### Graphing Inequalities in Two Variables

- To graph an inequality in two variables begin by graphing the corresponding line. It should be dotted if the inequality is strict, and solid if the inequality has “or equal.”
- To graph the line, remember you can use slope-intercept form with the y-intercept and the slope to graph it quickly.
- Shade on one side of the line, or the other. A test point can be used to determine which side of the line should be shaded.

### Graphing Inequalities in Two Variables

- Find slope and intercept

b = - 2

m = 1 - Determine line type
- Dashed since >
- Test point (0,0)

0 > 0 - 2

0 > - 2

Which is false

- Find slope and intercept

b = 0

m = 2 - Determine line type

Dashed since < - Test point (1,0)

0 < 2(1)

0 < 2

Which is true

- Find slope and intercept

b = 5m = 1 - Determine line type
- Solid since ≥
- Test point (0,0)

0 ≥ 5(0) + 1

0 ≥ 1

Which is false

- Find slope and intercept

b = 2

m = [1/3] - Determine line type

Solid since - Test point (0,0)

0 ≥ [0/3] + 2

0 ≥ 2

Which is false

- Find slope and intercept

b = 7

m = 1 - Determine line type
- Solid since £
- Test point (0,0)

0 ≤ 0 + 7

0 ≤ 7

Which is true

- Find slope and intercept

b = 5

m = 1 - Determine line type
- Dashed since >
- Test point (0,0)

0 - 0 > 5

0 > 5

Which is false

- Find slope and intercept

b = − 1

m = − [1/2] - Determine line type

Dashed since < - Test point (0,0)

2(0) + 0 < - 1

0 < - 1

Which is false

- Find 2 points using intercepts

x = 0 → (0,3)y = 0 → (3,0) - Determine line type

Dashed since > - Test point (0,0)

0 + 2(0) > 3

0 > 3

Which is false

- Find 2 points using intercepts

x = 0 → (0, − [5/3])y = 0 → (5,0) - Determine line type

Dashed since > - Test point (0,0)

3(0) - 0 > - 5

0 > - 5

Which is true

- Find 2 points using intercepts

x = 0 → (0,1)y = 0 → ([4/5],0) - Determine line type

Solid since ≤ - Test point (0,0)

[(4(0) + 5(0))/2] ≤ 2

[0/2] ≤ 2

0 ≤ 2

Which is true

*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

### Graphing Inequalities in Two Variables

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 0:07
- Graphing Inequalities in Two Variables 0:32
- Split Graph into Two Regions
- Graphing Inequalities 5:44
- Test Points
- Example 1 7:11
- Example 2 10:17
- Example 3 13:06

### Algebra 1 Online Course

### Transcription: Graphing Inequalities in Two Variables

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

*In this lesson we are going to look at graphing inequalities using two variables.*0002

*In order to make this process work out , we will first briefly review how to just simply graph lines.*0009

*We will see how this works with our inequalities which involved entire regions of our graph.*0016

*The important parts for highlighting these regions will be shading in the proper part of the graph.*0022

*Watch for these things to come into play.*0028

*Recall it when you are dealing with any inequality it is not just usually an isolated point, it is usually a whole range of solutions.*0035

*When we are working with graphing, and we have more than one variable to worry about we are dealing with points like x and y.*0043

*If we are looking for the x and y that will make an inequality true, we are not talking about just one point on a graph,*0052

*we are actually talking about an entire region of the graph, and all the little points in that region.*0058

*The way we do this is when you start off by graphing the equation, corresponding with the inequality.*0065

*Let me show you what I mean by that.*0071

*I’m looking at the inequality y > or = 6/5x + 7 rather than trying to tackle on the entire inequality first,*0074

*I looked at the corresponding equation like y = 6/5x +7.*0086

*In doing so, I can look at my techniques for graphing just that equation.*0093

*Maybe find the slope and find the y intercept.*0098

*We will see how that equation connects back to our inequality.*0106

*What we are going to find is that equation is going to split our entire graph into two regions.*0115

*In one of the regions I will involve our solution and the other region will not.*0121

*I have cooked up another example, so you can see how this works.*0126

*We are going to start in graphing y = 2x -4.*0131

*The first thing we wanted to do is look at the corresponding equation y = 2x – 4 and focus on that first.*0141

*We can see that it has a y intercept and it has a slope.*0151

*I'm going to end up graphing this line, and I'm going to graph it using a nice solid line.*0158

*I will get into that y in just a bit.*0163

*y intercepts way down here at -4 and a slope of 2.*0166

*From there up 1, up to over 1 and now we have our line.*0172

*This line is going to split the entire graph in to two regions.*0190

*We had this region on the left side of it or the region on the right side of it.*0195

*One of those regions will be our solution for our inequality.*0201

*The question is, in which one should it be?*0205

*Since this is the entire region, we can pick out an entire or just one little point from that and see if that region should be included or not.*0208

*To get all points inside of the region is either solution or all points inside the region are not a solution.*0217

*We need one to the find out.*0223

*I'm going to borrow a point, let us take this guy right here.*0226

*I’m going to take a point and test it into my inequality, y =4 and x = -2.*0236

*I will put that little question mark in there because I want to see if this is true or not.*0249

*Is 4 > or = 2, 2 × -2 = -4 + 4?*0253

*That is 4 > or = -8, yes it is, 4 > or = -8.*0260

*I know that, that particular point should be in my solution region.*0269

*To highlight that it is in my solution I will go ahead and shade everything on that side of the line.*0275

*Everything on that side, shade it.*0297

*Everything that is shaded in red and including that line on my graph represents a solution to the original inequality.*0301

*When you have an inequality that has the or equals to it, we usually use a solid line to show that everything on that line is actually one of our solutions.*0311

*Even if I use to pick like a point right on that line it should satisfy the inequality.*0322

*If it was a strict then we will use a nice dotted line to show that the line should not be included.*0328

*Now that we have the basics to graphing out one of these regions, let us give it a few more try.*0338

*We know that our equation will split the region up in the two regions and that the equation will either have to be solid or dash.*0348

*It is determined by the type of inequality used.*0358

*If we are dealing with a strict inequality like strictly less than or strictly greater than,*0363

*feel free to use a dotted line because you should not include the points on that line.*0367

*If you have the or equals to in there and use a nice solid line to show that the line is included.*0373

*To determine the region, shade on one side or the other, use a test point to help you out,*0381

*that determines is that point and the region it is in included or not.*0388

*Pick something inside the region, you do not want to pick something on the line.*0395

*If the point satisfies your inequality, go ahead and shade that entire region that it is in.*0399

*If it does not satisfies then feel free to shade the other side, which it is not in.*0405

*One last thing if I could add to this, go ahead and pick something easy.*0410

*Potentially you could pick any point in one of these regions and you have some freedom, pick something that is nice and easy to evaluate.*0420

*I think we have our tools, let us go ahead and go to one of our examples.*0428

*We want to graph the inequality 3x -2y > 6.*0435

*First, I'm going to look at this as if it was a line.*0440

*Let us use our little equal sign in here and that is the corresponding equation.*0446

*The tools that I will use on this one is, maybe I will look at the intercepts.*0450

*What happens when x = 0 and y = 0.*0456

*Plugging in 0 for x, I would have 0 × 3 =0 and that term would drop away.*0459

*Then I divide both sides by -2 giving me -3.*0470

*I have one of my points, let us go ahead and put that on there 0, -3.*0479

*Let us put in a 0 for y.*0487

*It looks good -2 × 0 = 0 so that is gone and dividing both sides by 3 we will get 2.*0495

*I have another point I can go ahead and put on this graph 2, 0.*0508

*I’m going to use these points to graph the line but I’m going to be very careful to make it a dotted line since our inequality is a strict inequality.*0514

*I do not actually want to include this.*0522

*This looks good, nice big dotted line.*0541

*We have to figure out which region we should shade, either on the left side of this or on the right of this.*0545

*To help out, I'm going to borrow a test point just to see what is going on.*0551

*We could choose anything on here but I’m going to choose something nice and simple.*0556

*We are going to choose the origin since they are 0 for x and 0 for y.*0561

*Let us see if this works.*0566

*0 for x and 0 for y is that greater than 6?*0568

*0 × 3 and 0 × 2 both of those would cancel.*0576

*Is 0 > 6? I am afraid not, it is not greater than 6.*0580

*That tells me that origin is not in my solution region and I should shade the other side of the line.*0586

*Let us do that then.*0593

*This entire region would be considered as our solution region and we have graphed the inequality.*0604

*Let us try another one of these.*0619

*This one is to 2y > or = 4x -2.*0620

*Let us start off like before, looking at the corresponding equation and see if we can graph that .*0626

*2y = 4x -2 this one is almost in slope intercept form, I’m going to divide everything by 2 and put it into that form y = 2x – 1.*0635

*Now I can see that it has a y intercept at -1 and a slope of 2.*0649

*Let us put those points on them.*0656

*-1 and from there I will go up to over 1 and get a few more points.*0661

*I can definitely graph out my line and I will make it a solid line, since my inequality has the or equals to.*0669

*There we are, almost done.*0687

*We just have to figure out what side we should shade on.*0690

*Let us pick a nice easy point that we can evaluate and see where we go from there.*0695

*I already used the origin last time, so I'm going to pick something a little bit farther away.*0699

*Let us try this point right here, -2, 2.*0706

*When y = 2 is this greater than or equal to 4 × -2 -2.*0715

*I’m going back to the original.*0730

*I have put in the point into x and y.*0731

*2 × 2 = 4, 4 × -2 = -8 -2, is 4 > or = -10.*0735

*4 is positive and -10 is negative, positive number should be greater than negative numbers.*0747

*I know that point is in my solution region, I should shade everything on that side of the line.*0752

*This one we are not only including the entire region that I'm shading in right now but we are also including everything on that line since it is nice and solid.*0763

*There is our inequality all graphed out.*0782

*In this last example, let us go ahead and work in the other direction.*0786

*You will notice that I have an inequality I already graphed.*0790

*We are going to try and create what that inequality was.*0793

*When we are going and building these from the ground up, we focus on the equation associated with it first.*0798

*We are going to start in the same spot.*0805

*When you look at this dotted line and I can see that it has a y intercept up here at 2.*0807

*I think I can also interpret it slope.*0813

*To get back on the line I have to go down 2 and to the right 3 to get back on the line.*0816

*I definitely know its slope as well.*0823

*The equation of the line be y = -2/3x + 2, that worked out pretty good.*0826

*I can see that it is dotted so whatever inequality I’m going to deal with next should be nice and strict, either less than or greater than.*0838

*The question is which one should I include?*0848

*I want to make sure that it has the proper solution region.*0850

*In order to help me figure this out, I’m going to borrow a point from inside that solution region.*0854

*Let us borrow the point 2, 2.*0860

*I’m going to plug it into both sides of the equation and figure out what inequality symbol should go in between and make the rest of the statement true.*0863

*I have 2 for y and I will put in 2 for x.*0872

*Let us just go ahead and simplify this and see how this turns out.*0879

*2 is already simplified, over in this side I have -4/3 + 2, I have 2.*0883

*-4/3 + 2 I will have 2/3 left.*0892

*What is greater 2 or 2/3?*0899

*That does not take too long to figure out, 2 is definitely greater.*0903

*We will go back and put it into all of the spots all the way back up here and that way we can actually build our inequality y > -2/3x + 2.*0908

*That is the one represented by this graph.*0924

*You can notice how I’m using a strict inequality because it is a dotted line.*0927

*Thank you for watching www.educator.com.*0931

1 answer

Last reply by: Professor Eric Smith

Wed Sep 3, 2014 1:36 PM

Post by sherman boey on August 13, 2014

for example 1 u did not use dotted line

1 answer

Last reply by: Professor Eric Smith

Sun Jul 6, 2014 2:36 PM

Post by patrick guerin on July 6, 2014

Thank you for the lecture!