Systems of Linear Inequalities

Find the solutions in the app
Exercises marked with requires Mathleaks premium to view it's solution in the app. Download Mathleaks app on Google Play or iTunes AppStore.
Sections
Exercise name Free?
Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Monitoring Progress 9
Exercises
Exercise name Free?
Exercises 1 There are two ways to know for sure that an ordered pair is a solution to a system of linear inequalities:If you graph the system, an ordered pair is a solution if it falls within the overlapping shaded area. If you substitute the ordered pair into all the inequalities of the system, it satisfies each inequality individually.
Exercises 2 Let's plot the given points on the same coordinate plane as the system of inequalities.When looking at the four points, the first thing we can notice is that (0,-4) and (-1,-6) lie squarely within the solution set, and the points (1,-2) and (2,-4) rest on the boundary lines. What does it mean when a line is dashed and when it is solid?A solid line tells us that the inequality is not strict. Therefore, the points on the boundary line are solutions to the inequality. A dashed line tells us that the inequality is strict. Therefore, the points on the boundary line are not solutions to the inequality.With the above in mind, we can conclude that (2,-4) is a solution to the system because it lies on a solid line. However, the point (1,-2), which lies on both a solid and a dashed line, is a solution to one of the inequalities but not a solution to the other. Therefore, (1,-2) does not belong with the others as it's not a solution.
Exercises 3 Let's graph the ordered pair (-4,3) and the system of inequalities on the same coordinate plane. If the ordered pair lies within the shaded area, it is a solution.The ordered pair (-4,3) is not a solution because it lies outside of the shaded area.
Exercises 4 Let's plot the point (-3,-1) on the given graph. If it lies in the shaded area, then the ordered pair is a solution of the system of inequalities.Since the point lies within the shaded area, the ordered pair (-3,-1) is a solution to the system of linear inequalities.
Exercises 5 If the given point is a solution to the system, it will lie in the shaded region of the graph. Let's plot the point (-2,5) on the same coordinate plane as the system.The point (-2,5) does not lie within the shaded region of the graph. Thus, it is not a solution.
Exercises 6 Let's plot the point (1,1) on the given graph. If it lies in the shaded area, then the ordered pair is a solution of the system of inequalities.Since the point does not lie within the shaded area, the ordered pair (1,1) is not a solution to the system of linear inequalities.
Exercises 7 When determining if a given point satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of inequalities, we can use the same method. However, substituting the point must create true statements in every inequality in the system. Let's test to see if (-5,2) is a solution to the given system. {y<4y>x+3​(I)(II)​x=-5, y=2⎩⎨⎧​(2)<?​4(2)>?​(-5)+3​(II):  Add terms{2<42>-2​ Since 2 is less than 4 and greater than -2, both statements are true. Therefore, the point (-5,2) is contained in the solution set of the system.
Exercises 8 When determining if a given point satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of inequalities, we can use the same method. However, substituting the point must create true statements in every inequality in the system. Let's test to see if (1,-1) is a solution to the given system. {y>-2y>x−5​(I)(II)​x=1, y=-1⎩⎨⎧​(-1)>?​-2(-1)>?​(1)−5​(II):  Subtract term{-1>-2-1>-4​ Since -1 is greater than -2 and -4, both statements are true. Therefore, the point (1,-1) is contained in the solution set of the system.
Exercises 9 When determining if a given point satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of inequalities, we can use the same method. However, substituting the point must create true statements in every inequality in the system. Let's test to see if (0,0) is a solution to the given system. {y≤x+7y≥2x+3​(I)(II)​x=0, y=0⎩⎨⎧​(0)≤?​(0)+7(0)≥?​2(0)+3​(II):  Zero Property of Multiplication⎩⎨⎧​0≤?​0+70≥?​0+3​(I), (II):  Add terms{0≤70≱3​ One of our resulting statements, 0≱3, is not true. The point (0,0) satisfies (I) but does not satisfy (II). Because both statements are not true, the point (0,0) is not contained in the solution set of the system.
Exercises 10 When determining if a given point satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of inequalities, we can use the same method. However, substituting the point must create true statements in every inequality in the system. Let's test to see if (4,-3) is a solution to the given system. {y≤-x+1y≤5x−2​(I)(II)​x=4, y=-3{(-3)≤-(4)+1(-3)≤5(4)−2​(I)(II)​(I), (II):  Multiply{-3≤-4+1-3≤20−2​(I)(II)​(I), (II):  Add and subtract terms{-3≤-3-3≤18​(I)(II)​ Since -3 is equal -3 and less than 18, both statements are true. Therefore, the point (4,-3) is contained in the solution set of the system.
Exercises 11 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {y>-3y≥5x​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system. Note that since Inequality (I) is in the form y>-3, the slope of its boundary line will be equal to 0. This boundary line will be horizontal.InformationInequality (I)Inequality (II) Given Inequalityy>-3y≥5x Boundary Line Equationy=-3y=5x Solid or Dashed?> ⇒ Dashed≥ ⇒ Solid y=mx+by=0x+(-3)y=5x+0 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (1,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityy>-3y≥5x Substitute (1,0)(0)>?​-3(0)≥?​5(1) Simplify0>-30≱5 Shaded Regionsameopposite For Inequality (I) we will shade the region containing our test point, or above the boundary line. For Inequality (II), we will shade the region opposite the test point, or above the boundary line.
Exercises 12 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {y<-1x>4​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Since Inequality (I) is in the form y<-1, the slope of its boundary line will be equal to 0. This boundary line will be horizontal. On the other hand Inequality (II) is in the form x>4, therefore the slope of its boundary line will be undefined. This line will be vertical.InformationInequality (I)Inequality (II) Given Inequalityy<-1x>4 Boundary Line Equationy=-1x=4 Solid or Dashed?< ⇒ Dashed> ⇒ Dashed y=mx+by=0x+-1x=4 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityy<-1x>4 Substitute (0,0)(0)<?​-1(0)>?​4 Simplify0≮-10≯4 Shaded Regionoppositeopposite For both inequalities we will shade the region opposite the test point. For Inequality (I) it will be to the right of the boundary line, however for Inequality (II) it will be below.
Exercises 13 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {y<-2y>2​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system. Note that since inequalities are in the form y<-2 and y>2, the slope of their boundary lines will be equal to 0. These lines will be horizontal.InformationInequality (I)Inequality (II) Given Inequalityy<-2y>2 Boundary Line Equationy=-2y=2 Solid or Dashed?< ⇒ Dashed> ⇒ Dashed y=mx+by=0x+(-2)y=0x+2 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityy<-2y>2 Substitute (0,0)(0)<?​-2(0)>?​2 Simplify0≮-20≯2 Shaded Regionoppositeopposite For both inequalities we will shade the region opposite our test point. For Inequality (I) it will be below the boundary line, however for Inequality (II) it will be above.Now that we have graphed the system, we see that there is no overlapping region. This means that there are no points that are a solution to the system, only points that are solutions to each individual inequality.
Exercises 14 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {y<x−1y≥x+1​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system.InformationInequality (I)Inequality (II) Given Inequalityy<x−1y≥x+1 Boundary Line Equationy=x−1y=x+1 Solid or Dashed?< ⇒ Dashed≥ ⇒ Solid y=mx+by=1x+(-1)y=1x+1 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityy<x−1y≥x+1 Substitute (0,0)(0)<?​(0)−1(0)≥?​(0)+1 Simplify0≮-10≱1 Shaded Regionoppositeopposite For both inequalities we will shade the region opposite our test point. For Inequality (I) it will be below the boundary line, however for Inequality (II) it will be above.Now that we have graphed the system, we see that there is no overlapping region. This means that there are no points that are a solution to the system, only points that are solutions to each individual inequality.
Exercises 15 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {y≥-5y−1<3x​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system. Note that since Inequality (I) is in the form y≥-5, the slope of its boundary line will be equal to 0. This boundary line will be horizontal.InformationInequality (I)Inequality (II) Given Inequalityy≥-5y−1<3x Boundary Line Equationy=-5y−1=3x Solid or Dashed?≥ ⇒ Solid< ⇒ Dashed y=mx+by=0x+(-5)y=3x+1 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityy≥-5y−1<3x Substitute (0,0)(0)≥?​-5(0)−1<?​3(0) Simplify0≥-5-1<0 Shaded Regionsamesame For both inequalities we will shade the region containing our test point. For Inequality (I) it will be above the boundary line, however for Inequality (II) it will be below.
Exercises 16 We have been given a system of linear inequalities and asked to graph the solution sets. Let's look at graph of each individual inequality first and then combine the resulting solution sets.x+y>4 The first thing we need to do in order to graph a linear inequality is draw the boundary line. In this case, our inequality is not in slope-intercept form so let's rewrite that first: x+y>4⇒y>-x+4. The boundary line is drawn by considering the inequality as though it were an equation and then choosing the line to be dashed or solid based on the symbol. This time, our boundary line is y=-x+4. It will be dashed because our symbol is > and points lying on the line are not solutions.Now we need to choose which side of the line to shade for our inequality. We can test a point in the original inequality to see if it's a solution. Let's use (0,0). x+y>4x=0, y=00+0>?​4Add terms0≯4 Since the point (0,0) does not satisfy the inequality, we should shade the region that does not contain that point.y≥23​x−9 Now we can go through the same process with the second inequality. We will start by graphing the boundary line. This time the line should remain solid because the symbol ≥ indicates that points lying on the line are valid solutions.Let's use (0,0) again to test for our shaded region. y≥23​x−9x=0, y=00≥?​23​⋅0−9Zero Property of Multiplication0≥-9 Since (0,0) satisfies the inequality, we should shade the region containing the point.Combined solution set Now we can combine the two solution sets for the individual inequalities to find the solution set for the system of inequalities.Finally, let's cut away the unnecessary information so that we can see exactly where the final solution set lies. We only need the overlapping section of the graph.
Exercises 17 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {x+y>1-x−y<-3​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system.InformationInequality (I)Inequality (II) Given Inequalityx+y>1-x−y<-3 Boundary Line Equationx+y=1-x−y=-3 Solid or Dashed?> ⇒ Dashed< ⇒ Dashed y=mx+by=-1x+1y=-1x+3 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityx+y>1-x−y<-3 Substitute (0,0)(0)+(0)>?​1-(0)−(0)<?​-3 Simplify0≯10≮-3 Shaded Regionoppositeopposite For both inequalities we will shade the region opposite the test point or above the boundary line.
Exercises 18 We have been given a system of linear inequalities and asked to graph the solution sets. Let's look at graph of each individual inequality first and then combine the resulting solution sets.2x+y≤5 The first thing we need to do in order to graph a linear inequality is draw the boundary line. In this case, our inequality is not in slope-intercept form so let's rewrite that first: 2x+y≤5⇒y≤-2x+5. The boundary line is drawn by considering the inequality as though it were an equation and then choosing the line to be dashed or solid based on the symbol. This time, our boundary line is y=-2x+5. It will be solid because our symbol is ≤ and points lying on the line are solutions.Now we need to choose which side of the line to shade for our inequality. We can test a point in the original inequality to see if it's a solution. Let's use (0,0). 2x+y≤5x=0, y=02⋅0+0≤?​5Add terms0≤5 Since the point (0,0) satisfies the inequality, we should shade the region that contains that point.y+2≥-2x Now we can go through the same process with the second inequality. Our boundary line will be: y+2≥-2x⇒y=-2x−2. This time the line should remain solid because the symbol ≥ indicates that points lying on the line are valid solutions. Once again, we can test a point to determine which region to shade. Let's use (0,0). y+2≥-2xx=0, y=00+2≥?​-2⋅0Zero Property of Multiplication0+2≥?​0Identity Property of Addition2≥0 Since (0,0) satisfies the inequality, we should shade the region containing the point. Let's add this inequality to our existing coordinate plane.Combined solution set Finally, let's cut away the unnecessary information so that we can see exactly where the final solution set lies. We only need the overlapping section of the graph.
Exercises 19 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. ⎩⎪⎨⎪⎧​x<4y>1y≥-x+1​(I)(II)(III)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I),(II) and (III).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Since Inequality (I) is in the form x<4, the slope of its boundary line will be undefined. This boundary line will be vertical. On the other hand Inequality (II) is in the form y>1, therefore the slope of its boundary line will be equal to 0. This boundary line will be horizontal.InformationInequality (I)Inequality (II)Inequality (III) Given Inequalityx<4y>1y≥-x+1 Boundary Line Equationx=4y=1y=-x+1 Solid or Dashed?< ⇒ Dashed> ⇒ Dashed≥ ⇒ Solid y=mx+bx=4y=0x+1y=-1x+1 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II)Inequality (III) Given Inequalityx<4y>1y≥-x+1 Substitute (0,0)(0)<?​4(0)>?​1(0)≥?​-(0)+1 Simplify0<40≯10≱1 Shaded Regionsameoppositeopposite For Inequality (I) we will shade the region containing our test point, or to the left of the boundary line. For inequalities (II) and (III), however, we will shade the region opposite the test point, or above the boundary line.
Exercises 20 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. ⎩⎪⎨⎪⎧​x+y≤10x−y≥2y>2​(I)(II)(III)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I),(II) and (III).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system. Note that since Inequality (III) is in the form y>2, the slope of its boundary line will be equal to 0. This boundary line will be horizontal.InformationInequality (I)Inequality (II)Inequality (III) Given Inequalityx+y≤10x−y≥2y>2 Boundary Line Equationx+y=10x−y=2y=2 Solid or Dashed?≤ ⇒ Solid≥ ⇒ Solid> ⇒ Dashed y=mx+by=-1x+10y=1x+(-2)y=0x+2 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II)Inequality (III) Given Inequalityx+y≤10x−y≥2y>2 Substitute (0,0)(0)+(0)≤?​10(0)−(0)≥?​2(0)>?​2 Simplify0≤100≱20≯2 Shaded Regionsameoppositeopposite For Inequality (I) we will shade the region containing our test point, or to the below of the boundary line. For inequalities (II) and (III), however, we will shade the region opposite the test point, which is below the boundary line for Inequality (II) and above for Inequality (III).
Exercises 21 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 inequalityIn this exercise, we have been given a system consisting of two linear inequalities. We'll tackle them one at a time and bring them together in a system at the end.The Strict Inequality Let's take a look at the graph of the strict inequality.We can see that the boundary line is horizontal and passes through (0,3), therefore the equation of the line is y=3. 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 ?​ 3​ To figure out what the symbol should be, let's substitute any point that lies within the solution set into the equation.We'll substitute (0,0) for this test, then make the inequality symbol fit the resulting statement. y ?​ 3y=00 ?​ 3 0 is less than 3, so the symbol will be either < or ≤. Since the boundary line in the given graph is dashed, the inequality is strict, and we can form the first inequality in the system. y<3​The Non-Strict Inequality Writing the inequality for this region of the graph will involve the same steps as above. We'll start by identifying two points.Notice that the boundary line is vertical and passes through (-1,0), therefore the equation of the line is x=-1. Once more, we replace the equals sign with a blank space. x ?​ -1​ We will need another point that lies within the solution set to determine the sign of this inequality.We'll substitute (0,0) for this test, then make the inequality symbol fit the resulting statement. x ?​ -1x=00 ?​ -1 0 is greater than -1, so the symbol will be either > or ≥. Since the boundary line in the given graph is solid, the inequality is not strict, and we can form the second inequality in the system. x≥-1​Writing the System To complete the system of inequalities, we will bring both of our inequalities together in system notation. {y<3x≥-1​​
Exercises 22 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 inequalityIn this exercise, we have been given a system consisting of two linear inequalities. We'll tackle them one at a time and bring them together in a system at the end.The First Inequality Let's take a look at the graph of the first inequality.We can see that the boundary line is vertical and passes through (2,0), therefore the equation of the line is x=2. 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. x ?​ 2​ To figure out what the symbol should be, let's substitute any point that lies within the solution set into the equation.We'll substitute (3,0) for this test, then make the inequality symbol fit the resulting statement. x ?​ 2x=33 ?​ 2 3 is greater than 2, so the symbol will be either > or ≥. Since the boundary line in the given graph is dashed, the inequality is strict, and we can form the first inequality in the system. x>2​The Second Inequality Writing the inequality for this region of the graph will involve the same steps as above.Notice that the boundary line is vertical and passes through (4,0), therefore the equation of the line is x=4. Once more, we replace the equals sign with a blank space. x ?​ 4​ We will need another point that lies within the solution set to determine the sign of this inequality.We'll substitute (3,0) for this test, then make the inequality symbol fit the resulting statement. x ?​ 4x=33 ?​ 4 3 is less than 4, so the symbol will be either < or ≤. Since the boundary line in the given graph is dashed, the inequality is strict, and we can form the second inequality in the system. x<4​Writing the System To complete the system of inequalities, we will bring both of our inequalities together in system notation. {x>2x<4​​
Exercises 23 We will write the inequalities that form the system of inequalities one at a time.Inequality (I) Let's start by writing the equation of the boundary line related to the first inequality. The line intercepts the y-axis at (0,2). Therefore, its y-intercept is 2. We can write a partial equation in slope-intercept form. y=mx+2​ We can see in the given graph that the point (1,-1) is also on the line. Therefore, it must satisfy its equation. To find the slope, we will substitute x=1 and y=-1 in the above equation, and solve for m. y=mx+2x=1, y=-1-1=m(1)+2 Solve for m Identity Property of Multiplication-1=m+2LHS−2=RHS−2-3=mRearrange equation m=-3 The slope of the line is -3. We can now write the full equation of the boundary line. y=-3x+2​ Let's write the inequality now. To do so, we will test a point within the shaded area. For simplicity, let's choose (1,1).Since the point lies on the shaded area, it must satisfy the inequality. Therefore, when its coordinates are substituted into the inequality, they must produce a true statement. Note that the line is solid and thus the inequality will not be strict. y ≥​ -3x+2x=1, y=11 ≥​ -3(1)+2Identity Property of Multiplication1 ≥​ -3+2Add terms1≥-1 The sign of the inequality is greater than or equal to. Inequality (I): y≥-3x+2​Inequality (II) We will follow the same procedure to find the equation of the second inequality. Note that, again, the line is solid and therefore the inequality will not be strict.y-intercept of the Boundary Lineb=-2 Slope of the Boundary Linem=32​ Boundary Liney=32​x−2 Test Point(1,1) Inequality Sign≥ Inequality (II)y≥32​x−2 System of Inequalities We can now write the system combining the obtained inequalities. {y≥-3x+2y≥32​x−2​(I)(II)​
Exercises 24 In order to write the system that is represented by the graph, we need to write two inequalities which represent the shaded region bounded by the dashed and solid line.Dashed Line The dashed line intercepts the x-axis at (2,0) and the point (0,-2). With this information, we can determine the slope and y-intercept.The slope and y-intercept are m=1 and b=-2. Also, since the boundary line is dashed and the shaded region lies above this line, the inequality for this region should be y>x−2.Solid Line From the graph, we can see that the slope and the y-intercept of the solid line are m=5 and b=1. With these pieces of information, we can graph the inequality as shown below.Also, since the boundary line is solid and the shaded region lies below this line, the inequality should be y≤5x+1.Conclusion Combining the inequalities found above, we can write the system of inequalities. The system represented by the graph is: {y>x−2y≤5x+1.​
Exercises 25 To write the system of inequalities that corresponds with this graph, we will analyze the graph of each inequality separately. Below we've added labels to each boundary line so we can refer to them more easily.Notice there isn't a shaded region. This means the system has no solutions because the solution sets of the individual inequalities don't overlap. This can be true only if the solution set of the inequalities look like below. Let's also add two points that are true for each of the inequalities. They will come in handy.We will now write the individual inequalities from this graph.Inequality A Let's begin by identifying the slope and y-intercept of the boundary line.From the graph we see that the boundary line has a slope of -2 and a y-intercept of -3. Hence the boundary line has the equation y=-2x−3. Next we simply need to replace the equal sign with an inequality symbol. But which one? At this point we have y 22​ -2x−3. Since the boundary line is dashed, the only options are < and >. To determine which of these symbols is correct, we choose the substitute the point into the (-3,-1) inequality which we already know is a solution. When substituting it the inequality should form a true statement. y 22​ -2x−3x=-3, y=-1-1 22​ -2(-3)−3 Simplify RHS Multiply-1 22​ 6−3Subtract terms -1 22​ 3 Since -1 is less than 3 the symbol must be <. Thus, Inequality A can be written as y<-2x−3.Inequality B To write Inequality B we follow the same strategy. From the given graph we see that the boundary lines are parallel which means they have the same slope. Additionally, the graph shows a y-intercept of -1. We get the inequality: y 22​ -2x−1 As was the case with Inequality A, the only options for the inequality symbol here are < and >. We will test the arbitrarily chosen point (1,-1) to determine which is the correct choice. y 22​ -2x−1x=1, y=-1-1 22​ -2⋅1−1 Simplify RHS Multiply-1 22​ -2−1Subtract terms -1 22​ -3 Since -1 is greater than -3, the symbol must be >. Thus, Inequality A can be written as y>-2x−1.Conclusion The system of inequalities shown by the given graph is {y<-2x−3y>-2x−1​
Exercises 26 To write the system of inequalities that corresponds with this graph, we will analyze the graph of each inequality separately. Below we've added labels to each boundary line so we can refer to them more easily.First, notice the shaded region of the graph. For this to be the solution set of the system, it must be true that for both Inequality A and Inequality B, the individual solution sets are the region above each boundary line. This is shown below. Let's also add two points that are true for each of the inequalities. They will come in handy.We will now find the individual inequalities.Inequality A Let's begin by identifying the slope and y-intercept of the boundary line.From the graph we can see that the boundary line has a slope of 21​ and a y-intercept of 0. Thus, we can write the equation of the boundary line as y=21​x. Next we simply need to replace the equal sign with an inequality symbol. But which one? At this point we have y 22​ 21​x. Since the boundary line is solid, the only options are ≤ and ≥. To determine which of these symbols is correct, we substitute (2,2) which we already know is a solution into the inequality. This should form a true statement. y 22​ 21​xx=2, y=22 22​ 21​⋅2 Simplify RHS b1​⋅a=ba​2 22​ 22​Calculate quotient 2 22​ 1 Since 2 is greater than 1 the symbol must be ≥. Thus, Inequality A can be written as y≥21​x.Inequality B To write Inequality B, we can follow the same strategy as we did for Inequality A. From the given graph, we can see that the boundary lines are parallel and that the inequality intercepts the y-axis at (0,-2) which means we get the following equation for the boundary line: y=21​x−2. Next we simply need to replace the equal sign with an inequality symbol. But which one? At this point we have y 22​ 21​x−2 As was the case with Inequality A, the only options for the inequality symbol here are < and >. We will test the arbitrarily chosen point (1,1) to determine which is the correct choice. y 22​ 21​x−2x=1, y=11 22​ 21​⋅1−2 Simplify RHS a⋅1=a1 22​ 21​−2Subtract terms 1 22​ -23​ Since -1 is greater than -23​, the symbol must be ≥. Thus, Inequality A can be written as y≥21​x−2.Conclusion The system of inequalities shown by the given graph is {y≥21​xy≥21​x−2​
Exercises 32 To determine the intersection of the half-planes of the given inequalities, we can graph them on the same coordinate plane and check for overlapping regions. To do that we have to identify their boundary lines which we can do by isolating y in both inequalities. {x−y≤4x−y≥4​(I)(II)​ (I), (II):  Isolate y (I), (II):  LHS−x=RHS−x{-y≤-x+4-y≥-x+4​(I), (II):  Flip inequality and change signs {y≥x−4y≤x−4​ Examining the inequalities we see that the only thing that separate them is the inequality symbol. This means they have the same boundary line y=x−4, which has a slope of 1 and a y-intercept of -4. Let's graph this boundary line twice, a red line and a blue line, representing each of the inequalities. Notice if you mix red and blue you get purple.To determine which region to shade, we will use (0,0) as a test point in each of the inequalities. If the inequality remains true after substituting the test point, we shade the side of the inequality that contains this test point. If not, we shade the opposite region. {y≥x−4y≤x−4​(I)(II)​(I), (II):  x=0, y=0⎩⎨⎧​0≥?​0−40≤?​0−4​(I), (II):  Subtract term{0≥-40≰-4​ The results tells us that for (I) we shade the region above the boundary line as it contains (0,0) and vice verse for (II).From the graph above, we can see that the only region both solutions sets have in common is the boundary line. Thus, the intersection is the boundary line y=x−4.
Exercises 33 aTo begin, we can plot the vertices on a coordinate plane. Connecting adjacent points will allow us to draw the rectangle. We will also label the sides.Notice that Sides (I) and (III) are vertical lines while Sides (II) and (IV) are horizontal. It follows then that the boundary line for the corresponding inequalities will take the following form. Side (I) & (III): Side (II) & (IV): ​x=…y=…​ In each equation, the variable will be equal to the value on the axis that the line passes through. For example, notice that Side (I) intersects the x-axis at x=-1. Thus the equation for that boundary line is x=-1. Examining the other lines as well we find that: Side (I):Side (II):Side (III):Side (IV):​x=-1y=-3x=6y=1​ To determine the actual inequalities all we need to do is to replace the equal signs in the above equations with the correct inequality symbol. Since the sides of the rectangle are solid, our options are either ≤ or ≥. We can use any point from inside the rectangle to determine which symbol is correct. Notice that (0,0) is a point inside the rectangle. That means, when we substitute that point into any of the equations, a true statement is made.SideInequalityx=0,y=0≤ or ≥ (I)x 22​ -10 22​ -10≥-1 (II)y 22​ -30 22​ -30≥-3 (III)x 22​ 10 22​ 1-0≤1- (IV)y 22​ 60 22​ 6-0≤6- Replacing 0 with the correct variable for each of the inequalities we get the following system of inequality: ⎩⎪⎪⎪⎨⎪⎪⎪⎧​x≥-1y≥-3x≤6y≤1​(I)(II)(III)(IV)​bThe area of a rectangle can be calculate using A=lw, where l is the length and w is the width. From the original graph, we see that l=7 and w=4.We can substitute l=7 and w=4 into the formula we get the area. A=lwl=7, w=4A=7⋅4MultiplyA=28 The area of the rectange is 28 square units.
Exercises 34 aLet's start by plotting the points on a coordinate plane and drawing the triangle.Inequality (I) The base of the triangle lies on a horizontal line. Therefore, the equation of this boundary line is y=k, where k is the y-coordinate of its points. Since the line passes through the points (-2,-3) and (6,-3), its equation is y=-3. Moreover, the shaded region is above the line and the line is solid. This means the inequality sign is greater than or equal to. y≥-3​Inequality (II) Let's treat one of the sides of the triangle as a line, paying close attention to the slope.To travel from (2,5) to (6,-3) on this boundary line, we move 4 units in the positive horizontal direction and 8 units in the negative vertical direction. Since we know the run is 4 and the rise is -8, we can find the slope. slope=runrise​run=4, rise=-8slope=4-8​Put minus sign in front of fractionslope=-48​Calculate quotientslope=-2 Now that we know the slope of the line is -2, we can partially write its equation in slope-intercept form. y=-2x+b​ To find the y-intercept b, we will use the fact that the line passes through (2,5). Let's substitute 2 and 5 for x and y, respectively, into the partial equation above and solve for b. y=-2x+bx=2, y=55=-2(2)+b Solve for b (-a)b=-ab5=-4+bLHS+4=RHS+49=bRearrange equation b=9 Now that we know that b=9, we can write the equation of the boundary line. y=-2x+9​ To determine the inequality sign, we will use a point that belongs to the shaded area.When substituted into the inequality, the point (3,-2) must produce a true statement. Note that, since the line is solid, the inequality will not be strict. y ≤​ -2x+9x=3, y=-2-2 ≤​ -2(3)+9(-a)b=-ab-2 ≤​ -6+9Add terms-2≤3 We can now obtain our second inequality by replacing the equals sign with the corresponding inequality sign. y≤-2x+9​Inequality (III) We'll find our third inequality following the same procedure as before.Slope2 y-intercept1 Boundary Liney=2x+1 Test Point(3,-2) Inequalityy≤2x+1 System of Inequalities The three obtained inequalities form a system of linear inequalities. ⎩⎪⎨⎪⎧​y≥-3y≤-2x+9y≤2x+1​(I)(II)(III)​bLet's consider the diagram one last time, paying attention to the base and the height of the triangle.We see that both the base and the height are 8. Let's substitute 8 for b and h into the formula for the area of a triangle. A=21​bhh=8, b=8A=21​(8)(8) Simplify right-hand side MultiplyA=21​(64)b1​⋅a=ba​A=264​Calculate quotient A=32
Exercises 37 Let's go through the differences and similarities one at a time.Differences There are a couple of main differences we can think off Solving a system of linear equations we are looking for the point when the graphs intersect. Conversely, when solving a system of linear inequalities, we are looking for a range of solutions. While the solutions to a system of linear equations can be illustrated algebraically such as {y=3x=5,​the solutions to a system of linear inequalities is better illustrated graphically like below.Similarities There aren't that many similarities between solving systems of equations and inequalities. The main similarity is that we are looking for solutions that makes all equations/inequalities true simultaneously in our system.
Exercises 38 In order for C to be a solution to the system of inequalities, we have to shade the region to the right of both inequalities. Let's do that and see what happens.Let's isolate the shading where both inequalities apply.At the moment, the shaded area includes both B and C as solutions as B is on the boundary line of y=-3x+4. To exclude B, we have to use either > or < for this boundary line. But which one! Let's find out by substituting (4,5), point C, into y 22​ -3x+4. This should form a true statement. y 22​ -3x+4x=4, y=55 22​ -3⋅4+4 Simplify RHS Multiply5 22​ -12+4Add terms 5 22​ -8 In order to form a true statement we need to use > which means this inequality is y>-3x+4 For y=2x+1 there are no solutions on its boundary line so we can either keep the line solid or dashed. Let's find out which way the inequality symbol should point by substituting point C into y 22​ 2x+1. Again, this should form a true statement when substituted. y 22​ 2x+1x=4, y=55 22​ 2⋅4+1 Simplify RHS Multiply5 22​ 8+1Add terms 5 22​ 9 For the second inequality we can either use < or ≤ for it to be true. To summarize the inequalities should be either {y >​ -3x+4y <​ 2x+1​or{y >​ -3x+4y ≤​ 2x+1​
Exercises 39 Absolute value inequalities can be solved using a compound inequality. Let's start by considering an example.Example Consider an absolute value inequality. ∣k∣<3​ The above means that the distance to 0 from k is less than 3.We see that, for this example, k must be greater than -3 and less than 3. ∣k∣<3⇔k>-3 and k<3​Writing the System Let's apply the same reasoning to the given inequality, where x>0. ∣y∣<x⇔y>-x and y<x​ We can combine the two inequalities that form the above compound inequality to form a system. {y>-xy<x​(I)(II)​Graphing the System Let's consider both of these inequalities individually before we combine them to find the solution set for the entire system.y<x First, we need to graph the boundary line, which will be dashed because the inequality is strict. Remember that we are also told that x>0, and must restrict our domain accordingly.Now we must choose which side of the line to shade. We can choose any arbitrary point within the domain. Let's test the point (1,4). If it produces a true statement, we shade the region which contains the point. Otherwise, we shade the opposite region. y<xx=1, y=44≮1 × We should shade the side of the line that does not contain this point. Recall, again, that x>0!y>-x Once again, we need to graph our boundary line first. This line is also dashed because we have a strict inequality. Remember that we must restrict our domain according to the fact that x>0.Now we must choose which side of the line to shade. We can choose any arbitrary point within the domain to test. Let's test the point (3,-2). y>-xx=3, y=-2-2>-3 ✓ We shade the side of the line that contains this point. Recall that x>0!Combined solution set Now we can combine the graphs to see where they overlap.Finally, we can cut away all the unnecessary parts and leave only the solution set for the system of inequalities.
Exercises 40 The solution to a system of inequalities is the intersection of the solution sets of the individual inequalities. If a system has no solution, it must be true that there is no intersection and there are no common values. Consider a system in which the boundary lines are parallel. Let the equations of the boundary lines of the system be {y=x−4y=x+2​ Depending on the inequality symbols in both equations, there are four different possibilities for the solution to the system. The combinations are shown in the table below. Notice that we will only consider strict inequalities.PossibilitySystem 1{y>x+2y<x−4​(I)(II)​ 2{y>x+2y>x−4​(I)(II)​ 3{y<x+2y>x−4​(I)(II)​ 4{y<x+2y<x−4​(I)(II)​ Graphing the systems from the table will allow us to see their solutions.Possibility 1Notice that the shaded regions do not overlap. This system has no solution.Possibility 2The overlap shows the solution to the system.Possibility 3The overlap shows the solution to the system.Possibility 4The overlap shows the solution to the system.Conclusion From the graphs above, we can see that only Possibility 1 does not have a solution. Thus, just because the boundary lines of a system are parallel, it is not necessarily true that it will have no solution. Our friend is incorrect.
Exercises 41 For the solutions to be all real numbers, the intersection of the inequalities has to cover the entire coordinate plane. Given this piece of information, we know that none of the inequalities can have the symbols < or > as this would exclude numbers that are on the boundary line. Let's explore what happens if we have a system of inequalities where the inequality symbols arePointing the same way Pointing the opposite way.Pointing the same way Consider the system of inequalities {x≥0x≥2​ Let's graph both of these inequalities and shade accordingly.As we are only looking for the intersection of the inequalities we have to cut away the shading where both inequalities don't apply.As you can tell x<2 is not a part of the solution set. No matter how you draw the inequalities, when the inequality symbols point in the same direction you will never be able to cover the entire coordinate plane as you will always an unshaded region either to the right/left or above/below the inequalities where they don't overlap.Pointing the opposite way Consider the system of inequalities {x≥0x≤2​ Let's graph both of these inequalities and shade accordingly.Again, as we are only looking for the intersection of the inequalities we have to cut away the shading where both inequalities don't apply.As you can tell x<0 and x>4 are not a part of the solution set. No matter how you draw the inequalities, when the inequality symbols point in the opposite direction you will never be able to cover the entire coordinate plane as you will only shade the area that is in between the inequalities.
Exercises 42 In Quadrant I, both the x- and y- values are positive. There are infinitely many systems of inequalities we could write such that their solutions lie in Quadrant I. Here we will show only one of them. Consider the following system. {y>1x>2​(I)(II)​​ Graphing the system, we can see that, although the solutions to the individual inequalities lie outside Quadrant I, the solution set to the system is only in Quadrant I.This is easier to see if we cut away the shading where the inequalities don't overlap.
Exercises 43 Note that in Quadrant II and in Quadrant IV, the x- and y-values have opposite signs. Therefore, to satisfy the required condition, the solutions to the system of inequalities must lie in either Quadrant II or IV. Consider the following system. {x<0y>0​(I)(II)​​ Graphing the system, we can see that the solution set to the system is only in Quadrant II.This is easier to see if we cut away the shading where the inequalities don't overlap.
Exercises 44 To begin, we should acknowledge that there are infinitely many solutions to this exercise. For the solution of a system to have no solution, it must be true that the individual solution sets do not have any values in common. This means the shaded region of each inequality should extend in different directions. Let's write the inequalities so that the boundary lines are parallel. Perhaps, {y=x+2y=x−2.​(I)(II)​ Graphing the boundary lines gives the following.From the graph above, we can see that we need to shade above Inequality (I) and below Inequality (II) to ensure the individual solution sets do not overlap. Thus, our system becomes {y>x+2y<x−2.​(I)(II)​ Shading the appropriate regions, we can see our system has no solutions.
Exercises 45 aWe will start by graphing the given inequality. Then we can write a second inequality that doesn't have any common solutions with the given one.Graphing -4x+2y>6 To begin, we will isolate y as this will make graphing the boundary line easier. -4x+2y>6 Solve for y LHS+4x>RHS+4x2y>4x+6LHS/2>RHS/2y>24x+6​Write as a sum of fractionsy>24x​+26​ca⋅b​=ca​⋅by>24​x+26​Calculate quotient y>2x+3 Let's now write the boundary line. To do so, we replace the inequality sign with an equals sign. Inequalityy>2x+3​​Boundary Liney=2x+3​ The boundary line has a slope of 2 and a y-intercept of 3. Notice that we have a strict inequality, and so the boundary line will be dashed.To decide which side to shade, we can use a test point that is not on the line. If the inequality holds true when substituting the coordinates of this point, we will shade the region that contains it. Otherwise, we will shade the opposite region. For simplicity, let's use (-2,2). y>2x+3x=-2, y=22>?​2(-2)+3a(-b)=-a⋅b2>?​-4+3Add terms2>-1 ✓ The inequality holds true. This means we should shade the side of the boundary line that contains the point.Writing the Second Inequality For our second inequality not to overlap with the one above, the boundary lines must be parallel. Otherwise the lines will inevitably intersect and the solution sets will overlap. Moreover, to also avoid overlapping solution sets, the y-intercept of our line must be less than 3. Let's arbitrarily choose a y-intercept of -1. y=2x−1​ Let's draw this line on the same coordinate plane.To shade the region below the second boundary line, we have to determine the inequality sign. To do so, we will use the point (2,0). When substituted into the inequality, it must produce a true statement. Note that we do not really care if the line is dashed or solid. For simplicity, we will leave it solid and thus the inequality will not be strict. x=2, y=0MultiplySubtract term0≤3 The inequality sign is less than or equal to. Therefore, our inequality is y≤2x−1. Let's shade the region!Finally, we form our system of inequalities by combining the inequalities. {y>2x+3y≤2x−1​bIf any system of inequalities has a solution, it does in fact, have infinitely many solutions. This is because any region on a coordinate plane contains infinitely many points. In Part A we created a system with no solution. {y>2x+3y≤2x−1​(I)(II)​ If we flip the inequality sign of Inequality (II) so it points in the opposite direction, we will get overlapping regions.Let's finally cut away the non-overlapping region.The shaded region, which is the solution to the system, contains infinitely many points. {y>2x+3y≥2x−1​​ The system above has infinitely many solutions.
Exercises 46 Let's suppose that the gift card we were given is loaded with \$120. We can let x be the cost of a t-shirt and y be the cost of a sweatshirt. A situation in which we are able to buy 9 t-shirts and 1 sweatshirt can be written as the inequality 9x+y≤120 because we can afford things that total to less than or equal to the amount on the gift card. Conversely, a situation where we are unable to purchase 3 t-shirts and 8 sweatshirts can be written as 3x+8y>120 because we cannot afford a purchase that totals to more than the value of our gift card. Combining these inequalities, we get the following system: {9x+y≤1203x+8y>120​ To graph this system, we need to first create boundary lines by writing the inequalities in slope-intercept form. The first inequality already has a coefficient of 1 for y so we can simply subtract 9x from both sides: 9x+y≤120⇒y≤-9x+120 Now let's solve the second inequality for slope-intercept form. 3x+8y>120LHS−3x>RHS−3x8y>-3x+120LHS/8>RHS/8y>-83​x+15 Our system can now be written as: {y≤-9x+120y>-83​x+15​ Next, we can graph our boundary lines. Remember that the first inequality will have a solid line and the second will have a dashed line.The overlapping area shows us the possible price combinations of t-shirts x and sweatshirts y such that we could purchase 9 t-shirts and 1 sweatshirt but we could not purchase 3 t-shirts and 8 sweatshirts. To view only the solution set for the system, we can cut away the non-overlapping area.
Exercises 47 Most systems of linear inequalities have infinitely many solutions or no solution, an area overlaps or there is no overlapping area. However, it is possible to create a system so that it has exactly one solution? We should note that there are infinitely many possible answers. What follows is only one example. For simplicity, we are only going to use vertical and horizontal lines. If we write a system with four inequalities, two with vertical boundary lines and two with horizontal boundary lines, the point at which the lines intersect will be the singular solution to the system.First pair of inequalities We will begin with two inequalities. Suppose we have: {x≥4y≥5​(I)(II)​ Graphing these inequalities, we have the following.Notice that, as it stands, the solution to the system is the upper right-hand corner of the graph. This is where the two inequality graphs overlap.Adding the second pair of inequalities We can add two inequalities to the system that have the same boundary lines as the other two but are shaded in the opposite direction. Our system becomes: ⎩⎪⎪⎪⎨⎪⎪⎪⎧​x≥4y≥5x≤4y≤5​(I)(II)(III)(IV)​ Let's first look at the coordinate plane with only the third and fourth inequalities shown.For this portion of the system, the solution set is the lower right-hand corner. Now we can place them all on the same coordinate plane.The only point on the entire graph that remains a solution for all four inequalities is their mutual point of intersection, (4,5).