Systems of Linear Inequalities

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Systems of Linear Inequalities

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Everything we've been doing so far has
involved systems of equations.
In this video, we're going to tackle the problem of systems
of linear inequalities.
What we're going to do is, essentially, figure out all of
the points on the xy plane that satisfy a system of
linear inequalities.
So let's start with our first linear inequality.
Let's say we have 2x minus 3y is less than or equal to 21.
And we've seen how to graph this before.
You can just graph this line, and then figure out what the
less than part means.
So let's just rearrange this into slope-intercept form.
So if you subtract 2x from both sides, we get negative
3y-- these cancel out --is less than or
equal to 21 minus 2x.
Or we could say negative 2x plus 21.
And then we can divide both sides by negative 3.
And remember, when you divide by a negative number, or you
multiply by a negative number, you swap the inequality.
So you get negative 3y over negative 3 is greater than or
equal to negative 2 over negative 3x plus 21 over
negative 3.
And so this equation becomes y is greater than or equal to
2/3 x minus 7.
So that's the first inequality right there.
And let's graph it.
I'll graph it in orange.
y-intercept is negative 7, so 1, 2, 3, 4, 5, 6, 7.
And it's slope is positive 2/3.
So when you run 3, you go up 2.
When you run 3, you go up 2.
And in this direction, if you go back 3, you go down 2, so
the line will look something like this.
The line will look like that.
Keep going.
And since we have y is greater than, or equal to,
essentially, this line, the area that satisfies this
inequality is going to be above the line.
At any point, for any x you pick, this is the
value 2/3 x minus 7.
So if y is going to be greater than that, it'll be all the
points above that.
So just this constraint right here, y is greater than 2/3 x
minus 7, or this constraint, which is the same thing, 2x
minus 3y is less than or equal to 21, will give you all of
this area up here.
I don't want to draw it too dark just yet.
So it'll give you all of this area all the
way above this line.
Now let's add another inequality to the mix.
Let me do that other inequality in blue.
Let's say I have x plus 4y.
Let's make it less than 6.
The problem I have written here, has less than or equal
to, but I want to mix it up with an equal sign and a
couple of them without an equal sign.
Just so we remember how to graph these things.
And so let's put this again into slope-intercept form, so
we can subtract x from both sides.
Then the left-hand side just becomes 4y.
The right-hand side becomes a negative x plus 6.
And then we can divide both sides by 4.
And we get y is less than negative 1/4 x plus 6.
So just like the first equation,
let's graph the line.
The y-intercept is positive 6.
So 1, 2, 3, 4, 5, 6.
And then the slope is a negative 1/4.
So every time we run by 4, 1, 2, 3, 4, we're
going to go down 1.
So you run by 4, 1, 2, 3, 4, you go down 1.
You run by 4, 1, 2, 3, 4, you're going to go down 1.
So this line is actually going off the table.
To go back by 4, 1, 2, 3, 4, you're going to go up by 1.
And since we're not actually including the line, we're
going to have to draw a dotted line here.
This is going to be a dotted line.
Try my best to draw a dotted line.
And actually, let me continue that previous line there
because I actually want show where they intersect.
Goes a little bit off of our graph paper.
But the graph of this equation is that right.
I did a dotted line because it's not less than or equal
to, it's just less than.
So this inequality, if I were to graph it, would be the area
below the line right there.
That's what satisfies this inequality.
For any x, this is the value of that line.
But the y's that satisfy it are the y's that
are below the line.
Now if I were to ask you, if I were to have a system, where I
have 2x minus 3y is less than, or equal to 21.
These aren't equations.
These are inequalities, they're not
equal to each other.
The other one is x plus 4y is less than 6.
And if I were to ask you what x's and y's satisfy both of
these any inequalities, well you say, it's just going to be
the overlap of their solution sets, right?
It's going to be the area in the xy plane that satisfies
both of them.
So in this case, it'll be the part that's
above the orange line.
Remember I shaded that in orange right there.
So all of this above the orange line, and the part
that's below the blue line, and the part that overlaps
that satisfies them both of these-- I'll do it in yellow
--the part that satisfies both of these is right in between
the lines there.
Now what if, just for fun, we were to add a third equation.
Let's add a third equation in green.
3x plus y is greater than or equal to negative 4.
Well, it's the same drill.
We put this in the slope-intercept form.
So you get y is greater than or equal to
negative 3x minus 4.
I just subtracted 3x from both sides.
We can graph this line.
It's y-intercept is 1, 2, 3, negative 4.
And it has a slope of negative 3.
So every time you go one to the right, you go down 3.
1, 2, 3.
If you go one to the left, if you go negative 1 in the
x-direction, you go up 3.
So you go negative 1, 1, 2, 3.
Negative 1, 1, 2, 3.
Negative 1, 1, 2, 3.
So this line right there's going to look like this.
It's going to look like that.
Trying my best to draw it.
That's that right there.
And I can have it in bold.
I don't have to do a dotted line because I have a greater
than or equal to and so what are all the y's that are
greater than this line?
Well, for any x value you pick, the value of negative 3x
minus 4 is going to be this.
And we care about the y values that are greater than that, so
it'll be above the line.
So for just this inequality it'll be
everything above the line.
And then if someone were to come along and say, all right
, tell me, show me, graph me, all of the points on the xy
plane that satisfy this system of inequalities.
What would you tell them?
Well, it would be the overlap of their solution sets?
So the part of the xy plane that exists in all of their
solutions, so what's above the orange line, right?
We have to be above the orange line, right?
And we graphed that before.
It was all in orange before, because we're above the line
to 2/3x minus 7.
It might trick you a little bit because
there's a less than here.
But when you put it in the slope-intercept form, you have
a greater than.
So it's everything above this line, which is all of that
over there.
Everything below this line-- below the blue line and not
including the blue line --and everything
above the green line.
So it's going to be this area right over here, is the
solution set.
And we're going to include this boundary right here, and
we're going to include that boundary right there.
And since I engineered the problem to have a less than
there, not a less than or equal to, up here is going to
be a dotted line.
Because you're not actually going to include that line.
Anyway, hopefully you found that useful.

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