Graphing two-variable inequalities (old)

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.