Solving Systems of Inequalities
Lesson 27

The warm-up for lesson 27 is to review graphing linear inequalities. We are going to review graphing linear inequalities because in lesson 27, we will be to be graphing systems of linear inequalities. Make sure to discuss with your students that they need to solve y=mx+b form in order to identify the slope and the y intercept. Also make sure that you discuss with your students the solid lines and dotted lines and how to shade. The solve problem for lesson 27, Brad has two summer jobs. He never works more than 30 hours in a week. He mows laws for $10 per hour and walks dogs for $5 per hour. If he must earn at least $100 per week, what graph shows the possible combinations of the number of hours he could work at each job? Our first step “S” study the problem, is to underline the question, what graph shows the possible combinations of the number of hours he could work at each job? Our second step is to answer the question, what is this question asking me to find? This problem is asking me to find the graph that shows the solutions to the inequalities. If we are going to graph the systems of inequalities, we will first look at our inequality, y is greater than -x + 3. We know that our slope is equal to -1, or -1 divided by 1. We also know that our y intercept is equal to 3 because the inequality is in y=mx+b form. From here we are going to create our line with the y intercept of 3 and a slope of -1 or down one, over one. We can connect our points but because of the inequality is greater than, we will use a dotted line. We have to decide which side of the line we are going to shade. If we test the point 0,0: 0 is greater than -0 + 3; is 0 greater than 3, no- that is not a true statement, so we will not shade the side with 0,0. We will shade this side. Our next inequality, y is less than or equal to x + 1 is also in slope intercept form. We know that our slope is 1, or 1 over 1. We also know that our y intercept or b is 1. We put a point on the y intercept. And we use our slope of up one, over one to create a line. Because it is less than or equal to, we can draw a solid line. We must shade a part of the graph. We can check the point 0,0 to see if it makes a true statement; is 0 less than or equal to 0 + 1; yes it is, so we will shade the side of the graph which contains 0,0. We see that we have created an overlapping section, where all of our solutions in the overlapping section will be solutions to both inequalities. If you chose a solution from the just orange section, it will not make a true statement in both inequalities. If you chose a point in the just purple section, it will not make a true statement in both inequalities. However, if you choose a solution from the overlapping section, it will make a true statement in both inequalities.

In example 2, we must first begin by converting each inequality to slope intercept form.

The first equation, we will subtract x from both sides, and then divide each term by -2 in order to isolate the y. When you divide an inequality by a negative, you must invert or flip the inequality. If we were going to graph, y is less than ½ x +1; our slope is equal to ½ ; and our y intercept is equal to 1. We will make a point on 1 and we will use our slope of up 1 and over 2 to graph other points. Because the inequality is less than, we will use a dotted line. We have to decide where to shade. If we plug 0,0 into the first inequality, is 0 greater than -2…the answer is yes, so we will shade this half of the graph, containing 0,0. In the next equation, we also have to isolate y, so we will subtract 2x from both sides; which leaves us with, y is greater than or equal to -2x -2; we have a y intercept of -2 and a slope of -2; which means we go down 2 and over 1, down 2 over 1. We use a solid line because the inequality is greater than or equal to. When we plug a point in to the original inequality, such as 0,0; 2 x 0 +0 is greater than or equal to -2; or 0+0 is greater than or equal to -2. That is a true statement, so we will shade the side of the inequality that contains 0,0. This creates three types of shading: the just purple shading, the purple and orange, and the just orange. Any of the points in the purple and orange are the solutions to this system of inequalities. You can plug any point in from the double shading and it will make a true statement in both inequalities. You cannot choose a point in the just purple or the just orange; it has to be from the purple and orange section.

In example 3, we will solve a system of inequalities. The first thing we must do is to change both of the inequalities so that they are in y=mx+b form, in order to graph the line. We will isolate the y by subtracting x from both sides; and dividing by -1; when you divide by -1, you have to flip the inequality; so we are left with, y is greater than x +3. To solve this, our slope is 1; and our y intercept is 3, we will rise one, run one in order to find other points. The inequality is less than so we must use a dotted line. And we will use the point 0,0 to check to see which side we should shade. 0-0 is less than -3; that is not true, 0 is not less than -3, so we will not shade the side of 0,0. To solve the next inequality, we will also isolate the y and divide by -1; y is less than or equal to x – 1. Our y intercept is at -1; and our slope of 1 and our y intercept of -1; our slope of 1 tells us we rise 1 and run 1; we go up one over one, up one over one, up one over one. You can draw a solid line because it is less than or equal to. And we check for point 0,0 to see if it makes a true statement; x-y is greater than or equal to 1; 0-0 is greater than or equal to 1; is 0 greater than or equal to 1…no, so we will not shade the side of 0,0. What is the solution to this system of inequalities? In the past two examples, the two shaded areas overlap, in this example the shaded areas do not overlap at all; therefore, the answer is no solution.

In our solve problem, we have already “S” the problem and we know that this problem is asking me to find, the solutions to the inequalities. In “O” organize the facts, we first identify the facts or strike the facts. Brad has two summer jobs. He never works more than 30 hours in a week. He mows laws for $10 per hour and walks dogs for $5 per hour. He must earn at least $100 per week. FACTS. We must now decide if the facts are necessary or unnecessary. Brad has two summer jobs. This fact is necessary. He never works more than 30 hours in a week. This fact is also necessary. He mows laws for $10 per hour, necessary, and walks dogs for $5 per hour. If he must earn at least $100 per week. We now can identify two variables to represent the number of hours he works at each job. Let x equal the number of hours mowing lawns; and let y equal the number of hours walking dogs. In “L” we line up a plan. We have to choose our operation or operations and because we are going to create both inequalities, and then graph them, we know that we can use addition, subtraction, multiplication, or division. In lining up our plan, we are going to create two inequalities, solve both inequalities for slope intercept form, graph the lines, shade each region, and find the intersection.

In V verify your plan with action we will first estimate our answer. We know that he is not going to work more than 30 hours a week between both types of jobs, so we know that both coordinates will be less than 30. We have to create two inequalities, we know that x = number of hours mowing lawns. And y= number of hours walking dogs. He never wants to work more than 30 hours. So x+y have to be less than or equal to 30 hours. A lot of times students get confused with the wording but he never wants to work more than 30, so that means x+y has to be less than or equal to 30. he also has to make at least 100 dollars per week he earns 10 dollars an hour mowing lawns and 5 dollars and hour walking dogs and he has to make this value has to be greater than or equal to at least 100 dollars. From here we will solve both inequalities for slope intercept form. We will subtract x from both sides and y is less than or equal to –x +30. We subtract -10x from both sides and divide all terms by 5 so y is greater than or equal to -2x +20.

Our first inequality that we are going to graph is y is less than or equal to –x +30. We have a slope of -1 and a y-intercept of 30. Put a point at 30 and if we have a slope of -1 that means we go down one over one that also means we can go down ten over ten. Because it is less than or equal to we connect the points with a solid line. And we must shade half the graph so we have to test a point I am going to test 0, 0 and 0 is less than or equal to 0 plus thirty... that is true so we shade the side that contains 0, 0. Our second inequality is y is greater than or equal to -2x+20. This means we have a slope of -2 and a y – intercept of 20. Put a point at 20 with a slope of -2 this means we go down two over one in this case we will go down twenty and over ten. It is greater than or equal to so we will connect our points with a solid line. And we have to decide where we are going to shade if we plug in 0, 0 is 0 greater than or equal to 0 + 20? No it is not so we shade the side that does not contain 0, 0 so you can see that your solution to the system of inequalities in right here where it is doubled over.

In E we will examine our results. Does your answer make sense? If we go back to our question, what graph shows the possible combinations of the number of hours he could work at each job. We did make a graph and we did find the intersection of the two inequalities, so our answer does make sense. Is your answer reasonable, our estimate says that most coordinates must be less than 30 we look at our graph we can see that all of the coordinates are less than thirty on the x and y axis in the shaded region. Is your answer accurate? We can choose a point from our shaded region such as 10, 10 to check and see if it is a solution to both inequalities. Write the original inequalities, and we are going to plug in the point 10, 10 10+10 less than or equal to 30 yes, 20 is less than or equal to 30 so it is a correct solution in the first inequality. 10 times 10 plus five times 10 is greater than or equal to 100, 10 times 10 is 100 plus five times ten is 50 100 plus 50 is a 150 is greater than 100. So our answer checks and it is accurate and now we are going to write our answer in a complete sentence. Any point in the overlapping shaded region such as (10.10) is a solution.

To close the lesson with the essential questions, number 1 why is it important to know if a line showing the solution to an inequality is solid or dotted? If the line is solid the points on the line are solutions, if the line is dotted, the points on the line are not solutions. Question 2 are all solutions of the first inequality solutions of the second inequalities? No, some points can be solutions for both inequalities but they don’t have to be.

In lesson 27, we can graph a system of inequalities. We will graph inequalities using the same method that we used to graph just one inequality, only this time we will use 2. You will press y equals; and on your screen, you can see that you can enter several equations, or in this case inequalities. We will enter our two inequalities: y= -x+3; in order to change it to y is greater than –x+3, we will arrow over to the left, to the line, if we press enter once it’s a dotted line, and if we press enter again, it is greater than. So now we have y is greater than –x+3. If we arrow down and enter our second equation as, y= x+1; we have to change it to the inequality by arrowing to the left to the line and you press enter once for a bold line, and you press enter twice for greater than, and if you press enter one more time, you will have less than. Please make sure your students know that using this method, we can not distinguish between a greater than or less than or greater than or equal to or less than or equal to. They must know which is the solid line and which is the dotted line. Once we have entered our equations, we press graph and we can see all the points in the double shaded area our solutions to the system of inequalities.