Thinking about solutions to systems
Independent and Dependent Systems Independent and Dependent Systems
⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.
- Is the system of linear equations below
- dependent or independent, and they gave us
- two equations right here.
- Before I tackle this specific problem,
- let's just do a little bit of review of what
- dependent or independent means.
- And actually, I'll compare that to
- consistent and inconsistent.
- So, just to start off with, if we're dealing
- with systems of linear equations in two dimensions,
- there's only three possibilities
- that the lines, or the equations,
- can have relative to each other. So let me
- draw the three possibilities. Let me draw
- three coordinate axes. So that's my first x-axis
- and y-axis, so x and y. Let me draw another
- one. And that is x, and that is y. Let me draw
- one more, because there's only three possibilities
- in two dimensions, x and y, if we're dealing with
- linear equations, x and y. So you can have
- the situation where the lines just intersect in
- one point, so you can have one line like that,
- and maybe the other line does something
- like that, and they intersect at one point.
- You could have the situation where the two
- lines are parallel. So you could have the situation
- actually let me draw it over here, where
- you have one line that goes like that, and
- the other line has the same slope, but
- it's shifted, it has a different y-intercept,
- so maybe it looks like this, and you have no points
- of intersection. And then you could have the
- situation where they're actually the same line,
- so that both lines have the same slope
- and the same y-intercept. So really, they are
- the same line, they intersect on infinite number
- of points. Every point on either of those lines
- is also a point on the other line. So just to give
- you a little bit of the terminology here, and
- we learned this in the last video, this type of
- system, where they don't intersect,
- where you have no solutions, this is an
- inconsistent system.
- And by definition, or I guess just taking the
- opposite of inconsistent, both of these
- would be considered consistent. Both of these
- are consistent.
- But then, within consistent, there's obviously
- a difference, here we only have one
- solution, these are two different lines
- that intersect at one place. And here,
- they're essentially the same exact line.
- And so we differentiate between these two
- scenarios by calling this one over here
- independent, and this one over here dependent.
- So independent, both lines are doing their own
- thing, they're not dependent on each other,
- they're not the same line. They will intersect
- at one place. Dependent, they're
- the exact same line, any point that satisfies
- one line will satisfy the other.
- Any point that satisfies one equation,
- will satisfy the other. So with that said,
- let's see if this system of linear equations
- right here is dependent or independent.
- So they're kinda having us assume that
- it's going to be consistent. That we're either
- going to intersect at one place, or we're
- going to intersect at an infinite number
- of places. And the easiest way to do this,
- we already have this second equation here,
- it's already in slope-intercept form,
- we know the slope is -2, the y-intercept is 8.
- Let's put this first equation up here in
- slope-intercept form and see if it has
- a different slope or a different intercept,
- or maybe it's the same line. So we have
- 4x plus 2y is equal to 16.
- We can subtract 4x from both sides,
- what we want to do is isolate the y on the
- left hand side. So let's subtract 4x from
- both sides. The left hand side we are
- just left with a 2y, and then the right hand side,
- we have a -4x plus 16. I
- just wrote the -4 in front of the 16
- just so that we have it in the traditional
- slope-intercept form. And now we can
- divide both sides of this equation by 2,
- so that we can isolate the y on the left-hand side.
- Divide both sides by 2, we are left with
- y is equal to -4 divided by 2 is -2x
- plus 16 over 2, plus 8. So all I did is
- algebraically manipulated this top equation
- up here, and when I did that, when I solved
- essentially for y, I got this right over here,
- which is the exact same thing as the
- second equation. We have the exact same
- slope, -2, -2, and we have the exact
- same y-intercept, 8 and 8. If I were
- to graph these equations, that's my
- x axis, and that is my y-axis,
- both of them have a y-intercept at 8
- and have a slope of -2.
- So they look something, I'm just drawing
- an approximation of it, but they would look
- something like that. So maybe this is the graph of
- this equation right here, this first equation.
- And then the second equation will be
- the exact same graph, it has the exact
- same y-intercept and the exact same slope.
- So clearly these two lines are dependent.
- Dependent. They have an infinite number
- of points that are common to both of them
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
|
Have something that's not a question about this content? |
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
abuse
- disrespectful or offensive
- an advertisement
not helpful
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
wrong category
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site
Share a tip
Suggest a fix
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.