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.
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- 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)
- How do you know if the line will be solid or dotted? Does it matter?(3 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'.(3 votes)
- Can systems of inequalities be solved with subsitution or elimination? or only by graphing? thanks! :)(2 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)
- 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?(2 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)
- 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)
- Without Graphing, would you be able to solve a system like this:
- 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)
- What is a "boundary line?" also, we are setting the > and < signs to 0? than plotting them right? I'm confused.(2 votes)
- It's the line forming the border between what is a solution for an inequality and what isn't.(1 vote)
- At3:14Sal said that y is equal to 5-x even though it said y is less then 5-x?(1 vote)
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.