slope-intercept form. y=mx+b⇒y=21x+2Forming the Inequality To finish forming the inequality, we need to determine the inequality symbol. This means replacing the equals sign with a blank space, since it is still unknown to us. y ? 21x+2 To figure out what the symbol should be, we need a test point that lies within the solution set.We'll substitute (-2,2) for this test, then make the inequality symbol fit the resulting statement. y ? 21x+2x=-2, y=22 ? 21(-2)+2Multiply2 ? -1+2Add terms2 ? 1 Two is greater than 1. We can infer that, of the four inequality symbols, only two would make this a true statement, > or ≥. Returning to the given graph one last time, we see that the boundary line is dashed, not solid. This implies that the inequality is strict. We can now form our final inequality. y>21x+2 | |

Exercises 37 There are two major steps to writing an inequality when given its graph.Write an equation for the boundary line Determine the inequality symbol and complete the inequalityLet's get started by focusing on the boundary line.Writing the Boundary Line Equation It only takes two points to create a unique equation for any line, so let's identify two points on the boundary line.Here we've identified two points, (0,-2) and (2,-3), and indicated the horizontal and vertical changes between them. This gives us the "rise" and "run" of the graph, which will give us the slope m. runrise=2-1⇔m=-21 One of the points we selected, (0,-2), is also the y-intercept. This means that we can combine the slope m and the y-intercept at the point (0,b) to write an equation for the boundary line in slope-intercept form. y=mx+b⇒y=-21x+(-2)Forming the Inequality To finish forming the inequality, we need to determine the inequality symbol. This means replacing the equals sign with a blank space, since it is still unknown to us. y ? -21x−2 To figure out what the symbol should be, we need a test point that lies within the solution set.We'll substitute (-2,-3) for this test, then make the inequality symbol fit the resulting statement. y ? -21x−2x=-2, y=-3-3 ? -21(-2)−2-a(-b)=a⋅b-3 ? 1−2Subtract term-3 ? -1 -3 is less than -1. We can infer that, of the four inequality symbols, only two would make this a true statement, < or ≤. Returning to the given graph one last time, we see that the boundary line is solid, not dashed. This implies that the inequality is not strict. We can now form our final inequality. y≤-21x−2 | |

Exercises 38 There are two major steps to writing an inequality when given its graph.Write an equation for the boundary line Determine the inequality symbol and complete the inequalityLet's get started by focusing on the boundary line.Writing the Boundary Line Equation It only takes two points to create a unique equation for any line, so let's identify two points on the boundary line.Here we've identified two points, (0,-3) and (1,0), and indicated the horizontal and vertical changes between them. This gives us the "rise" and "run" of the graph, which will give us the slope m. runrise=13⇔m=3 One of the points we selected, (0,-3), is also the y-intercept. This means that we can combine the slope m and the y-intercept at the point (0,b) to write an equation for the boundary line in slope-intercept form. y=mx+b⇒y=3x+(-3)Forming the Inequality To finish forming the inequality, we need to determine the inequality symbol. This means replacing the equals sign with a blank space, since it is still unknown to us. y ? 3x−3 To figure out what the symbol should be, we need a test point that lies within the solution set.We'll substitute (2,1) for this test, then make the inequality symbol fit the resulting statement. y ? 3x−3x=2, y=11 ? 3(2)−3Multiply1 ? 6−3Subtract term1 ? 3 One is less than 3. We can infer that, of the four inequality symbols, only two would make this a true statement, < or ≤. Returning to the given graph one last time, we see that the boundary line is solid, not dashed. This implies that the inequality is not strict. We can now form our final inequality. y≤3x−3 | |

Exercises 39 | |

Exercises 40 aTo match the inequality with the correct graph, we should graph it. When graphing an inequality we first have to identify its boundary line. It can be found by replacing the inequality symbol with an equals sign. Inequality:Boundary Line:3x−2y≤63x−2y=6 To draw this line, we will first rewrite the equation in slope-intercept form. 3x−2y=6 Write in slope-intercept form LHS−3x=RHS−3x-2y=-3x+6LHS⋅(-1)=RHS⋅(-1)2y=3x−6LHS/2=RHS/2y=23x−6Write as a difference of fractionsy=23x−26Calculate quotienty=23x−3ca⋅b=ca⋅b y=23x−3 Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept b. y=23x−3⇔y=23x+(-3) We will plot the y-intercept at (0,-3), and then use the slope m=23 to plot another point on the line. Note that the boundary line must be solid because the inequality is not strict.To decide which region of the coordinate plane to shade, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. 3x−2y≤6x=0, y=03(0)−2(0)≤?6Zero Property of Multiplication0−0≤?6Subtract term0≤6 ✓ Since the substitution of the test point created a true statement, we will shade the region that contains the point.The graph of the inequality matches with option C.bNote that we have almost the same inequality as in Part A. The only difference is that the inequality sign, this time, is strict. Therefore, the shaded region and the boundary line will be the same, but the boundary line will be dashed.The graph of the inequality matches with option A.cOnce again, the boundary line is the same as in Parts A and B. However, since the inequality sign is reversed with respect to the previous parts, the shaded half-plane will be the region below the line. Since the inequality is strict, the line will be dashed.The graph of the inequality matches with option D.dFinally, this last inequality is almost the same as the one in Part C. The only difference is that this is a non-strict inequality. Therefore, the boundary line and the region will be the same as in Part C, but the line will be solid.The graph of the inequality matches with option B. | |

Exercises 41 Let's suppose we are graphing the inequality y≥x, whose boundary line is y=x.Let's now use the point (0,0), which is on the boundary line, to determine the region we should shade. If substituting the point in the inequality produces a true statement, then the region to be shaded is the one which contains the point. Otherwise, we should shade the opposite region. y≥xx=0, y=00≥0 ✓ The test point produced a true statement. We should shade the half-plane that contains this point. But… what is this region? We are not able to determine the region, because the test point lies on the boundary line. Choosing a test point on the boundary line does not help us to determine the half-plane we should shade.Let's test the point (-2,2), which is not on the boundary line. y≥xx=-2, y=22≥-2 ✓ We obtained a true statement, and we should shade the region that contains the point. This time, it should not be hard to identify the half-plane, since (-2,2) is not on the boundary line. | |

Exercises 42 Before we graph the points on a coordinate plane we will name them: Solutions: Not solutions: A:(-1,3), B:(0,-1), C:(0,1)D:(1,1), E:(0,0), F:(0,-1) The points that are solutions we color red and the points that aren't we color blue.For the red points to be included in the solution set, they have to be a part of the shaded area of our inequality. Let's draw a line through D with a negative slope that includes A, B and C when we shade the region above the line.The three points A, B and C are definitely part of the solution set to this inequality. But so is D as it sits on the boundary line which is solid. To exclude D we can simply dash the boundary line.This inequality satisfies the two properties. To write it algebraically we need the equation of the boundary line in slope-intercept form: y=mx+c We see that it intercepts the y-axis at (0,1) so c=1. To find the slope, we count the number of steps the line decreases as we move on step to the right on the x-axis.As we move one step to the right, the graph falls by 1 which means the slope is -1. Since the shading is above the boundary line, we are going to use the symbol >. Pulling it all together, we get the inequality: y>-x+1. | |

Exercises 43 The most common point to use as a test point is (0,0). However, this point cannot be used as a test point when it is on the boundary line. For example, let's look at the graph of y<x.If the point (0,0) is part of the line, it will always satisfy a strict inequality and will never satisfy a non-strict inequality. In this case, another point that is not on the line will need to be used. There are an infinite number of other points to choose, as long as they are not found on the boundary line. | |

Exercises 44 Let's start by writing the equation of the boundary line. First we calculate the slope using the given points. m=x2−x1y2−y1Substitute (2,5) & (-3,-5)m=2−(-3)5−(-5) Simplify RHS a−(-b)=a+bm=2+35+5Add termsm=510Calculate quotient m=2 Next we substitute the slope and one of the points into the point-slope form and solve for y. y−y1=m(x−x1)Substitute valuesy−5=2(x−2) Solve for y Distribute 2y−5=2x−2⋅2Multiplyy−5=2x−4LHS+5=RHS+5 y=2x+1 To find the inequality symbol we should use, we graph the boundary line along with the remaining two points that are solutions.To include (6,5) as a solution of the inequality, we have to shade everything below the boundary line which means we need to use the inequality symbol that signals less than, <. So far we have: y<2x+1. The second point (-2,-3) lies on the boundary line, which means the boundary line has to be included in the solution set. The inequality symbol we are looking for is less than or equal to, ≤. Our final inequality becomes: y≤2x+1. Let's graph the inequality including all the points. | |

Exercises 45 To begin, we can make sense of the given information.The points (-7,-16) and (1,8) lie on the boundary line. The boundary line must pass through these points. The points (-7,0) and (3,14) are not solutions of the inequality. These points cannot lie in the shaded region on the graph.Let's plot the given points on a coordinate plane. We will also name the points to make them easier to discuss.To write the inequality, we must complete the following.Draw the boundary line. Shade the region containing the solutions.Drawing the boundary line As was established above, we must draw the boundary line so that points A and B lie on the line. We will determine the equation of the line that connects A and B. First, we will determine the slope of the line using the slope formula. m=x2−x1y2−y1 Calculate slope Substitute (-7,-16) & (1,8)m=1−(-7)8−(-16)-(-a)=am=1+78+16Add termsm=824Calculate quotient m=3 If we write our equation in the form y=mx+b, we now have y=3x+b. To determine the value of b, we can substitute the coordinates of A or B into the above equation for x and y and solve. We will arbitrarily choose to use the coordinates of B. y=3x+b Solve for b x=1, y=88=3⋅1+bMultiply8=3+bLHS−3=RHS−35=bRearrange equation b=5 Thus, the equation for the boundary line is y=3x+5. Now, we can add the boundary line to the graph above.We can see that A and B lie on the boundary line. However, point D also lies on the line. Since D is not a solution of the inequality, we must dash the boundary line. This means the inequality symbol will either be < or >. We will determine that next.Now, we can determine which region of the coordinate plane to shade.Shading the inequality At this point, we have the following inequality. y > 3x+5 We need to determine whether < or > is the correct inequality symbol. From the graph above, we can see that point C lies to the left of the boundary line. Since C is not a solution, we want to shade the region to the right. To determine which inequality symbol will accomplish this, we can choose any point from the region on the right and substitute it into the above inequality. We will arbitrarily choose the point (0,0). y > 3x+5x=0, y=00 > 3⋅0+5Zero Property of Multiplication0 > 5 To ensure that (0,0) is a solution of the inequality, the symbol must be <. Thus, our inequality is y<3x+5. We can graph the inequality to ensure that we have satisfied the requirements. | |

Exercises 46 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we see that the change from one term to the next is adding 8. 0→+88→+816→+824→+832 To find the next three terms in the sequence, we will extend this pattern three times. 0→+88→+816→+824→+832→+840→+848→+856 | |

Exercises 47 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we see that the change from one term to the next is subtracting 3. -5→−3-8→−3-11→−3-14→−3-17 To find the next three terms in the sequence, we will extend this pattern three times. -5→−3-8→−3-11→−3-14→−3-17→−3-20→−3-23→−3-26 | |

Exercises 48 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we see that the change from one term to the next is adding 1. -23→+1-21→+121→+123→+125 To find the next three terms in the sequence, we will extend this pattern three times. -23→+1-21→+121→+123→+125→+127→+129→+1211 |

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##### Other subchapters in Solving Systems of Linear Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Solving Systems of Linear Equations by Graphing
- Solving Systems of Linear Equations by Substitution
- Solving Systems of Linear Equations by Elimination
- Solving Special Systems of Linear Equations
- Quiz
- Solving Equations by Graphing
- Systems of Linear Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment