Main content

## Integrated math 1

### Course: Integrated math 1 > Unit 7

Lesson 2: Graphing two-variable inequalities- Intro to graphing two-variable inequalities
- Graphing two-variable inequalities
- Graphs of inequalities
- Two-variable inequalities from their graphs
- Two-variable inequalities from their graphs
- Intro to graphing systems of inequalities
- Graphing systems of inequalities
- Systems of inequalities graphs
- Graphing inequalities (x-y plane) review

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Graphing systems of inequalities

Sal graphs the solution set of the system "y≥2x+1 and y<2x-5 and x>1.". Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Do you have an easier way to know which side to shade ?(30 votes)
- My method is to pick a point which will definitely lie on one side or the other (not on the line) and determine if it fits the equation. You can pick a point which is really easy; usually the origin is a good one.

For example, if we start with:

7y < (3/2)x + 5

It seems annoying. Sub in the origin (0,0) and we get:

0 < 0 + 5, or 0 < 5.

That is, the xs and ys just disappear! Easy sauce! This is true, (0 is less than 5), so the side with the origin should be shaded.

Just remember to be careful with sign. For example:

7y < (3/2)x - 5

becomes:

0 < -5

Obviously false - don't shade this side. But it is easy on a quick glance to forget that 0 is actually more than -5. Sounds silly, but it's one of those silly mistakes I make - a LOT.

Hope this helps!(50 votes)

- im confused on how you new which way the coordinate of x>1, at about 3:2(16 votes)
- x=1 would be graphed as a vertical line that is on crosses the x axis at 1. Skip the rest of this paragraph if that already clicks for you. If not, you could also think of it as taking any y, the x coordinate =1, so pick any two y such as 2 and 3. Since you know x always equal 1, then you get the two points (1,2) and (1,3). If you graph the line through these two points, You will see that you get the vertical line going through the point (1,0).

So now since the inequality is > and not greater than or equal to, you use a dashed vertical line.

And not for what you asked. To figure out which side to shade, when x > 1, you can choose any point where x is greater than 1 such as (3,3) or (2,-1) and graph that point. Since that is a point you want to include, and you see that point is on the right, you would shade the area on the right. After a couple times it will just click that x > any number is a dashed vertical line at that the point (0,that number) shaded on the right.

I hope that helps.(26 votes)

- Why is my graphing calculator making X>1 different than the way your doing? It's making a line on Y 1. Please help if this makes any sense to anyone who reads this.(12 votes)
- I believe that you have to type it in a different way.(2 votes)

- How do you tell which side of the line that you shade?(12 votes)
- For Example:

y is equal to or GREATER than 2x+1

since y is greater than the line itself or the points on the line, you would shade up.

x is equal to or LESS than 1

since we are talking about s values, we should shade right or left not up or down. Also since x is LESS than one we should shade everything to the left of one because everything to the left of one is less than 1.

Hope that helps :)(13 votes)

- Is there a way to solve a system of inequalities without graphing?(6 votes)
- What if y has a number next to it like for example 3y, but has the other variable without a number...like 3y < -x-1 ....what you do then(6 votes)
- Just divide both sides by 3 to get rid of the y's coefficient. :) So...3y < -x-1 would be y < (-x-1)/3(5 votes)

- So we just memorize what goes on top and bottom? Any tips/ tricks?(4 votes)
- 1015809,

Yes. Memorize these facts:

If the inequality is < or > (with no*equal to*), the line is dashed.

If the inequality is <= or >= (contains*equal to*), the line is solid.

If the inequality is < or <=, shade*below*the line.

If the inequality is > or >=, shade*above*the line.(6 votes)

- how do you know if you shade above or below?(4 votes)
- Try one "test" point and see if it works. If it does, you shade the side that point is on. If it doesn't, you shade the other side. For example, if you have y>5, then if your test point is y =6, you find 6>5, which is true, so you shade that side. If you chose y = 4 for your test point, then you have 4 >5, which is not true, so you shade the other side.(6 votes)

- I still don't understand which part of the graph to shade..heellpp!(3 votes)
- if y is greater than mx+b, you shade the higher side and if the slope is nearly vertical, shade the right. if y is greater and equal, it is a solid line, if y is just greater than, it is a striped or dotted line(6 votes)

- How would u graph a problem with the equation of 3x<y(2 votes)
- The equation " 3x < y " would have the following graph:

It would have a y-intercept of 0 and increase at a rate of 3/1.

All the values higher then the line would be filled in.(7 votes)

## Video transcript

We're asked to determine the
solution set of this system, and we actually have three
inequalities right here. A good place to start is just to
graph the solution sets for each of these inequalities and
then see where they overlap. And that's the region of the
x, y coordinate plane that will satisfy all of them. So let's first graph y is equal
to 2x plus 1, and that includes this line, and then
it's all the points greater than that as well. So the y-intercept
right here is 1. If x is 0, y is 1, and
the slope is 2. If we move forward in the
x-direction 1, we move up 2. If we move forward 2, we'll
move up 4, just like that. So this graph is going to look
something like this. Let me graph a couple more
points here just so that I make sure that I'm drawing
it reasonably accurately. So it would look something
like this. That's the graph of y is
equal to 2x plus 1. Now, for y is greater than or
equal, or if it's equal or greater than, so we have to put
all the region above this. For any x, 2x plus 1 will be
right on the line, but all the y's greater than that
are also valid. So the solution set of that
first equation is all of this area up here, all of the area
above the line, including the line, because it's greater
than or equal to. So that's the first inequality
right there. Now let's do the second
inequality. The second inequality is y
is less than 2x minus 5. So if we were to graph 2x minus
5, and something already might jump out at you
that these two are parallel to each other. They have the same slope. So 2x minus 5, the y-intercept
is negative 5. x is 0, y is negative 1, negative 2, negative
3, negative 4, negative 5. Slope is 2 again. And this is only less than,
strictly less than, so we're not going to actually
include the line. The slope is 2, so it will
look something like that. It has the exact same slope
as this other line. So I could draw a bit of a
dotted line here if you like, and we're not going to include
the dotted line because we're strictly less than. So the solution set for this
second inequality is going to be all of the area
below the line. For any x, this is 2x minus 5,
and we care about the y's that are less than that. So let me shade that in. So before we even get to this
last inequality, in order for there to be something that
satisfies both of these inequalities, it has to be in
both of their solution sets. But as you can see, their
solutions sets are completely non-overlapping. There's no point on the x, y
plane that is in both of these solution sets. They're separated by this kind
of no-man's land between these two parallel lines. So there is actually
no solution set. It's actually the null set. There's the empty set. Maybe we could put an empty set
like that, two brackets with nothing in it. There's no solution set
or the solution set of the system is empty. We could do the x is
greater than 1. This is x is equal to 1, so
we put a dotted line there because we don't want
include that. So it would be all
of this stuff. But once again, there's nothing
that satisfies all three of these. This area right here satisfies
the bottom two. This area up here satisfies
the last one and the first one. But there's nothing that
satisfies both these top two. Empty set.