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# Graphing two-variable inequalities (old)

## Video transcript

Graph the inequality y minus
4x is less than negative 3. So the first thing
we could do is we could kind of put
this in mx plus b form, or slope-intercept form,
but as an inequality. So we're starting with y minus
4x is less than negative 3. We can add 4x to both
sides of this inequality. So let's add 4x to both
sides of this inequality, and then we'll just have
a y on the left-hand side. These guys cancel out. So you have y is
less than 4x minus 3. We could have had
negative 3 plus 4x, but we want to write the 4x
first just because that's a form that we're
more familiar with. So it's less than 4x minus 3. And now we can
attempt to graph it. But before I graph it, I want
to be a little bit careful here. So let me draw our axes. So this is the x-axis,
and is that is the y-axis. And we want to be
careful, because this says y is less than 4x
minus 3, not less than or equal to 4x minus 3, or
not y is equal to 4x minus 3. So what we want to do is kind
of create a boundary at y is equal to 4x minus 3, and
the solution to this inequality will be all of the area below
that, all of the y values less than that. So let's try to do it. So the boundary line would look
like-- so let me write it over here-- so we have a boundary
at y is equal to 4x minus 3. Notice this isn't
part of the solution. This isn't less than or equal. It's just less than. But this will at
least help us draw, essentially, the boundary. So we could do it two ways. If you know slope
and y-intercept, you know that 4 is our
slope and that negative 3 is our y-intercept. Or you can literally
just take two points, and that'll help you
define a line here. So you could say, well, when
x is equal to 0, what is y? You get 4 times 0 minus 3, you
get y is equal to negative 3. And we knew that because
it was the y-intercept. So you have 0, and then you
have 1, 2, 0, and negative 3. And then you have
the point, let's say, when x is equal
to-- I don't know-- let's say when x is equal to 2. When x is equal to 2, what is y? We have 4 times
2 is 8 minus 3, y is then going to be equal to 5. So then you go 1, 2, and
you go 1, 2, 3, 4, 5. And so you have that
point there as well. And then we can just
connect the dots. Or you could say, look,
there's a slope of 4. So every time we move
over 1, every time we move 1 in the x direction, we
move up 4 in the y direction. So we could draw it like that. So the line will look
something like this. And I'm just going to
draw it in a dotted line because, remember, this
isn't part of the solution. Actually, let me draw
it a little bit neater because that point should
be right about there, and this point should
be right about there. And then this boundary line I'm
going to draw as a dotted line. So it's going to look
something like that. I draw it a dotted line
to show that it's not part of the solution. Our solution has the
y's less than that. So for any x, so you pick an x
here, if you took 4x minus 3, you're going to
end up on the line. But we don't want the y's
that are equal to that line. We want for that
particular x, the y's that are less than the line. So it's going to be all
of this area over here. We're less than the line, and
we're not including the line, and that's why I put
a dotted line here. You can also try values out. You can say, well, this line
is dividing our coordinate axes into, essentially,
the region above it and the region below it,
and you can test it out. Let's take something
that's above it. Let's take the
point 0, 0 and see if that satisfies
our inequality. If we have y is 0 is
less than 0 minus 3, or we get 0 is less
than negative 3. This is definitely not the case. This is not true. And it makes sense
because that 0, 0 is not part of the solution. Now, we could go on the other
side of our boundary line. And we could take the
point, I don't know, let's take the point 3 comma 0. So let's say that this is the
point-- well, that's right. There's a point 2 comma 0. Let's take the point 3
comma 0 right over here. This should work because
it's in the region less than. But let's verify
it for ourselves. So we have y is 0. 0 is less than 4
times 3 minus 3. 0 is less than 12 minus 3. 0 Is less than 9, which
is definitely true. So that point does
satisfy the inequality. So in general, you
want to kind of look at this as an equal to
draw the boundary line. We did it. But we drew it as a
dotted line because we don't want to include
it because this isn't less than or equal to. It's just less than. And then our solution
to the inequality will be the region
below it, all the y's less than the line
for x minus 3.