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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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Graphing two-variable inequalities
Sal graphs the inequality y<3x+5. Created by Sal Khan and Monterey Institute for Technology and Education.
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When graphing, how do you know if you should use a solid line or a dotted line? Does the sign determine if it should be a solid or dotted line?(34 votes)- if you have a less than or equal to/greater than or equal to the line will be solid
if you just have a greater than/less than then the graph of the line will be dashed(31 votes)
how do you remember when to flip the sign? Like, sometimes it needs to flipped from < to > or vice versa.(12 votes)
Try to remember that if you divide by a negative value then the inequality flips. For example, the equation: -3x < 12 becomes x > -4 because you have divided both sides by a negative value. To check if this is correct, replace x with a value that is less than -4 such as -5. Then evaluate the original equation to see if it makes sense. In this case if you insert a -5 in place of x in the equation -3x < 12 you get 15 < 12. This condition is not true so the solution can only be values of x that are > -4. Also, if you replace x with -4 in the original equation you get 12 < 12. This condition is not true either. So only values > -4 make the condition true. If you insert a -3 in place of x then the equation evaluates to 9 < 12. This condition is true and you know the inequality is facing in the right direction. Hope this helps. Jeff(17 votes)
on this video I'm having a serious issue at seeing the graph. can it be made bigger by any chance?(8 votes)
If you hold the CTRL key and press + on your numpad, it will zoom in.(14 votes)
i still don't understand shading!(8 votes)
I have a question regarding2:40. How do you know which area will satisfy the inequality y < 3x+ 5 ? How do you know whether it's the area to the left of the dotted line or the area to the right of the dotted line?(6 votes)
In the case that y<ax+b, then the shaded area will be "below" the dotted line, in the case that y>ax+b, then the shaded area will be "above" the dotted line.(7 votes)
how does he know the shading(4 votes)
The shading is determined by the inequality.
If the inequality is less than: y < mx+b, you shade below the line. If the shading is greater than: y > mx+b, you shade above the line.
If you want to confirm that you have shaded the correct side, pick a point from the side where you shaded. Then, test that ordered pair in the inequality. If it makes the inequality be "true", then you have shaded the correct side.
Hope this helps.(7 votes)
- So if y < something, I draw below the dotted line, and if y > something, I draw above the dotted line? Thanks in advance...(5 votes)
- Yes, that is one way of thinking about it(4 votes)
- how can you tell whether the line is shaded up or down? Im so confused(2 votes)
- If the inequality is in slope intercept form: y<mx+b or y>mx+b, then you can tell where to shade based upon the inequality.
-- If the inequality is "<", you shade below the line.
-- If the inequality is ">", you shade above the line.
You can also always pick a point on either side of the line and test it in the inequality.
-- If the point makes the inequality be true, then you shade that side.
-- If the point makes the inequality false, then you shade the opposite side.
Hope this helps.(9 votes)
What if you had to graph in point slope form or another way besides slope intercept form? And how would you figure out what x equals in the first place?(3 votes)
If you have it in point slope form, you have a point that you can graph, then use the slope to find additional points. In standard form, you can set x=0 to find y intercept and y=0 to find x intercept, then graph these two points.
If it is a linear function or linear inequality, the domain is all real numbers, so x could be anything.(5 votes)
- how can I graph something like absolute value of (x-3)= 2
where the left side is an absolute value but right side isn't.
Would I say: let x -3= y1
and let 2=y2
then graph those two and the number of points of intersections will give me the number of solutions and the solutions themselves to solve this absolute value.(3 votes)
2:40Video transcript
We're asked to graph the
inequality y is less than 3x plus 5. So if you give us any x-- and
let me label the axes here. So this is the x-axis. This is the y-axis. So this is saying you
give me an x. So let's say we take x is
equal to 1 right there. 3 times 1 plus 5. So 3 times x plus 5. So 3 times 1 is 3 plus 5 is 8. So one, two, three, four,
five, six, seven, eight. This is saying that y
will be less than 8. y will be less than
3 times 1 plus 5. So the y-values that satisfy
this constraint for that x are going to be all of these
values down here. Let me do it in a
lighter color. It'll be all of these values. For x is equal to 1, it'll be
all the values down here, and it would not include
y is equal to 8. Y has to be less than 8. Now, if we kept doing that, we
would essentially just to graph the line of y is equal to
3x plus 5, but we wouldn't include it. We would just include everything
below it, just like we did right here. So we know how to graph just
y is equal to 3x plus 5. Let me write it over here. So if I were to write y is equal
to 3x plus 5, we would say, OK, 3 is the slope. Slope is equal to 3, and then
5 is the y-intercept. Now, I could just graph the
line, but because that won't be included in the y's that
satisfy this constraint, I'm going to graph it as
a dotted line. So we'll start with the
y-intercept of 5. So one, two, three,
four, five. That's the y-intercept. And the slope is 3. So if you go over to
the 1, you go up 3. So let me do that in that
darker purple color. So it'll look like this. It will look like that. That point would be on it. That point would be on it. If you go back, you're going
to go down by 3. So that point will be on it,
that point, and that point, and I'll just connect the
dots with a dotted line. That dotted line is the graph
of y is equal to 3x plus 5, but we're not going
to include it. So that's why I made it a dotted
line because we want all of the y's that are
less than that. So for any x-- so
you pick an x. Let's say x is equal
to negative 1. If you evaluate 3x plus 5 for
that x, you'd get here. But we only care about
the y's that are strictly less than that. So you don't include the line. It's everything below it. So for any x you pick, it's
going to be below that line. You take the x, go up to that
line and everything below it. So for all of the x's, it's
going to be this entire area. Let me draw it a little
bit neater than that. It's going to be this entire
area that's under the line. I'll do it in this orange. It's a little bit
easier to see. So this entire area under the
line is y is less than 3x.