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## Algebra 1

### Course: Algebra 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

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# Intro to graphing systems of inequalities

CCSS.Math:

Learn how to graph systems of two-variable linear inequalities, like "y>x-8 and y<5-x.". Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- How do you know if the line will be solid or dotted? Does it matter?(10 votes)
- It does matter.

It will be solid if the inequality is less than OR EQUAL TO (≤) or greater than OR EQUAL TO ≥.

It will be dotted if the inequality is less then (<) or greater then (>).

Think of a simple inequality like x > 5.

x can be ANY value greater then 5, but not exactly 5. x could be 5.000000000001, but not 5. They put the dotted line because its saying 'this is where the inequality will work, except right on this line'.(23 votes)

- Can systems of inequalities be solved with subsitution or elimination? or only by graphing? thanks! :)(12 votes)
- the best method is cross multiplication method or the soluton using cramer rule...... it might seem lengthy but with practice it is the easiest of all and always reliable..(5 votes)

- If the slope was 2 would the line go 2 up and 2 across, 2 up and 1 across, or 1 up and 2 across?? Thanks!(3 votes)
- If the slope was 2 it would go up two and across once.

Slope = y / x

2 = 2/1

2 = y ( Vertical )

1 = x ( Horizontal )(12 votes)

- Why is the slope not a fraction3:21? How do I know I have to only go over 1 on the x axis if there isn't a number to specify that I have to?(4 votes)
- All integers can be written as a fraction with a denominator of 1. Since that concept is taught when students learn fractions, it is expected that you have remembered that information for lessons that come later (like this one).

So, any slope that is a number like 5 or -3 should be written in fraction form as 5/1 or -3/1.

Hope this helps.(3 votes)

- how do you graph an inequality if the inequality equation has both "x" and "y" variables?(5 votes)
- It depends on what sort of equation you have, but you can pretty much never go wrong just plugging in for values of x and solving for y.(3 votes)

- how do you know its a dotted line?(3 votes)
- Hint: to get ≥ hold down ALT button and put in 242 on number pad, ≤ is ALT 243. Makes it easier than words(4 votes)

- wait if you were to mark the intersection point,would the intersection point be inclusive of exclusive if one of the lines was dotted and the other was not(2 votes)
- The intersection point would be exclusive. The easiest way to see this is with an example:

If we had the two lines x >= 3 and y < 6, the intersection point (3,6) wouldn't be a solution, because to be a solution, it would have to fulfill both equations:

3 >= 3

6 !< 6

Since 6 is not less than 6, the intersection point isn't a solution.(4 votes)

- On the example above: y = -1x + 5

I got a little confused on this one because the instructor went to positive 5 on x axis. When I saw the -1 the first thing I did on my paper was start moving to the left on a -1 slope. Was I wrong? I am making mistakes on (+,-) so I'm trying to stay consistent. My line was essentially the same but my method was different from the instructor(2 votes)- Remember, lines move both left and right. A slope of -1 can be interpreted as:

-1/1 = move down 1 as you move right 1

1/(-1) = move up 1 as you move left 1

Both versions work.

Also, you can see if you solve for the x-intercept using y=0, that you will calculate x=5.

Hope this helps.(3 votes)

- Without Graphing, would you be able to solve a system like this:

Y+x^2-2x+1

Y=x+1(3 votes)- Yes! I think you meant to write y = x^2 - 2x + 1 instead of y + x^2 - 2x + 1. So, if: y = x^2 - 2x + 1, and

y = x + 1, using substitution we get, x + 1 = x^2 - 2x + 1, subtracting 1 from each side we get,

x = x^2 - 2x, adding 2x to each side we get 3x = x^2, dividing each side by x we get, 3 = x, so y = 4. So, yes, you can solve this without graphing.(2 votes)

- but Sal but we plot the x intercept it gives the equation like 8>x and when we reverse that it says that x<8?? then how do we shade the graph when one point contradicts all the other points!(2 votes)
- If 8>x then you have a dotted vertical line on the point (8,0) and shade everything to the left of the line. If it's 8<x, then you shade to the right side of the line.

Hope this helps, God bless!(2 votes)

## Video transcript

Graph the solution
set for this system. It's a system of inequalities. We have y is greater than x
minus 8, and y is less than 5 minus x. Let's graph the solution set
for each of these inequalities, and then essentially
where they overlap is the solution set for the
system, the set of coordinates that satisfy both. So let me draw a
coordinate axes here. So that is my x-axis, and
then I have my y-axis. And that is my y-axis. And now let me draw
the boundary line, the boundary for this
first inequality. So the boundary line
is going to look like y is equal to x minus 8. But it's not going
to include it, because it's only
greater than x minus 8. But let's just graph x minus 8. So the y-intercept
here is negative 8. When x is 0, y is
going to be negative 8. So just go negative 1,
negative 2, 3, 4, 5, 6, 7, 8. So that is negative 8. So the point 0, negative
8 is on the line. And then it has a slope of 1. You don't see it right there,
but I could write it as 1x. So the slope here
is going to be 1. I could just draw a line
that goes straight up, or you could even say that it'll
intersect if y is equal to 0, if y were equal to 0,
x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. If it has a slope
of 1, for every time you move to the right 1,
you're going to move up 1. So the line is going to
look something like this. And actually, let me not
draw it as a solid line. If I did it as a
solid line, that would actually be this
equation right here. But we're not going
to include that line. We care about the y values that
are greater than that line. So what we want to do
is do a dotted line to show that that's just
the boundary, that we're not including that in
our solution set. Let me do this in a new color. So this will be the
color for that line, or for that inequality,
I should say. So that is the boundary line. And this says y is
greater than x minus 8. So you pick an x,
and then x minus 8 would get us on
the boundary line. And then y is greater than that. So it's all the y values above
the line for any given x. So it'll be this region above
the line right over here. And if that confuses
you, I mean, in general I like to just
think, oh, greater than, it's going to be above the line. If it's less than, it's
going to be below a line. But if you want
to make sure, you can just test on either
side of this line. So you could try the
point 0, 0, which should be in our solution set. And if you say, 0 is
greater than 0 minus 8, or 0 is greater than
negative 8, that works. So this definitely should
be part of the solution set. And you could try something
out here like 10 comma 0 and see that it doesn't work. Because you would have 10 minus
8, which would be 2, and then you'd have 0. And 0 is not greater than 2. So when you test
something out here, you also see that it won't work. But in general, I
like to just say, hey look, this is
the boundary line, and we're greater than the
boundary line for any given x. Now let's do this one over here. Let's do this one. The boundary line for
it is going to be y is equal to 5 minus x. So the boundary line is
y is equal to 5 minus x. So once again, if x
is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a
slope of negative 1. We could write this as y is
equal to negative 1x plus 5. That's a little bit
more traditional. So once again, y-intercept at 5. And it has a slope
of negative 1. Or another way to think
about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move
to the right one, we go down one because we
have a negative 1 slope. So it will look like this. And once again, I want
to do a dotted line because we are-- so
that is our dotted line. And I'm doing a
dotted line because it says y is less than 5 minus x. If it was y is
equal to 5 minus x, I would have included the line. If it was y is less than
or equal to 5 minus x, I also would have
made this line solid. But it's only less than,
so for any x value, this is what 5
minus x-- 5 minus x will sit on that boundary line. But we care about the y values
that are less than that, so we want everything
that is below the line. And once again, you can test
on either side of the line. 0, 0 should work for this
second inequality right here. 0 is indeed less than 5 minus 0. 0 is less than 5. And then you could try
something like 0, 10 and see that it doesn't
work, because if you had 10 is less than 5 minus
0, that doesn't work. So it is everything
below the line like that. And like we said,
the solution set for this system are all
of the x's and y's, all of the coordinates that
satisfy both of them. So all of this shaded
in purple satisfies the second inequality. All of this shaded in green
satisfies the first inequality. So the stuff that satisfies
both of them is their overlap. So it's all of this
region in blue. Hopefully this isn't
making it too messy. All of this region in blue
where the two overlap, below the magenta dotted
line on the left-hand side, and above the
green magenta line. That's only where they overlap. So it's only this
region over here, and you're not including
the boundary lines.