### Systems of Inequalities

- To graph a system of inequalities, graph each inequality separately.
- The solution region of the system is where the individual graphs overlap.
- You may use a test point from the overlapping region in the original system to see if the solution region is correct.
- Look for word problems that involve a range of possibilities to determine if an inequality would be more appropriate when setting it up.

### Systems of Inequalities

y â‰¤ x

y <âˆ’ x + 2

- Graph y â‰¤ x

b = 0m = 1

Use test point (1.0) - Graph y <âˆ’ x + 2b = 2m = âˆ’ 1

Use test point (0,0)

y > x y <âˆ’ 2x

- Graph y > xb = 0m = 1

Use test point (1.0) - Graph y <âˆ’ 2xb = 0m = âˆ’ 2

Use test point (1,0)

y > 3x + 1

y â‰¥ x âˆ’ 1

- Graph y > 3x + 1b = 1m = 3

Use test point (0.0) - Graph y â‰¥ x âˆ’ 1b = âˆ’ 1m = 1

Use test point (1,0)

y â‰¤ 3

x â‰¥ 4

- Graph y â‰¤ 3

Horizontal line at y = 3

Use test point (0.0) - Graph x â‰¥ 4

Vertical line at x = 4

Use test point (1,0)

y â‰¤ [x/2] âˆ’ 3

y < 5

- Graph y â‰¤ [x/2] âˆ’ 3b = âˆ’ 3m = [1/2]

Use test point (0.0) - Graph y < 5

Horizontal line at y = 5

Use test point (1,0)

y â‰¤ 3x âˆ’ 9

y â‰¥ âˆ’ [x/3]

- Graph y â‰¤ 3x âˆ’ 9b = âˆ’ 9m = 3

Use test point (0.0) - Graph y â‰¥ âˆ’ [x/3]b = 0m = âˆ’ [1/3]

Use test point (1,0)

y â‰¤ âˆ’ 4

y > 1

- Graph y â‰¤ âˆ’ 4

Horizontal line at y = 4

Use test point (0.0) - Graph y > 1

Horizontal line at y = 1

Use test point (1,0)

No Solutions

y > 4x âˆ’ 2

y > x + 7

- Graph y > 4x - 2

b = - 2

m = 4

Use test point (0.0 - Graph y > x + 7

b = 7

m = 1

Use test point (1,0)

y + x < 7

2y + 2x >âˆ’ 4

- Graph y + x < 7x = 0 â†’ (0,7)y = 0 â†’ (7,0)

Use test point (0.0) - Graph 2y + 2x >âˆ’ 4x = 0 â†’ (0, âˆ’ 2)y = 0 â†’ ( âˆ’ 2,0)

Use test point (0.0)

2y âˆ’ 3x â‰¤ âˆ’ 6

6y + 4x > 12

- Graph 2y âˆ’ 3x â‰¤ âˆ’ 6x = 0 â†’ (0, âˆ’ 3)y = 0 â†’ (2,0)

Use test point (0.0) - Graph 6y + 4x > 12x = 0(0,2)y = 0(3,0)

Use test point (0.0)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Systems of Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:08
- Systems of Inequalities 0:24
- Test Points
- Steps to Solve Systems of Inequalities
- Example 1 2:23
- Example 2 7:28
- Example 3 12:51

### Algebra 1 Online Course

### Transcription: Systems of Inequalities

*Welcome back to www.educator.com.*0000

*In this lesson we are going to wrap things up with inequalities by looking at systems of inequalities.*0002

*We do not have a whole lot to cover but we will look at how you can satisfy a system of inequalities rather just individual inequalities.*0011

*That will bring us to looking at overlapping solution regions.*0019

*Recall it when we are usually dealing with the system, such as a system of equations,*0027

*we want to make sure that all of the equations in the system are satisfied.*0031

*When we are dealing with the system of inequalities, rather than equations, but it is the same goal.*0037

*We want to make sure that every single inequality in here gets satisfied.*0042

*This makes it a little unusual because rather than just looking for individual points that would satisfy inequality*0048

*remember that each of these is a region on our graph.*0054

*What we are looking for then is where these regions for each of them would end up overlapping.*0059

*That overlapping region would be where the entire system is satisfied.*0065

*Like before, after we are done identifying an overlapping region, we can use a test point to go ahead*0070

*and figure out whether that is the correct overlapping region.*0077

*It is not about process, we just have to get into figuring out what those regions are.*0081

*What we are going to do to basically figure out the overlapping region is, first we will graph each individual inequality in that system.*0090

*That breaks it down into figuring out the equation and figuring what side is shaded on that line.*0099

*Be very carefully that you are using either a dotted or dash line depending on the type of inequality in there.*0107

*We are using the dotted for our strict inequalities and we are using the nice solid line for our or equals to.*0113

* Since the line could be included as a solution.*0122

*Individually, we will make sure we shade on one side or the other and we can check that out using a test point.*0126

*Here is the big one that we have for our system, look for where the two regions overlap so we know where the solution to the system is.*0132

*That is a little bit of background, let us see these examples and see how this works in practice.*0145

*In my system of inequalities, I have 2x + y < or = 1 and I have x > 2y so individually let us take a look at these.*0150

*The corresponding equation to the first inequality would be 2x + y = 1.*0161

*If you want to graph that one out, you could get that one into slope intercept form by moving the 2x to the other side by subtracting a 2x.*0169

*From this I can see it has a y intercept of 1 and the slope of -2.*0181

*From that point I go down 2/1, down 2/1.*0186

*Looking at the original I can see it involves an or equals to so I'm going to use a nice solid line.*0192

*Now we have our equation, we need to figure out what side to shade it on.*0212

*We usually grab a test point to figure out what it should be and it is good to grab a nice test point that is to easy to evaluate.*0216

*Iâ€™m going to borrow the point 0, 0 here and see if it works in the original.*0224

*2x is 0, y is 0 is that less than or equal to 1, let us find out.*0231

*2 Ã— 0 + = 0 and 0 < or = 1.*0240

*I know I should shade on that side of the line.*0248

*This entire region that we can see right now is only for the first inequality in our system.*0255

*Simply just going to go through the same process for the second inequality and figure out where both of their regions overlap.*0265

*Let us get some space and let us see this process for the second equation.*0278

*The second equation is x > 2y.*0289

*Maybe I will find this one in its slope intercept form so I will move the two together side by dividing everything by 2.*0293

*I could look at this as the equation y = Â½ x, it has a y intercept of 0 and a slope of 1/2.*0305

*It goes through the origin 0 up 1/2, up 1/2, up 1/2, up Â½.*0317

*Let us use a dotted line since the inequality is strict and see what we have.*0325

*We just need to shade on one side or the other.*0348

*We could test point for this particular line, let us just choose one way out here.*0353

*This one is at 3, 3.*0358

*Plug 3 in for y and plug 3 in for x and we will see if it is true not.*0363

*Is 3 < 3/2? That is not looking so good, that is not true.*0376

*What that means is we should actually shade on the other side of that line.*0386

*Let us shade everything below it.*0390

*Now comes the very important part of this.*0402

*We can see where each individual inequality is being satisfied but since we are interested in the region where the entire system is being satisfied,*0406

*we are looking at the overlapping regions.*0418

*Iâ€™m going to highlight that.*0421

*This region right down here is where the two overlap and I can see along one of the borders, it is dotted and along the other border it is solid.*0424

*This yellow area and its borders would be considered the solution to the system.*0440

*Let us try another of these and be careful as we walk through the process so we can see where that overlap is.*0450

*In this one we are dealing with the system 3x - 2y > 12 and 5x - y < 6.*0455

*Let us grab the first one and look at its corresponding equation.*0463

*The form of this ones in would be nice if we just use our intercepts to go ahead and see where it crosses the x and y axis.*0477

*I will put in a 0 for x and 0 for y.*0486

*If x = 0 what is y?*0491

*This term would go away for sure and I can divide both sides by -2 so y = -6.*0496

*Let us go ahead and put that on our graph, 0, -1, 2, 3, 4, 5, 6 right there.*0507

*We will go ahead and put in a 0 for y.*0515

*Let us see how this one turns out, 2 Ã— 0 = 0, 3x = 12 divide both sides by 3 now and we will get x = 4.*0524

*4, 0 is our other point.*0536

*Now that we have 2 points, let us go ahead and connect it.*0541

*We will connect it using a dotted line and then we will figure out what side we should shade this on.*0544

*To determine what side to shade it on, let us borrow a nice, good test point.*0564

*Let us borrow the origin add 0, 0 and see if it is true or not.*0569

*(3 Ã— 0) â€“ (2 Ã— 0) > 12? That would simplify to 0.*0574

*Is 0 > 12? No.*0583

*Let us shade the other side.*0587

*That entire side would be our solution region or least for the first one.*0602

*Let us get some space and do the same exact process for our second equation.*0608

*The corresponding equation would be 5x - y = 6.*0622

*This looks like maybe be easier to graph it if it was in slope intercept form.*0629

*Iâ€™m going to move the 5x to the other side and then multiply everything by -1.*0634

*Okay, my y intercept is way down there at -6 and it has a slope of 5 so up 1, 2, 3, 4, 5/1, up 1, 2, 3, 4, 5/1.*0649

*I will graph this one using a dotted line.*0669

*I have a few regions that it might be if we want to figure out until we shade one side of the blue line or the other.*0692

*Let us grab a test point for it.*0701

*I think origin 0, 0 is going to work out very nicely for this one as well.*0702

*(5 Ã— 0) â€“ 0 is that less than 6?*0708

*The entire left side simplifies to 0 and 0 < 6.*0713

*Let us go ahead and shade everything on that side of the line.*0720

*There is a lot of region over here to shade.*0734

*Now comes the important part where exactly do both of these overlap?*0744

*This is a little tough to tell, but if you look near the bottom you can see that there is the small triangle here where both of them overlap.*0751

*We would have that as our solution region.*0761

*As long as it is in the overlapping portion we are good to go.*0765

*For this last example I wanted something that was a little bit more like a word problem.*0773

*You could see how inequalities have a lot of good applications in them*0777

*and how some more problems lead to the rise of a system of inequalities rather than a system of equations.*0781

*What we want to do for this one is graph the solution region for the system and interpret what that means in the context of the problem.*0789

*What we are looking to do is recreate what it means when a person's heart rate,*0797

*or a person's maximum heart rate 220 â€“ x, where x is the personâ€™s age.*0802

*This one is good as long as it is for people between 20 and 70 years old.*0808

*When a person exercises and are trying to lose some weight, it is recommended that the person strive for heart rate*0815

*that is at least 60% of the maximum and at most 70% of the maximum.*0820

*There is a little bit of a range for that maximum heart rate that we are aiming for.*0826

*Let us see if we can interpret some of the things in here and create an inequality for what we have.*0832

*First of all, let us talk about a person's age.*0839

*Earlier it says that x is our person's age and that we are only interested when x is in between 20 and 70.*0844

*Let us say x > 20 and x < 70.*0856

*There are two inequalities that we can put in our system.*0864

*We also want to look at a person losing weight.*0868

*Also, it recommends that they strive for 60% of their maximum and no more than 70% of their maximum.*0873

*What exactly is their maximum?*0880

*It is way up here at this expression, 220 - x.*0883

*60% of their maximum 3.60 (220 â€“ x) and we want to make sure their heart rate is greater than that.*0890

*You do not want to stress them too much, so we will make sure that it is less than .70 of the maximum heart rate.*0915

*What we now have here are 1, 2, 3, 4 different inequalities in our system that we will be looking to satisfy.*0924

*And we will graph this out so we can see the region it creates.*0934

*Let us end up rewriting this just so we have them handy.*0943

*x must be greater than 20, x is less than 70,*0947

*and we also have two more, H is greater than (.6 Ã— 220) - x and H is less than (.7 Ã— 220) â€“ x.*0955

*If I'm going to graph these other two here I want to get them in a better form.*0974

*It is tough to see maybe what their slope is and what their y intercepts is.*0980

*Let us do a little bit of work with distributing that .6 and .7 in so we have a better idea.*0985

*This one will be H > .6 Ã— 220 = 132 - .6x and for this other one we have 154 -.7x, that gives us a much better idea.*0995

*I think Iâ€™m going to write them again so I can put that x term first and see that is my slope.*1020

*We will end up graphing these two inequalities in our system as well as these two on our system.*1039

*Let us see how we can do that.*1046

*With the first two that have to deal with the age of the person, we will use that along our x axis.*1048

*We want the age of the person to be between 20 and 70.*1059

*Let us use a nice vertical line and Iâ€™m going to make this dotted because my inequality is strict.*1068

*Let us do the same thing for our 70 here.*1080

*If I was looking to just satisfy those first two then I would have to be between of these two lines.*1092

*Let us work in what we want their heart rate to be and see how that fits into this.*1104

*In the first one I can see it has a y intercept of 132 and the other one has a y intercept of 154.*1113

*My slope on the first one is a -.6x so since it is negative it should be going down, maybe something like this.*1127

*In the other one is very similar its slope is negative as well, but it is a little bit steeper since its .7 and we will put that on.*1153

*Now where should the shading for this one be?*1174

*We want our heart rate to be more than the .6x + 132.*1177

*We want to think of that as our minimum heart rate that we are aiming for in terms of exercise.*1183

*The other one would be a .7x + 157 we want them to be no greater than that value.*1189

*We are looking in between these 2 lines again.*1195

*From looking at these two solution regions, it is this little box looking thing where they end up overlapping.*1202

*We can interpret what that solution region means in terms of the problem.*1216

*If you we are to just pick a random point inside that box, what we are getting is maybe an age of a person,*1221

*let us say 65 and we are also getting what the heart rate should be when they exercise to try and lose weight.*1231

*As long as they stay inside that box they should do some good things to their heart.*1239

*If we try and fall outside of the solution region that means either the formula is for our maximum heart rate no longer apply*1245

*or we would end up overtaxing the heart and either one we definitely do not want.*1254

*When setting up a system of inequalities look at them individually and graph out each of the regions.*1261

*Then look at the overlap and see what would be included in both.*1267

*Thank you for watching www.educator.com.*1271

2 answers

Last reply by: Sage Stark

Mon Jul 24, 2017 2:41 PM

Post by Sage Stark on July 24 at 12:35:00 PM

Is there any algebraic way of solving systems of inequalities, or is graphing the only way?

1 answer

Last reply by: Professor Eric Smith

Sat Aug 10, 2013 1:05 PM

Post by Emily Engle on August 10, 2013

How would you find the numeric values of the solution of the inequalities ? Â