Permutation formula

WEBVTT
Kind: captions
Language: en

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- [Voiceover] We know that
if we had five people,

00:00:02.588 --> 00:00:05.456
let's say person A, person B, person C,

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person D, and person E,

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and we wanted to put
them in five different,

00:00:11.492 --> 00:00:14.012
let's say, positions, or chairs,

00:00:14.012 --> 00:00:16.879
so position one, position
two, position three,

00:00:16.879 --> 00:00:20.177
position four, and position five.

00:00:20.177 --> 00:00:23.020
If we wanted to count
the number of scenarios,

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or we could say the number of permutations

00:00:25.319 --> 00:00:29.000
of putting these five
people in these five chairs,

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we could say, well, we
have five different ...

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If we seated people in order,

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which we might as well do,

00:00:36.214 --> 00:00:37.782
we could say, look, five different people

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could sit in chair one.

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So, for each of those scenarios,

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four different people
could sit in chair two.

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Now, for each of these scenarios, now,

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so we have 20 scenarios.

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Five times four, we have 20 scenarios

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where we've seated seat one and seat two,

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how many people could we
now seat in seat three

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for each of those 20 scenarios?

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Well, three people haven't sat down yet,

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so there's three possibilities there.

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So now, there's five times
four times three scenarios

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for seating the first three people.

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How many people are left for seat four?

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Well, two people haven't sat down yet,

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so there's two possibilities.

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So now there's five times four times three

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times two scenarios of
seating the first four seats.

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For each of those, how many possibilities

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are there for the fifth seat?

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Well, one.

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For each of those scenarios,
we only have one person

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who hasn't sat down left,
so there's one possibility.

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So the number of permutations,
the number of ...

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Let me write this down.

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The number of permutations, permutations,

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of seating these five
people in five chairs

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is five factorial.

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Five factorial, which is
equal to five times four

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times three times two times one,

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which, of course, is equal to, let's see,

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20 times six, which is equal to 120.

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We have already covered
this in a previous video.

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But now let's do something
maybe more interesting,

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or maybe you might find
it less interesting.

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Let's say that we still
have these five people,

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but we don't have as many chairs,

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so not everyone is going
to be able to sit down.

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Let's say that we only have three chairs.

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We have chair one, we have chair two,

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and we have chair three.

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How many ways can you have five people,

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where only three of them
are going to sit down

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in these three chairs, and we care

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which chair they sit in?

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I encourage you to pause the
video and think about it.

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I am assuming you have had your go at it.

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Let's use the same logic.

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If we seat them in order,
and we might as well,

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how many different people,

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if we haven't sat anyone yet,

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how many different people
could sit in seat one?

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Well, we could have, if no one sat down,

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we had five different people,

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five different people
could sit in seat one.

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For each of these
scenarios where one person

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has already sat in seat one,

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how many people could sit in seat two?

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In each of these scenarios,
if one person has sat down,

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there's four people left
who haven't been seated,

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so four people could sit in seat two.

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So we have five times four scenarios

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where we've seated seats one and seat two.

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For each of those 20 scenarios,

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how many people could sit in seat three?

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Well, we haven't sat,
we haven't seaten or sat

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three of the people yet,

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so for each of these
20, we could put three

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different people in seat three,

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so that gives us five times
four times three scenarios.

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So this is equal to five times four

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times three scenarios, which is equal to,

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this is equal to 60.

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So there's 60 permutations

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of sitting five people in three chairs.

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Now this, and my brain, whenever I start

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to think in terms of permutations,

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I actually think in these ways.

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I just literally draw it out

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because I don't like formulas.

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I like to actually conceptualize

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and visualize what I'm doing.

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But you might say, hey, when we just did

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five different people in
five different chairs,

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and we cared which seat they sit in,

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we had this five factorial.

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Factorial is kind of neat
little operation there.

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How can I relate factorial
to what we did just now?

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It looks like we kind of did factorial,

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but then we stopped.

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We stopped at, we didn't
go times two times one.

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So one way to think
about what we just did,

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is we just did five times four times three

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times two times one, but of course

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we actually didn't do the two times one,

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so you could take that and you

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could divide by two times one.

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If you did that, this two times one

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would cancel with that two times one

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and you'd be left with five
times four times three.

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The whole reason I'm writing this way

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is that now I can write
it in terms of factorial.

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I could write this as five factorial,

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five factorial, over two factorial,

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over two factorial.

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But then you might have the question

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where did this two come from?

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I have three seats.

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Where did this two come from?

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Well, think about it.

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I multiplied five times four times three,

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I kept going until I had that many seats,

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and then I didn't do the remainder.

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So the things that I left out,

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the things that I left out,

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that was essentially the number of people

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minus the number of chairs,

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so I was trying to put five
things in three places.

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Five minus three, that
gave me two left over.

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So I could write it like this.

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I could write it as five ...

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Let me use the same colors.

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I could write it as five factorial

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over, over

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five minus three, which of course is two,

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five minus three factorial.

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Another way of thinking about it,

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if we wanted to generalize,

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is if you're trying to figure
out the number of permutations

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and there's a bunch of
notations for writing this,

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if you're trying to figure
out the number of permutations

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where you could put n people in r seats,

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or the number of permutations

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you could put n people in r seats,

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and there's other notations as well,

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well, this is just going to be n factorial

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over n minus r factorial.

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Here n was five, r was three.

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Five minus three is two.

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Now, you'll see this in a
probability or a statistics class,

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and people might memorize this thing.

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It seems like this kind of daunting thing.

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I'll just tell you right now,

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the whole reason why I
just showed this to you

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is so that you could connect it

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with what you might see in your textbook,

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or what you might see in a class,

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or when you see this type of formula,

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you see that it's not
some type of voodoo magic.

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But I will tell that for me, personally,

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I never use this formula.

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I always reason it through,

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because if you just memorize the formula,

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you're always going, wait,
does this formula apply there?

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What's n? What's r?

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But if you reason it through,

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it comes out of straight logic.

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You don't have to memorize anything.

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You don't feel like you're just memorizing

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without understanding.

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You're just using your
deductive reasoning, your logic.

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That's especially valuable
because as we'll see,

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not every scenario's going fit
so cleanly into what we did.

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There might be some tweaks on this,

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where like maybe only person B

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likes sitting in one of the chairs,

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or who knows what it might be?

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Then your formula is going to be useless.

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So I like reasoning through it like this,

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but I just showed you this

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so that you could connect to it a formula

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that you might see in a
lecture or in a class.


Introduction to combinations

WEBVTT
Kind: captions
Language: en

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- So let's keep thinking
about different ways

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to sit multiple people in
the certain number of chairs.

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So let's say we have six people.

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We have person A, we have
person B, we have person C,

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person D, person E, and we have person F.

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So we have six people.

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And for the sake of this
video, we're going to say

00:00:19.666 --> 00:00:21.300
oh we want to figure
out all the scenarios,

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all the possibilities,
all the permutations,

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all the ways that we could
put them into three chairs.

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So there's chair number
one, chair number two

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and chair number three.

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This is all review.

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This is covered in the permutations video.

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But it'll be very instructive as we move

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into a new concept.

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So what are all the
permutations of putting

00:00:38.466 --> 00:00:41.300
six different people into three chairs?

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Well, like we've seen before,

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we can start with the first chair.

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And we can say look if no one's sat--

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If we haven't seated anyone yet,

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how many different people could
we put in chair number one?

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Well there's six different
people who could be

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in chair number one.

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Let me get a different color.

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There are six people who
could be in chair number one.

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Six different scenarios for
who sits in chair number one.

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Now for each of those six scenarios,

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how many people, how many different people

00:01:08.566 --> 00:01:10.500
could sit in chair number two?

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Well each of those six
scenarios we've taken

00:01:12.266 --> 00:01:14.900
one of the six people to
sit in chair number one.

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So that means you have
five out of the six people

00:01:17.166 --> 00:01:19.700
left to sit in chair number two.

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Or another way to think about it is

00:01:21.200 --> 00:01:23.899
there's six scenarios of
someone in chair number one

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and for each of those six,
you have five scenarios

00:01:25.933 --> 00:01:27.600
for who's in chair number two.

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So you have a total of 30 scenarios

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where you have seated six
people in the first two chairs.

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And now if you want to say

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well what about for the three chairs?

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Well for each of these 30 scenarios,

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how many different people could you put

00:01:41.400 --> 00:01:43.766
in chair number three?

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Well you're still going
to have four people

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standing up not in chairs.

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So for each of these 30 scenarios,

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you have four people who you could put

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in chair number three.

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So your total number of scenarios,

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or your total number of permutations

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where we care who's sitting in which chair

00:01:58.800 --> 00:02:01.733
is six times five times four,

00:02:01.733 --> 00:02:06.733
which is equal to 120 permutations.

00:02:06.733 --> 00:02:09.300
Permutations.

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Now lemme, permutations.

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Now it's worth thinking about

00:02:12.300 --> 00:02:14.167
what permutations are counting.

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Now remember we care,
when we're talking about

00:02:15.633 --> 00:02:18.033
permutations, we care about who's sitting

00:02:18.033 --> 00:02:18.966
in which chair.

00:02:18.966 --> 00:02:23.466
So for example, for example
this is one permutation.

00:02:23.466 --> 00:02:26.833
And this would be counted
as another permutation.

00:02:26.833 --> 00:02:30.967
And this would be counted
as another permutation.

00:02:30.967 --> 00:02:33.100
This would be counted
as another permutation.

00:02:33.100 --> 00:02:35.033
So notice these are all
the same three people,

00:02:35.033 --> 00:02:36.566
but we're putting them
in different chairs.

00:02:36.566 --> 00:02:37.500
And this counted that.

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That's counted in this 120.

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I could keep going.

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We could have that, or we could have that.

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So our thinking in the permutation world.

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We would count all of these.

00:02:46.533 --> 00:02:49.733
Or we would count this as
six different permutations.

00:02:49.733 --> 00:02:52.000
These are going towards this 120.

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And of course we have other permutations

00:02:53.400 --> 00:02:54.766
where we would involve other people.

00:02:54.766 --> 00:02:59.766
Where we have, it could
be F, B, C, F, C, B,

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F, A, C, F, F.

00:03:05.367 --> 00:03:06.566
Actually let me do it this way.

00:03:06.566 --> 00:03:08.166
I'm going to be a little
bit more systematic.

00:03:08.166 --> 00:03:10.433
F, uh lemme do it.

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B, B, F, C, B, C, F.

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And obviously I could keep doing.

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I can do 120 of these.

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I'll do two more.

00:03:19.466 --> 00:03:22.966
You could have C, F, B.

00:03:22.966 --> 00:03:25.833
And then you could have C, B, F.

00:03:25.833 --> 00:03:28.234
So in the permutation world, these are,

00:03:28.234 --> 00:03:31.633
these are literally 12
of the 120 permutations.

00:03:31.633 --> 00:03:33.933
But what if we, what if all we cared about

00:03:33.933 --> 00:03:36.033
is the three people we're choosing

00:03:36.033 --> 00:03:38.066
to sit down, but we don't care

00:03:38.066 --> 00:03:40.034
in what order that they're sitting,

00:03:40.034 --> 00:03:41.567
or in which chair they're sitting.

00:03:41.567 --> 00:03:45.133
So in that world, these would all be one.

00:03:45.133 --> 00:03:47.966
This is all the same set of three people

00:03:47.966 --> 00:03:50.933
if we don't care which
chair they're sitting in.

00:03:50.933 --> 00:03:54.400
This would also be the
same set of three people.

00:03:54.400 --> 00:03:56.000
And so this question.

00:03:56.000 --> 00:03:59.000
If I have six people
sitting in three chairs,

00:03:59.000 --> 00:04:02.266
how many ways can I choose
three people out of the six

00:04:02.266 --> 00:04:04.300
where I don't care
which chair they sit on?

00:04:04.300 --> 00:04:05.933
And I encourage you to pause the video,

00:04:05.933 --> 00:04:09.966
and try to think of what that
number would actually be.

00:04:09.966 --> 00:04:12.700
Well a big clue was when we essentially

00:04:12.700 --> 00:04:15.000
wrote all of the permutations
where we've picked

00:04:15.000 --> 00:04:16.200
a group of three people.

00:04:16.200 --> 00:04:19.566
We see that there's six ways
of arranging the three people.

00:04:19.566 --> 00:04:21.500
When you pick a certain
group of three people,

00:04:21.500 --> 00:04:23.766
that turned into six permutations.

00:04:23.766 --> 00:04:25.933
And so if all you want to do is care about

00:04:25.933 --> 00:04:27.866
well how many different
ways are there to choose

00:04:27.866 --> 00:04:30.100
three from the six?

00:04:30.100 --> 00:04:32.200
You would take your whole permutations.

00:04:32.200 --> 00:04:34.800
You would take your
number of permutations.

00:04:34.800 --> 00:04:37.133
You would take your
number of permutations.

00:04:37.133 --> 00:04:39.533
And then you would divide
it by the number of ways

00:04:39.533 --> 00:04:41.933
to arrange three people.

00:04:41.933 --> 00:04:46.933
Number of ways to arrange,
arrange three people.

00:04:51.700 --> 00:04:54.100
And we see that you can
arrange three people,

00:04:54.100 --> 00:04:55.466
or even three letters.

00:04:55.466 --> 00:04:57.633
You can arrange it in six different ways.

00:04:57.633 --> 00:05:02.300
So this would be equal
to 120 divided by six,

00:05:02.300 --> 00:05:05.166
or this would be equal to 20.

00:05:05.166 --> 00:05:07.567
So there are 120 permutations here.

00:05:07.567 --> 00:05:09.767
If you said how many different
arrangements are there

00:05:09.767 --> 00:05:12.567
of taking six people and
putting them into three chairs?

00:05:12.567 --> 00:05:13.966
That's 120.

00:05:13.966 --> 00:05:15.466
But now we're asking another thing.

00:05:15.466 --> 00:05:19.100
We're saying if we start with 120 people,

00:05:19.100 --> 00:05:21.533
and we want to choose.

00:05:21.533 --> 00:05:25.734
Sorry if we're starting with six people

00:05:25.734 --> 00:05:27.533
and we want to figure out how many ways,

00:05:27.533 --> 00:05:29.967
how many combinations,
how many ways are there

00:05:29.967 --> 00:05:32.800
for us to choose three of them?

00:05:32.800 --> 00:05:36.866
Then we end up with 20 combinations.

00:05:36.866 --> 00:05:39.733
Combinations of people.

00:05:39.733 --> 00:05:41.800
This right over here, once again,

00:05:41.800 --> 00:05:43.600
this right over here is
just one combination.

00:05:43.600 --> 00:05:46.167
It's the combination, A, B, C.

00:05:46.167 --> 00:05:47.833
I don't care what order they sit in.

00:05:47.833 --> 00:05:49.033
I have chosen them.

00:05:49.033 --> 00:05:50.933
I have chosen these three of the six.

00:05:50.933 --> 00:05:52.566
This is a combination of people.

00:05:52.566 --> 00:05:54.166
I don't care about the order.

00:05:54.166 --> 00:05:55.600
This right over here
is another combination.

00:05:55.600 --> 00:05:57.966
It is F, C, and B.

00:05:57.966 --> 00:05:59.667
Once again I don't care about the order.

00:05:59.667 --> 00:06:02.067
I just care that I've
chosen these three people.

00:06:02.067 --> 00:06:03.933
So how many ways are there to choose

00:06:03.933 --> 00:06:05.733
three people out of six?

00:06:05.733 --> 00:06:07.300
It is 20.

00:06:07.300 --> 00:06:08.933
It's the total number of permutations.

00:06:08.933 --> 00:06:11.900
It's 120 divided by the number of ways

00:06:11.900 --> 00:06:15.133
you can arrange three people.


Counting pot and flower scenarios

WEBVTT
Kind: captions
Language: en

00:00:00.403 --> 00:00:01.880
- [Voiceover] You're at a
florist store and you're

00:00:01.880 --> 00:00:05.016
interested in buying some
type of a potted flower.

00:00:05.016 --> 00:00:08.561
And you ask the florist, "What
type of flowers do you sell?"

00:00:08.561 --> 00:00:12.118
And he says, "Well we sell
four types of flowers, we sell

00:00:12.118 --> 00:00:15.500
"roses, tulips, sunflowers, and lilies."

00:00:15.500 --> 00:00:17.759
And you say, "What type of
pots could I put them in?"

00:00:17.759 --> 00:00:20.159
And he says, "Well you could
pick any flower, and then you

00:00:20.159 --> 00:00:22.359
"could pick any of our three
pots, we have brown pots,

00:00:22.359 --> 00:00:27.359
"we have yellow pots,
and we have green pots."

00:00:27.499 --> 00:00:29.972
So, the question that I ask to you is,

00:00:29.972 --> 00:00:34.619
how many types of, I guess,
flower and pots put together can

00:00:34.619 --> 00:00:38.019
you walk out of this florist store with?

00:00:38.019 --> 00:00:40.479
For example, you could get
a rose and a brown pot,

00:00:40.479 --> 00:00:42.699
you could get a rose and a green pot.

00:00:42.699 --> 00:00:45.199
Or you could get a yellow pot
that has a sunflower in it.

00:00:45.199 --> 00:00:47.359
Or a yellow pot that has a lily in it.

00:00:47.359 --> 00:00:50.379
So, how many scenarios could you walk out

00:00:50.379 --> 00:00:52.219
of that store with?

00:00:52.219 --> 00:00:54.719
And like always, I'll encourage
you to pause the video

00:00:54.719 --> 00:00:57.741
and try to figure it out on your own.

00:00:57.741 --> 00:00:59.151
Let's think through it.

00:00:59.151 --> 00:01:03.399
I'll just write the first
letters, just to visualize

00:01:03.399 --> 00:01:05.180
or just so I don't have
to write down everything.

00:01:05.180 --> 00:01:10.180
So, you could have a brown pot,
you could have a yellow pot,

00:01:12.479 --> 00:01:13.939
or you could have a green pot.

00:01:13.939 --> 00:01:15.709
You difinitely have to pick a
pot, so you're going to have

00:01:15.709 --> 00:01:16.899
one of those.

00:01:16.899 --> 00:01:19.828
And then for each of these
three, there's four possible

00:01:19.828 --> 00:01:21.223
flowers you could have.

00:01:21.223 --> 00:01:24.159
You could have a rose with the brown pot,

00:01:24.159 --> 00:01:25.539
you could have a rose with the yellow pot,

00:01:25.539 --> 00:01:27.767
you could have a rose with the green pot.

00:01:27.767 --> 00:01:31.599
You could have a tulip with the brown pot,

00:01:31.599 --> 00:01:34.779
a tulip with the yellow pot,
a tulip with the green pot.

00:01:34.779 --> 00:01:39.779
You could have a sunflower
with each of the three pots,

00:01:40.359 --> 00:01:44.659
or you could have a
lily with the brown pot,

00:01:44.659 --> 00:01:48.929
a lily with the yellow pot,
and a lily with a green pot.

00:01:48.929 --> 00:01:51.839
So, how many scenarios
are we talking about?

00:01:51.839 --> 00:01:56.839
Well, we had three pots, so
we have three pots right over

00:01:57.459 --> 00:02:00.419
here, and we have four possible
flowers to put in the pots,

00:02:00.419 --> 00:02:04.434
and so we see that we have
four possible flowers for each

00:02:04.434 --> 00:02:05.168
of the three pots.

00:02:05.168 --> 00:02:10.108
It's going to be three times
four possibilites, or 12.

00:02:10.108 --> 00:02:11.749
You see them right over here.

00:02:11.749 --> 00:02:15.739
This is brown with rose, brown with tulip,

00:02:15.739 --> 00:02:17.739
brown with sunflower, brown with lily.

00:02:17.739 --> 00:02:20.411
Yellow with rose, yellow with
tulip, yellow with sunflower,

00:02:20.411 --> 00:02:20.895
yellow with lily.

00:02:20.895 --> 00:02:23.239
So, if we count these, one,
two, three, four, five, six,

00:02:23.239 --> 00:02:26.939
seven, eight, nine, ten, eleven, twelve.

00:02:26.939 --> 00:02:29.699
Twelve possible pot, flower scenarios

00:02:29.699 --> 00:02:33.699
to walk out of that florist store with.


Balancing more complex chemical equations

WEBVTT
Kind: captions
Language: en

00:00:00.510 --> 00:00:02.739
- [Voiceover] Let's now see
if we can balance a chemical

00:00:02.739 --> 00:00:07.000
equation with slightly
more complex molecules.

00:00:07.243 --> 00:00:10.044
So, here we have a chemical equation,

00:00:10.044 --> 00:00:11.428
describing a chemical reaction.

00:00:11.428 --> 00:00:14.029
This is actually a combustion reaction.

00:00:14.029 --> 00:00:17.256
You have some ethylene right over here,

00:00:17.256 --> 00:00:20.089
in the presence of oxygen,
and you need to get a

00:00:20.089 --> 00:00:22.643
little bit of energy to get
this going, but then you're

00:00:22.643 --> 00:00:24.315
going to have this reaction
that's actually going to

00:00:24.315 --> 00:00:26.845
release energy as well, but
we're not accounting for

00:00:26.845 --> 00:00:28.787
the energy, at least the
way we've written it.

00:00:28.787 --> 00:00:32.154
Right over here, you have
some ethylene, and this little

00:00:32.154 --> 00:00:35.126
g in parentheses, says
it's in the gas form

00:00:35.126 --> 00:00:39.498
or gaseous form, so
gaseous ethylene plus some

00:00:40.390 --> 00:00:43.879
dioxygen molecule, which is
the most prevalent form of

00:00:43.879 --> 00:00:47.600
oxygen molecule that you
would find in the atmosphere.

00:00:47.600 --> 00:00:49.411
And so, that's also in the gas form.

00:00:49.411 --> 00:00:52.754
Put them together, you end
up with some carbon dioxide

00:00:52.754 --> 00:00:55.610
gas and some liquid water.

00:00:55.610 --> 00:00:58.629
This is the classic combustion reaction.

00:00:58.629 --> 00:01:01.160
But now let's think about,
how do we balance this thing?

00:01:01.160 --> 00:01:03.157
Let's make sure we have the same number

00:01:03.157 --> 00:01:07.104
of each atom on both sides.

00:01:07.104 --> 00:01:09.519
And when you see something
more complicated like this,

00:01:09.519 --> 00:01:13.445
where, you know, here I have
an oxygen and two different

00:01:13.476 --> 00:01:17.215
molecules over here, and a
lot of these molecules have

00:01:17.215 --> 00:01:19.049
multiple elements in it.

00:01:19.049 --> 00:01:20.187
It might be very daunting.

00:01:20.187 --> 00:01:21.278
Where do I start?

00:01:21.278 --> 00:01:23.661
And this is where the
art of balancing chemical

00:01:23.661 --> 00:01:26.122
equations starts to come into play.

00:01:26.122 --> 00:01:28.444
The general idea is,

00:01:28.444 --> 00:01:30.828
Try to balance the...

00:01:30.828 --> 00:01:33.599
try to balance the
molecules that have multiple

00:01:33.599 --> 00:01:37.476
elements in them first, and leave the...

00:01:37.476 --> 00:01:40.462
molecules that only have one
element in them for last.

00:01:40.462 --> 00:01:43.039
And the idea there is,
is that these are harder.

00:01:43.039 --> 00:01:44.851
They're going to have all
sorts of implications,

00:01:44.851 --> 00:01:46.592
and then, at the end of
the day, you can just set

00:01:46.592 --> 00:01:50.887
a number here for the number of dioxygens.

00:01:50.887 --> 00:01:54.779
If you saved, say the ethylene for last,

00:01:54.779 --> 00:01:57.728
then every time, and you're
trying to balance the carbons,

00:01:57.728 --> 00:01:59.647
you try to change the number
of carbons, you're going to

00:01:59.647 --> 00:02:00.613
change the number of hydrogens,

00:02:00.613 --> 00:02:01.913
which is going to change the...

00:02:01.913 --> 00:02:03.112
You're going to have to balance
over and over, and you're

00:02:03.112 --> 00:02:05.712
going to go into this really
really really confusing circle.

00:02:05.712 --> 00:02:08.336
So, the best thing to do, try to balance

00:02:08.336 --> 00:02:11.981
the complex molecules first, and then

00:02:11.981 --> 00:02:15.023
save the single element
molecules for last.

00:02:15.023 --> 00:02:16.857
So let's do that.

00:02:16.857 --> 00:02:19.644
So, let's start with the carbons.

00:02:19.644 --> 00:02:22.639
So, over here, I have two carbons.

00:02:22.639 --> 00:02:26.725
Over here, I only have one carbon.

00:02:26.725 --> 00:02:28.420
I only have one carbon.

00:02:28.420 --> 00:02:31.558
So, it seems like the best
way to balance it is, I should

00:02:31.558 --> 00:02:35.456
have two molecules of carbon dioxide,

00:02:35.456 --> 00:02:37.464
and I haven't even thought
about the oxygens yet.

00:02:37.464 --> 00:02:39.855
By putting that two there,
that's going to change the

00:02:39.855 --> 00:02:42.154
number of oxygens I have
on the righthand side.

00:02:42.154 --> 00:02:44.499
But at least it balances my carbons.

00:02:44.499 --> 00:02:47.332
I now have two carbons
on the lefthand side,

00:02:47.332 --> 00:02:49.886
and I have two carbons
on the righthand side.

00:02:49.886 --> 00:02:52.208
I’m no longer magically destroying

00:02:52.208 --> 00:02:55.296
a carbon atom, all right.

00:02:55.296 --> 00:02:57.780
Now, let's move on to the
hydrogens, and remember,

00:02:57.780 --> 00:03:00.450
what I said is, let's wait
to do the oxygens last,

00:03:00.450 --> 00:03:03.864
because we have a molecule

00:03:03.864 --> 00:03:06.232
that only contains oxygen right over here,

00:03:06.232 --> 00:03:08.298
so we'll save oxygen for last.

00:03:08.298 --> 00:03:11.804
So, let's do hydrogen next.

00:03:11.804 --> 00:03:15.241
So, hydrogen, right over
here, we have four hydrogens.

00:03:15.241 --> 00:03:18.761
And on the righthand side,
we have two hydrogens.

00:03:18.761 --> 00:03:21.492
So, it seems like the easiest
thing to do to balance

00:03:21.492 --> 00:03:25.254
the hydrogens is to have two
of these water molecules.

00:03:25.254 --> 00:03:27.520
Now I have four hydrogens here,

00:03:27.520 --> 00:03:30.655
and I have four hydrogens there.

00:03:30.655 --> 00:03:33.023
Now, let's do the oxygen.

00:03:33.023 --> 00:03:34.788
Now, let's do the oxygen.

00:03:34.788 --> 00:03:36.854
I've balanced the carbons
and the hydrogens.

00:03:36.854 --> 00:03:38.633
And the reason why oxygen's
going to be interesting,

00:03:38.633 --> 00:03:41.466
I can just count the amount
of oxygen I now have here,

00:03:41.466 --> 00:03:45.660
after changing the amount
of molecules I have.

00:03:45.660 --> 00:03:48.182
And then I can adjust this
accordingly, because this is

00:03:48.182 --> 00:03:50.272
only going to affect the number of oxygens

00:03:50.272 --> 00:03:52.338
that I have on the lefthand side.

00:03:52.338 --> 00:03:54.939
Right now, on the lefthand
side, I have two oxygens,

00:03:54.939 --> 00:03:57.911
and on the righthand
side, let me count this,

00:03:57.911 --> 00:04:00.200
I have two O two's, really.

00:04:00.200 --> 00:04:02.940
So, this is going to be four oxygens here,

00:04:02.940 --> 00:04:05.216
and then I have, each of
these water molecules has

00:04:05.216 --> 00:04:07.421
one oxygen, but I have
two water molecules,

00:04:07.421 --> 00:04:11.485
so this is going to be two
oxygens, two oxygens here.

00:04:11.485 --> 00:04:15.609
So, on the righthand side, I
have four plus two oxygens.

00:04:15.609 --> 00:04:17.620
So, I have six oxygens
on the righthand side.

00:04:17.620 --> 00:04:19.919
I need six oxygens on the lefthand side.

00:04:19.919 --> 00:04:21.962
I need this number to be six.

00:04:21.962 --> 00:04:23.193
So, how do I do that?

00:04:23.193 --> 00:04:24.795
Well, I just need three
of these molecules.

00:04:24.795 --> 00:04:27.395
If I have three molecules,
each of them have two oxygens,

00:04:27.395 --> 00:04:29.833
I'm going to have a total of six oxygens.

00:04:29.833 --> 00:04:31.946
And just like that, we have balanced this

00:04:31.946 --> 00:04:36.084
combustion reaction,
this chemical equation.


Balancing another combustion reaction

WEBVTT
Kind: captions
Language: en

00:00:00.558 --> 00:00:03.332
- All right now we have
another combustion reaction.

00:00:03.332 --> 00:00:08.325
Instead of ethylene, we
now have ethane, C2H6,

00:00:08.932 --> 00:00:11.985
has two carbons and six hydrogen

00:00:11.985 --> 00:00:15.026
atoms in each molecule of ethane,

00:00:15.026 --> 00:00:16.154
and it is reacting.

00:00:16.154 --> 00:00:19.508
It's ethane gas, it is
reacting with molecular oxygen

00:00:19.508 --> 00:00:22.747
in gaseous form and they combust to

00:00:22.747 --> 00:00:26.507
form carbon dioxide gas and liquid water,

00:00:26.507 --> 00:00:28.504
and like we've seen in previous examples,

00:00:28.504 --> 00:00:31.210
this chemical equation is not balanced.

00:00:31.210 --> 00:00:32.405
How can we tell?

00:00:32.405 --> 00:00:35.436
Well here on the left hand
side we have two carbons.

00:00:35.436 --> 00:00:37.130
Here on the right hand
side we have one carbon.

00:00:37.130 --> 00:00:38.953
Here on the left hand side
we have six hydrogens.

00:00:38.953 --> 00:00:42.447
Here on the right hand side
we only have two hydrogens.

00:00:42.447 --> 00:00:46.126
Here on the left hand
side we have two oxygens.

00:00:46.126 --> 00:00:49.715
On the right hand side we
have two plus, three oxygens,

00:00:49.715 --> 00:00:53.476
so none of the elements here are balanced.

00:00:53.476 --> 00:00:56.691
But like we did in the
example of ethylene,

00:00:56.691 --> 00:00:58.925
whenever you see this
where you have several

00:00:58.925 --> 00:01:01.107
somewhat complex molecules involved,

00:01:01.107 --> 00:01:05.194
it's good to save the element that

00:01:05.194 --> 00:01:08.665
is in a molecule by itself for last,

00:01:08.665 --> 00:01:10.023
because you can just tweak this

00:01:10.023 --> 00:01:11.474
to change the number of oxygens

00:01:11.474 --> 00:01:13.425
without it having any other side effects

00:01:13.425 --> 00:01:15.467
on the number of carbons or hydrogens.

00:01:15.467 --> 00:01:17.515
So what I'm going to do is
I'm going to first balance,

00:01:17.515 --> 00:01:18.512
like we've done before,

00:01:18.512 --> 00:01:20.239
the carbons and the hydrogens,

00:01:20.239 --> 00:01:22.503
which are going to have
implications on the oxygens

00:01:22.503 --> 00:01:23.803
because if I change the number here,

00:01:23.803 --> 00:01:25.080
it's going to change
the number of oxygens.

00:01:25.080 --> 00:01:26.717
If I change the number here it's going to

00:01:26.717 --> 00:01:28.273
change the number of oxygens.

00:01:28.273 --> 00:01:30.780
But lucky for me I have
this dioxygen molecule

00:01:30.780 --> 00:01:33.717
on the left hand side that
I can just tweak at the end

00:01:33.717 --> 00:01:36.433
to balance the entire chemical equation.

00:01:36.433 --> 00:01:38.499
So let's start with, you know last time

00:01:38.499 --> 00:01:40.775
we started with carbon, let's
start with hydrogen this time.

00:01:40.775 --> 00:01:42.041
Just for kicks.

00:01:42.041 --> 00:01:44.652
So over here I have six hydrogens

00:01:44.652 --> 00:01:46.534
on the left hand side, of
the entire left hand side

00:01:46.534 --> 00:01:48.356
I only have six hydrogen atoms.

00:01:48.356 --> 00:01:50.620
On the right hand side I
only have two right now,

00:01:50.620 --> 00:01:55.413
so if I want to have six I would
multiply these two by three

00:01:55.717 --> 00:01:57.655
so now I have three water molecules,

00:01:57.655 --> 00:01:59.757
each of them have two hydrogen atoms,

00:01:59.757 --> 00:02:03.274
so I'm going to have
six, six hydrogen atoms

00:02:03.274 --> 00:02:04.991
on the right hand side.

00:02:04.991 --> 00:02:06.013
Fair enough.

00:02:06.013 --> 00:02:07.198
Now let's go to the carbon.

00:02:07.198 --> 00:02:09.523
Remember I'm saving oxygen for last.

00:02:09.523 --> 00:02:12.669
Carbon on the left hand
side I have two carbons.

00:02:12.669 --> 00:02:14.503
How many carbons do I have
on the right hand side?

00:02:14.503 --> 00:02:16.303
Well right now I only have one,

00:02:16.303 --> 00:02:18.031
but I can change that very easily.

00:02:18.031 --> 00:02:20.109
Instead of having one
molecule of carbon dioxide,

00:02:20.109 --> 00:02:23.022
I can have two molecules
of carbon dioxide.

00:02:23.022 --> 00:02:24.972
And so now my carbons are balanced,

00:02:24.972 --> 00:02:26.806
two carbons, two carbons.

00:02:26.806 --> 00:02:29.383
And now let's go to the oxygens.

00:02:29.383 --> 00:02:34.350
So right now, right now on,
I'll do this in a mauve color,

00:02:34.561 --> 00:02:38.148
right now on the left hand
side I have two oxygens,

00:02:38.148 --> 00:02:40.622
but on the right hand side what do I have?

00:02:40.622 --> 00:02:42.629
Let's see, I have two times two,

00:02:42.629 --> 00:02:45.660
so this right over here is four oxygens,

00:02:45.660 --> 00:02:47.447
and then I have three water molecules,

00:02:47.447 --> 00:02:49.606
each of them have one oxygen atom,

00:02:49.606 --> 00:02:52.706
so three times one, so
this is going to be three

00:02:52.706 --> 00:02:54.029
right over here.

00:02:54.029 --> 00:02:58.139
So on the entire right hand
side I have seven oxygen atoms

00:02:58.139 --> 00:03:00.427
and on the left I only have two.

00:03:00.427 --> 00:03:01.530
So what can I do here?

00:03:01.530 --> 00:03:05.139
What can multiply by two to get to seven?

00:03:05.139 --> 00:03:07.983
Two times what is equal to seven?

00:03:07.983 --> 00:03:11.408
Well two times three-and-a-half
is equal to seven.

00:03:11.408 --> 00:03:14.032
So two times three-and-a-half
is equal to seven.

00:03:14.032 --> 00:03:17.189
Remember I have two here,
I'm saying two times

00:03:17.189 --> 00:03:19.454
something is equal to seven.

00:03:19.454 --> 00:03:22.425
I want to get to the four plus the three.

00:03:22.425 --> 00:03:25.979
Well I multiply it by three point five,

00:03:25.979 --> 00:03:29.763
and now I have seven, seven oxygen atoms

00:03:29.763 --> 00:03:32.480
on both sides of my chemical equation.

00:03:32.480 --> 00:03:34.803
But like we've seen in previous videos,

00:03:34.803 --> 00:03:37.264
it is not standard to just
leave a three-and-a-half here,

00:03:37.264 --> 00:03:38.913
it's kind of this weird notion of

00:03:38.913 --> 00:03:40.909
three-and-a-half molecules,

00:03:40.909 --> 00:03:42.825
we like to have whole numbers here.

00:03:42.825 --> 00:03:45.924
So how do we make sure
we have all whole number

00:03:45.924 --> 00:03:48.351
coefficients in front of our molecules?

00:03:48.351 --> 00:03:50.452
Well we could just
multiply everything by two,

00:03:50.452 --> 00:03:52.402
then this thing is
going to become a seven,

00:03:52.402 --> 00:03:53.727
this thing is going to become a two,

00:03:53.727 --> 00:03:55.027
this is going to become a four,

00:03:55.027 --> 00:03:57.232
this is going to become a six,

00:03:57.232 --> 00:03:58.102
so let's just do that.

00:03:58.102 --> 00:04:00.122
I'll write the whole reaction over again.

00:04:00.122 --> 00:04:03.210
So I have my ethane, and I won't write

00:04:03.210 --> 00:04:06.485
actually what state it's
in just to save some time,

00:04:06.485 --> 00:04:09.202
plus some molecular oxygen.

00:04:09.202 --> 00:04:11.569
They combust, they yield,

00:04:11.569 --> 00:04:12.788
so these are the reactants,

00:04:12.788 --> 00:04:15.424
the products are carbon dioxide gas

00:04:15.424 --> 00:04:19.695
and liquid water, and liquid water.

00:04:19.695 --> 00:04:22.808
So let's see, you multiply,
if you say that there

00:04:22.808 --> 00:04:24.316
was a one out here before,

00:04:24.316 --> 00:04:26.871
we're going to multiply
that by two to get two.

00:04:26.871 --> 00:04:30.540
We had a three point five
here, multiply that by two,

00:04:30.540 --> 00:04:32.503
you get a seven here.

00:04:32.503 --> 00:04:36.763
We had a two right over
here, multiply that by two

00:04:36.763 --> 00:04:38.017
you get a four.

00:04:38.017 --> 00:04:40.002
Once again I'm just multiplying
all the coefficients

00:04:40.002 --> 00:04:41.802
by two just like you did
in algebraic equations

00:04:41.802 --> 00:04:43.763
in your algebra classes.

00:04:43.763 --> 00:04:47.479
And then finally three times two is six.

00:04:47.479 --> 00:04:48.663
And we're all balanced.

00:04:48.663 --> 00:04:50.079
We were balanced here but we didn't

00:04:50.079 --> 00:04:51.519
have whole number coefficients.

00:04:51.519 --> 00:04:53.516
Multiplying everything by two gave us

00:04:53.516 --> 00:04:56.521
the whole number coefficients
and we are much happier.


Balancing chemical equations

WEBVTT
Kind: captions
Language: en

00:00:00.375 --> 00:00:02.516
- Balancing chemical
equations is one of those

00:00:02.516 --> 00:00:07.087
concepts in chemistry that
often confuses people.

00:00:07.087 --> 00:00:11.624
But I think we'll see that
if we work through this

00:00:11.624 --> 00:00:13.673
carefully and methodically,
and we also appreciate

00:00:13.673 --> 00:00:15.797
the art of balancing chemical equations,

00:00:15.797 --> 00:00:18.120
that it's actually not too bad.

00:00:18.120 --> 00:00:20.722
So first of all, what
is a chemical equation?

00:00:20.722 --> 00:00:22.670
Well this is a chemical
equation right over here.

00:00:22.670 --> 00:00:24.776
It's describing a reaction.

00:00:24.776 --> 00:00:28.788
So if I take an atom of
aluminum and I add it

00:00:28.788 --> 00:00:32.070
to a dioxygen molecule, so a molecule that

00:00:32.070 --> 00:00:36.458
has two oxygens with it, under
the appropriate conditions

00:00:36.458 --> 00:00:40.054
they will react to form aluminum oxide.

00:00:40.054 --> 00:00:43.153
And the aluminum oxide molecule has

00:00:43.153 --> 00:00:47.064
two aluminum atoms and three oxygen atoms.

00:00:47.064 --> 00:00:48.601
And so you might say, "Okay, well what's

00:00:48.601 --> 00:00:49.960
"the balancing business all about?

00:00:49.960 --> 00:00:51.496
"I have a chemical reaction.

00:00:51.496 --> 00:00:53.059
"What do I have to balance?"

00:00:53.059 --> 00:00:55.342
Well if you look carefully,
you might notice that

00:00:55.342 --> 00:01:00.024
you don't have the same number
of each atom on both sides.

00:01:00.024 --> 00:01:02.341
For example, right over
here on the left-hand side,

00:01:02.341 --> 00:01:04.126
how many aluminums do we have?

00:01:04.126 --> 00:01:06.901
Well on the left-hand
side, we have one aluminum.

00:01:06.901 --> 00:01:09.151
How many do we have on
the right-hand side?

00:01:09.151 --> 00:01:12.303
Well on the right-hand
side, we have two aluminums.

00:01:12.303 --> 00:01:14.985
And so aluminum just can't
appear out of thin air

00:01:14.985 --> 00:01:17.312
by virtue of some magical reaction.

00:01:17.312 --> 00:01:20.076
You have to have the same amount
of aluminums on both sides,

00:01:20.076 --> 00:01:22.499
and the same thing is
true for the oxygens.

00:01:22.499 --> 00:01:26.400
Over here on the left-hand
side, we have two oxygens.

00:01:26.400 --> 00:01:31.138
They form one dioxygen molecule
that has two oxygen atoms.

00:01:31.138 --> 00:01:33.779
And then over here in the
aluminum oxide molecule,

00:01:33.779 --> 00:01:35.538
we have three.

00:01:35.538 --> 00:01:38.486
We have three oxygen atoms.

00:01:38.486 --> 00:01:41.208
So once again, we can't
just have miraculously

00:01:41.208 --> 00:01:44.838
an oxygen atom appear out of nowhere.

00:01:44.838 --> 00:01:47.567
So we have to balance the number
of aluminums on both sides,

00:01:47.567 --> 00:01:49.733
this number and this
number should be the same,

00:01:49.733 --> 00:01:51.638
and we have to balance
the number of oxygens,

00:01:51.638 --> 00:01:55.127
this number and that
number should be the same.

00:01:55.127 --> 00:01:56.552
So how do we do that?

00:01:56.800 --> 00:01:58.856
Well one thing might be
to say, "Okay, if I've got

00:01:58.856 --> 00:02:02.238
"two aluminums here and
I have one aluminum here,

00:02:02.238 --> 00:02:03.964
"well why don't I just double the number

00:02:03.964 --> 00:02:06.294
"of aluminums right over here?"

00:02:06.294 --> 00:02:08.157
I could just write a two in front of it,

00:02:08.157 --> 00:02:09.747
so now this has two aluminums,

00:02:09.747 --> 00:02:11.592
so I no longer have one aluminum here.

00:02:11.592 --> 00:02:14.102
I now have two aluminums,
and so it looks like

00:02:14.102 --> 00:02:17.624
the aluminums are balanced,
and they are indeed balanced.

00:02:17.624 --> 00:02:20.363
But still we have an
issue with the oxygens.

00:02:20.363 --> 00:02:22.079
Over here I have two oxygens.

00:02:22.079 --> 00:02:24.432
Over here I have three oxygens.

00:02:24.647 --> 00:02:25.893
So one thing that you might say is,

00:02:25.893 --> 00:02:27.894
"Okay, well how do I go from two to three?

00:02:27.894 --> 00:02:30.112
"I could multiply by 1.5."

00:02:30.112 --> 00:02:34.647
So I could multiply by 1.5,

00:02:34.647 --> 00:02:37.336
and if I multiply 1.5 times two,

00:02:37.336 --> 00:02:38.906
that's going to be three.

00:02:38.906 --> 00:02:42.560
So now I have three
oxygen atoms on this side

00:02:42.560 --> 00:02:45.219
and three oxygen atoms on this side.

00:02:45.219 --> 00:02:47.414
But the convention is that we don't like

00:02:47.414 --> 00:02:50.106
writing "1.5 molecules."

00:02:50.106 --> 00:02:52.404
We don't like having this
notion of a half molecule,

00:02:52.404 --> 00:02:54.329
which is kind of this bizarre notion.

00:02:54.329 --> 00:02:56.869
We want whole number molecules.

00:02:56.869 --> 00:02:58.900
So what can we do?

00:02:59.623 --> 00:03:01.748
Well, you can imagine that
this makes it very similar

00:03:01.748 --> 00:03:04.305
to what you did in algebra,
an algebraic equation.

00:03:04.305 --> 00:03:08.344
We just can multiply both
sides by the same number

00:03:08.344 --> 00:03:12.479
that gets rid of having this
fraction or this decimal here.

00:03:12.479 --> 00:03:14.165
So if we multiplied both sides

00:03:14.165 --> 00:03:15.992
by two, we're going to do that.

00:03:15.992 --> 00:03:19.236
This is going to be a four,
this is going to be a three,

00:03:19.236 --> 00:03:21.217
this is going to be a two right over here.

00:03:21.217 --> 00:03:22.249
So let me do that.

00:03:22.249 --> 00:03:25.139
Let me multiply both sides by two.

00:03:25.139 --> 00:03:26.738
So instead of two aluminum atoms,

00:03:26.738 --> 00:03:28.330
let me have four aluminum...

00:03:28.330 --> 00:03:30.855
Actually, let me just write
the chemical equation first

00:03:30.855 --> 00:03:32.346
in the form that it was before.

00:03:32.346 --> 00:03:36.509
So I had aluminum plus dioxygen,

00:03:36.709 --> 00:03:40.508
a molecule of two oxygens,
yielding in the reaction --

00:03:40.508 --> 00:03:44.315
these are the reactants,
this is the product --

00:03:44.315 --> 00:03:49.295
aluminum, aluminum, aluminum oxide.

00:03:49.728 --> 00:03:51.993
So what I'm saying here
is to get rid of this 1.5,

00:03:51.993 --> 00:03:55.030
to turn it into a whole number,
let's multiply all of these,

00:03:55.030 --> 00:03:57.810
all of the number of molecules by two.

00:03:57.810 --> 00:04:01.773
And here, there's implicitly a one...

00:04:01.773 --> 00:04:03.293
Let me do this in a different color.

00:04:03.692 --> 00:04:06.380
There is implicitly a one right over here.

00:04:06.380 --> 00:04:08.394
So let's multiply all of these by two.

00:04:08.394 --> 00:04:10.680
So two times two is...

00:04:10.680 --> 00:04:12.622
Let me do that same color.

00:04:13.022 --> 00:04:16.363
Two times two is four...

00:04:16.640 --> 00:04:18.500
That's not the same color.

00:04:18.838 --> 00:04:21.346
Two times two is four.

00:04:21.346 --> 00:04:25.117
1.5 times two is three.

00:04:25.117 --> 00:04:27.316
And then one times two,

00:04:27.316 --> 00:04:29.235
one times two is two.

00:04:29.235 --> 00:04:31.141
And now you can verify how many aluminums

00:04:31.141 --> 00:04:32.991
do we have on each side?

00:04:32.991 --> 00:04:37.050
Well I have four aluminum
atoms on the left-hand side,

00:04:37.050 --> 00:04:39.015
and how many do I have
on the right-hand side?

00:04:39.015 --> 00:04:40.976
I have four aluminum atoms.

00:04:40.976 --> 00:04:43.935
How many oxygens do I have
on the left-hand side?

00:04:43.935 --> 00:04:48.382
I have three molecules of dioxygen.

00:04:48.382 --> 00:04:50.623
Each molecule has two oxygen atoms,

00:04:50.623 --> 00:04:53.656
so I have six oxygens on the left,

00:04:53.656 --> 00:04:57.218
and I have two times three oxygens

00:04:57.218 --> 00:04:58.908
on the right, or I have six oxygens.

00:04:58.908 --> 00:05:02.108
So my chemical equation is now balanced.


Visually understanding balancing chemical equations

WEBVTT
Kind: captions
Language: en

00:00:00.289 --> 00:00:01.113
- [Voiceover] We've now seen a couple

00:00:01.113 --> 00:00:03.622
of examples of balancing
chemical equations.

00:00:03.622 --> 00:00:05.909
And we've seen that, okay, if let's say,

00:00:05.909 --> 00:00:08.094
we're trying to balance this
equation right over here

00:00:08.094 --> 00:00:09.624
and we started with the carbons.

00:00:09.624 --> 00:00:12.062
I have two carbons on the reactant side.

00:00:12.062 --> 00:00:14.490
They're both sitting in
this ethylene molecule.

00:00:14.490 --> 00:00:17.905
And so I would want two
carbons on the product side.

00:00:17.905 --> 00:00:20.283
And right now, I only have one
carbon on the product side.

00:00:20.283 --> 00:00:23.214
And so what we've done
is let's just put a two

00:00:23.383 --> 00:00:25.881
out front here and so now we have two

00:00:26.466 --> 00:00:28.168
for every molecule of ethylene

00:00:28.168 --> 00:00:30.873
and we're not done balancing
this chemical equation yet.

00:00:30.873 --> 00:00:35.538
We're now producing two
molecules of carbon dioxide.

00:00:35.690 --> 00:00:37.014
But one thing that you
might have been thinking,

00:00:37.014 --> 00:00:38.906
"Why put this big two out front

00:00:38.906 --> 00:00:41.054
"of the entire carbon dioxide,

00:00:41.054 --> 00:00:43.782
"I like the way these little
subscript little twos look,

00:00:43.782 --> 00:00:48.203
"so why not put a two right over there. "

00:00:48.765 --> 00:00:50.158
And the reason why you can't do that

00:00:50.158 --> 00:00:53.561
is that's actually changing the molecule.

00:00:53.561 --> 00:00:54.921
It's no longer carbon dioxide,

00:00:54.921 --> 00:00:56.466
it's now this bizarre thing that

00:00:56.466 --> 00:00:59.401
doesn't really exist in nature.

00:00:59.611 --> 00:01:02.049
Which is a C2O2 thing.

00:01:02.049 --> 00:01:03.664
You're actually changing the reaction

00:01:03.664 --> 00:01:04.802
when you're doing that.

00:01:04.802 --> 00:01:06.459
When you're balancing chemical reactions,

00:01:06.459 --> 00:01:08.671
the reaction itself is,

00:01:08.671 --> 00:01:10.415
even before it's balanced is describing

00:01:10.415 --> 00:01:12.550
something that happens.

00:01:12.765 --> 00:01:14.547
When it's unbalanced, it just doesn't

00:01:14.547 --> 00:01:17.101
have the numbers right in terms

00:01:17.101 --> 00:01:18.703
of number of molecules.

00:01:18.703 --> 00:01:19.818
So the only thing that you can

00:01:19.818 --> 00:01:21.585
change when you're changing these

00:01:21.585 --> 00:01:23.441
is the number of molecules.

00:01:23.441 --> 00:01:25.867
You can't change the number of

00:01:25.867 --> 00:01:27.680
constituents within the molecule.

00:01:27.680 --> 00:01:31.566
And that's why we do not
change these subscripts.

00:01:31.582 --> 00:01:34.336
And if you want to visualize
it a little bit differently,

00:01:34.812 --> 00:01:37.665
let's draw each of these molecules

00:01:37.665 --> 00:01:40.366
So ethylene looks like this.

00:01:40.485 --> 00:01:42.631
Double bond carbons.

00:01:42.866 --> 00:01:46.021
Then each carbon is
bonded to two hydrogens.

00:01:46.373 --> 00:01:49.562
Notice you have two
carbons and four hydrogens.

00:01:50.423 --> 00:01:53.268
Molecular oxygen, O2.

00:01:53.268 --> 00:01:54.255
Looks like this.

00:01:54.255 --> 00:01:56.879
It's oxygen doubled bonded to oxygen.

00:01:56.879 --> 00:02:01.337
And then you have carbon dioxide.

00:02:01.675 --> 00:02:06.398
Carbon dioxide is a carbon, double bonded,

00:02:06.398 --> 00:02:09.382
to two oxygens each.

00:02:09.382 --> 00:02:11.181
And then finally, you have water.

00:02:11.181 --> 00:02:14.470
Finally, water, actually I'm going to do

00:02:14.470 --> 00:02:16.408
this in a slightly different color.

00:02:16.408 --> 00:02:17.802
Your water right over here,

00:02:17.802 --> 00:02:21.222
this is an oxygen bonded to two hydrogens.

00:02:21.363 --> 00:02:23.905
So let me write the plus signs there.

00:02:23.917 --> 00:02:26.008
So plus is right over there.

00:02:26.008 --> 00:02:30.050
So if you were to
somehow write a subscript

00:02:30.050 --> 00:02:32.486
of two right over there, you would somehow

00:02:32.486 --> 00:02:35.483
be changing the structure,
you would be changing

00:02:35.483 --> 00:02:38.024
what this molecule is.

00:02:38.024 --> 00:02:39.767
As opposed to that, which
we don't want to do,

00:02:39.767 --> 00:02:42.498
we want to say, we're definitely
producing carbon dioxide.

00:02:42.498 --> 00:02:43.538
We're definitely producing carbon dioxide,

00:02:43.538 --> 00:02:45.664
but how many carbon dioxides do we produce

00:02:45.664 --> 00:02:49.694
for each molecule of ethylene?

00:02:49.694 --> 00:02:51.170
And so that's where we say,

00:02:51.170 --> 00:02:52.272
"Okay, we have two carbons here,

00:02:52.272 --> 00:02:53.817
"we want two carbons here."

00:02:53.817 --> 00:02:55.789
We don't want to change
the actual structure here,

00:02:55.789 --> 00:02:57.925
so let's write the two out front.

00:02:57.925 --> 00:03:00.735
So we're going to have two of these.

00:03:00.735 --> 00:03:01.817
Or if you want to do it,

00:03:01.817 --> 00:03:03.699
if you want to have it more visual,

00:03:03.699 --> 00:03:04.240
you could write it,

00:03:04.240 --> 00:03:06.118
okay, we're just going to
have another molecule here.

00:03:06.118 --> 00:03:09.564
We're just going to have another molecule.

00:03:09.624 --> 00:03:11.296
So now we've balanced the carbons,

00:03:11.296 --> 00:03:13.316
two carbons on the reactant side.

00:03:13.316 --> 00:03:16.345
Two carbons on the product side.

00:03:16.686 --> 00:03:18.206
And then you go to the hydrogens.

00:03:18.206 --> 00:03:19.760
Say, "okay, we have four hydrogens here,

00:03:19.760 --> 00:03:21.118
"we only have two here."

00:03:21.118 --> 00:03:23.556
Well, what if we had two
of these water molecules?

00:03:23.556 --> 00:03:27.095
So, let's draw another
water molecule here.

00:03:27.447 --> 00:03:30.104
So O H, and actually let me write

00:03:30.104 --> 00:03:31.984
this right over here, so
now we have two of these.

00:03:31.984 --> 00:03:36.060
And now we have two of these.

00:03:36.097 --> 00:03:38.278
And now we want to balance the oxygens.

00:03:38.278 --> 00:03:40.649
So let's see, on this side
we only have two oxygens.

00:03:40.649 --> 00:03:41.986
On this side we have one, two,

00:03:41.986 --> 00:03:44.968
three, four, five, six oxygens.

00:03:46.167 --> 00:03:46.934
In order to balance it, we're going to

00:03:46.934 --> 00:03:49.207
have six oxygens on the reactant side.

00:03:49.207 --> 00:03:51.666
We need three of these molecules.

00:03:52.051 --> 00:03:57.051
So, another one and another
one right over there.

00:03:57.162 --> 00:03:58.418
And now we're all balanced.

00:03:58.418 --> 00:04:00.403
We have two carbon atoms on both sides.

00:04:00.403 --> 00:04:02.063
Carbons, carbons.

00:04:02.063 --> 00:04:05.663
We have four hydrogen atoms on both sides.

00:04:05.663 --> 00:04:08.796
Here they're in the ethylene,
here they're in the waters.

00:04:08.927 --> 00:04:12.445
And then we have six
oxygen atoms on both sides.

00:04:12.445 --> 00:04:15.767
And here some of the oxygens
are in the carbon dioxides,

00:04:15.767 --> 00:04:17.518
some of them are in the waters.

00:04:17.518 --> 00:04:20.026
And here, they're all in
the molecular oxygens.

00:04:20.026 --> 00:04:22.165
But notice, we didn't change the

00:04:22.165 --> 00:04:25.045
actual structure of the molecules.

00:04:25.045 --> 00:04:27.923
We just changed the number
of molecules we have and

00:04:27.923 --> 00:04:30.223
that's what these coefficients in front of

00:04:31.868 --> 00:04:35.320
these molecular formulas represent.


Balancing chemical equation with substitution

WEBVTT
Kind: captions
Language: en

00:00:00.571 --> 00:00:02.566
- [Voiceover] All right,
let's see what's going on

00:00:02.566 --> 00:00:04.133
in this chemical reaction.

00:00:04.200 --> 00:00:05.666
On the reactant side,

00:00:06.033 --> 00:00:08.433
we have an iron oxide.

00:00:08.433 --> 00:00:10.767
This is ferric oxide right over here,

00:00:10.866 --> 00:00:13.866
reacting with sulfuric acid,

00:00:14.166 --> 00:00:15.501
and it's producing,

00:00:15.933 --> 00:00:20.067
this is ferric sulfate and water.

00:00:20.533 --> 00:00:22.600
We want to balance this chemical equation.

00:00:22.600 --> 00:00:25.900
I encourage you to pause this
video and try to balance this.

00:00:27.000 --> 00:00:28.567
I'm assuming you've had a go at it,

00:00:28.567 --> 00:00:30.600
and you might have been able
to successfully balance it.

00:00:30.873 --> 00:00:32.166
If you didn't, if you weren't able

00:00:32.166 --> 00:00:33.400
to successfully balance it,

00:00:33.400 --> 00:00:37.267
one theory of why you
weren't able to is because

00:00:37.267 --> 00:00:39.700
this sulfate group made
things really confusing.

00:00:39.900 --> 00:00:42.300
You have four oxygens, three oxygens here,

00:00:42.300 --> 00:00:43.534
you have seven on this side,

00:00:43.768 --> 00:00:46.200
then you have four oxygens
in the sulfate group here,

00:00:46.200 --> 00:00:47.467
but you have three sulfate groups.

00:00:47.467 --> 00:00:49.533
This is 12 oxygens here,

00:00:49.666 --> 00:00:51.033
and then you have another oxygen here.

00:00:51.033 --> 00:00:52.906
This seems really, really,
really, really confusing,

00:00:53.133 --> 00:00:55.300
especially with all this
sulfate group business.

00:00:55.767 --> 00:00:58.900
The key here is to appreciate
that the sulfate group

00:00:58.900 --> 00:01:00.600
is kind of staying together.

00:01:00.600 --> 00:01:01.700
You can kind of treat it ...

00:01:01.700 --> 00:01:03.534
Instead of just saying,
"Hey, let's try to balance

00:01:03.534 --> 00:01:04.901
"all of the oxygens," you can say,

00:01:04.901 --> 00:01:07.133
"Let's balance the oxygens

00:01:07.133 --> 00:01:09.300
"that are outside of the
sulfate groups separately,

00:01:09.300 --> 00:01:11.733
"and let's balance the
sulfate groups separately."

00:01:12.067 --> 00:01:14.166
To help our brains grapple with that,

00:01:14.166 --> 00:01:15.867
I'm going to rewrite
this chemical equation

00:01:15.867 --> 00:01:16.933
with a substitution.

00:01:17.267 --> 00:01:18.033
I'm going to say,

00:01:18.033 --> 00:01:21.833
let's say that x is equal to a sulfate,

00:01:22.233 --> 00:01:23.466
a sulfate group.

00:01:23.500 --> 00:01:24.820
Let me rewrite all of this business.

00:01:25.666 --> 00:01:26.633
I have ...

00:01:27.166 --> 00:01:29.067
It's nice to have, I guess
this is close to a rust color,

00:01:29.067 --> 00:01:30.368
which seems appropriate.

00:01:30.400 --> 00:01:32.300
Let's say I have some ferric oxide.

00:01:32.300 --> 00:01:35.433
I'm just rewriting what I
have above right over here.

00:01:35.433 --> 00:01:39.167
Ferric oxide plus some sulfuric acid.

00:01:39.733 --> 00:01:41.200
But instead of writing H2,

00:01:41.233 --> 00:01:42.667
and then writing the sulfate group,

00:01:42.667 --> 00:01:44.533
I'm going to write H2 and then x.

00:01:45.266 --> 00:01:49.600
H2 and then x is going to yield ...

00:01:50.200 --> 00:01:51.567
Is going to yield

00:01:52.233 --> 00:01:53.501
ferric sulfate.

00:01:54.067 --> 00:01:56.133
Ferric sulfate has three sulfate groups.

00:01:57.067 --> 00:01:58.833
So x is a sulfate group.

00:01:59.100 --> 00:02:00.333
It's going to have three of them.

00:02:00.675 --> 00:02:03.866
Ferric sulfate plus molecular,

00:02:04.366 --> 00:02:06.166
plus molecular water.

00:02:06.633 --> 00:02:08.933
Now, even though x
represents an entire group,

00:02:09.201 --> 00:02:11.333
let's treat it like an
element and just ...

00:02:11.333 --> 00:02:13.433
make sure we have the same number of x's,

00:02:13.433 --> 00:02:15.733
or the same number of
sulfate groups on both sides.

00:02:15.800 --> 00:02:17.466
Let's balance this chemical equation.

00:02:18.400 --> 00:02:20.133
Let's start with the iron.

00:02:21.501 --> 00:02:23.033
Over here, I have two irons.

00:02:23.933 --> 00:02:26.100
And over here, on the right-hand
side, I have two irons.

00:02:26.300 --> 00:02:28.366
It doesn't seem like I have
to tweak the irons at all.

00:02:28.966 --> 00:02:31.267
Now let's move on to

00:02:32.082 --> 00:02:33.133
the oxygens.

00:02:34.000 --> 00:02:35.333
I have three oxygens here,

00:02:36.466 --> 00:02:37.366
on the left-hand side.

00:02:37.366 --> 00:02:39.066
On the right-hand side,
I only have one oxygen.

00:02:39.071 --> 00:02:40.267
But I can change that by saying,

00:02:40.267 --> 00:02:42.869
"Let's have three water molecules."

00:02:43.617 --> 00:02:45.300
Then this is going to be

00:02:45.733 --> 00:02:47.166
three, right over here.

00:02:47.534 --> 00:02:49.242
Now let's focus on the hydrogens.

00:02:50.269 --> 00:02:52.401
Focus on the hydrogens, I
have two hydrogens here,

00:02:52.933 --> 00:02:56.033
and I have six hydrogens right over here.

00:02:56.667 --> 00:02:59.566
If I have six hydrogens right over there,

00:02:59.966 --> 00:03:01.366
how do I get six hydrogens here?

00:03:01.766 --> 00:03:05.701
Well, I'll have three
molecules of sulfuric acid.

00:03:05.701 --> 00:03:07.700
Each of them have two hydrogen atoms.

00:03:07.700 --> 00:03:09.366
Now I have six hydrogens.

00:03:10.166 --> 00:03:12.433
My hydrogens, my irons, and my oxygens

00:03:12.433 --> 00:03:14.933
that are not part of the sulfate group

00:03:14.940 --> 00:03:15.966
are all balanced.

00:03:15.966 --> 00:03:16.903
Now let's see

00:03:17.111 --> 00:03:18.366
if we can balance the sulfate groups.

00:03:19.036 --> 00:03:21.400
On the left-hand side, I
have three sulfate groups.

00:03:21.833 --> 00:03:23.333
Let me do that in that magenta color.

00:03:24.468 --> 00:03:26.366
I have three sulfate groups.

00:03:26.468 --> 00:03:29.766
On the right-hand side, I also
have three sulfate groups.

00:03:30.238 --> 00:03:31.700
I'm all balanced.

00:03:32.033 --> 00:03:33.866
And if we want to un-substitute,

00:03:33.900 --> 00:03:35.100
we just go back up here.

00:03:35.133 --> 00:03:36.435
Okay, we didn't change the coefficient

00:03:36.435 --> 00:03:39.100
on this molecule, on the ferric oxide.

00:03:39.667 --> 00:03:42.467
We did change the coefficient
on the sulfuric acid.

00:03:42.566 --> 00:03:46.333
We say we have, for every
molecule of ferric oxide,

00:03:46.366 --> 00:03:48.766
we have three molecules of sulfuric acid.

00:03:49.305 --> 00:03:50.200
We didn't change this.

00:03:50.200 --> 00:03:51.555
They're going to yield

00:03:51.555 --> 00:03:53.967
one for every one
molecule of ferric oxide,

00:03:54.133 --> 00:03:55.867
and three molecules of sulfuric acid.

00:03:56.166 --> 00:03:57.800
It's going to yield,

00:03:58.639 --> 00:03:59.866
the product,

00:04:00.102 --> 00:04:03.000
are going to be one
molecule of ferric sulfate

00:04:03.266 --> 00:04:04.146
and

00:04:05.200 --> 00:04:08.136
three molecules of water.

00:04:08.633 --> 00:04:09.939
And we are done.


Factoring algebraic expressions using the distributive property

WEBVTT
Kind: captions
Language: en

00:00:00.171 --> 00:00:02.982
- In earlier mathematics
that you may have done,

00:00:02.982 --> 00:00:05.789
you probably got familiar
with the idea of a factor.

00:00:05.789 --> 00:00:06.926
So for example, let me just pick

00:00:06.926 --> 00:00:09.382
an arbitrary number, the number 12.

00:00:09.382 --> 00:00:14.382
We could say that the
number 12 is the product of

00:00:14.552 --> 00:00:18.730
say two and six; two
times six is equal to 12.

00:00:18.730 --> 00:00:19.958
So because if you take the product of

00:00:19.958 --> 00:00:21.517
two and six, you get 12, we could say

00:00:21.517 --> 00:00:23.837
that two is a factor of
12, we could also say

00:00:23.837 --> 00:00:25.151
that six is a factor of 12.

00:00:25.151 --> 00:00:27.752
You take the product of
these things and you get 12!

00:00:27.752 --> 00:00:32.236
You could even say that
this is 12 in factored form.

00:00:32.236 --> 00:00:33.683
People don't really talk that way

00:00:33.683 --> 00:00:35.815
but you could think of it that way.

00:00:35.815 --> 00:00:39.417
We broke 12 into the things
that we could use to multiply.

00:00:39.417 --> 00:00:41.283
And you probably remember
from earlier mathematics

00:00:41.283 --> 00:00:43.040
the notion of prime factorization,

00:00:43.040 --> 00:00:45.042
where you break it up into
all of the prime factors.

00:00:45.042 --> 00:00:46.492
So in that case you could break the six

00:00:46.492 --> 00:00:48.294
into a two and a three, and you have

00:00:48.294 --> 00:00:53.294
two times two times three is equal to 12.

00:00:53.697 --> 00:00:55.625
And you'd say, "Well, this would be 12

00:00:55.625 --> 00:01:00.175
"in prime factored form or the
prime factorization of 12,"

00:01:00.175 --> 00:01:04.300
so these are the prime factors.

00:01:04.300 --> 00:01:08.095
And so the general idea,
this notion of a factor

00:01:08.095 --> 00:01:10.798
is things that you can multiply together

00:01:10.798 --> 00:01:12.748
to get your original thing.

00:01:12.762 --> 00:01:14.852
Or if you're talking about factored form,

00:01:14.852 --> 00:01:17.275
you're essentially taking the
number and you're breaking

00:01:17.275 --> 00:01:19.029
it up into the things
that when you multiply

00:01:19.029 --> 00:01:21.697
them together, you get
your original number.

00:01:21.697 --> 00:01:23.076
What we're going to do now is extend

00:01:23.076 --> 00:01:25.743
this idea into the algebraic domain.

00:01:25.743 --> 00:01:29.196
So if we start with an expression,

00:01:29.196 --> 00:01:31.527
let's say the expression is

00:01:31.527 --> 00:01:35.552
two plus four X,

00:01:35.552 --> 00:01:38.362
can we break this up into the product

00:01:38.362 --> 00:01:41.499
of two either numbers or two expressions

00:01:41.499 --> 00:01:44.579
or the product of a
number and an expression?

00:01:44.579 --> 00:01:47.183
Well, one thing that might jump out at you

00:01:47.183 --> 00:01:50.486
is we can write this as

00:01:50.486 --> 00:01:55.265
two times one plus two X.

00:01:55.265 --> 00:01:57.094
And you can verify if you like

00:01:57.094 --> 00:02:00.126
that this does indeed
equal two plus four X.

00:02:00.126 --> 00:02:02.588
We're just going to distribute the two.

00:02:02.588 --> 00:02:07.422
Two times one is two, two times two X

00:02:07.422 --> 00:02:11.820
is equal to four X, so plus four X.

00:02:11.820 --> 00:02:15.992
So in our algebra brains, this will often

00:02:15.992 --> 00:02:18.487
be reviewed as or referred to as

00:02:18.487 --> 00:02:22.644
this expression factored
or in a factored form.

00:02:22.644 --> 00:02:23.935
Sometimes people would say that

00:02:23.935 --> 00:02:26.893
we have factored out the two.

00:02:26.893 --> 00:02:28.305
You could just as easily say that

00:02:28.305 --> 00:02:30.372
you have factored out a one plus two X.

00:02:30.372 --> 00:02:35.266
You have broken this thing
up into two of its factors.

00:02:35.266 --> 00:02:37.194
So let's do a couple of examples

00:02:37.194 --> 00:02:38.882
of this and then we'll
think about, you know,

00:02:38.882 --> 00:02:40.787
I just told you that we
could write it this way

00:02:40.787 --> 00:02:42.730
but how do you actually figure that out?

00:02:42.730 --> 00:02:44.731
So let's do another one.

00:02:44.731 --> 00:02:47.147
Let's say that you had, I don't know,

00:02:47.147 --> 00:02:51.832
let's say you had, six, let
me just in a different color,

00:02:51.847 --> 00:02:56.127
let's say you had six X

00:02:56.127 --> 00:03:01.127
six X plus three,

00:03:01.396 --> 00:03:03.204
no, let's write it

00:03:03.204 --> 00:03:06.484
six X plus 30, that's interesting.

00:03:06.484 --> 00:03:08.549
So one way to think about it is

00:03:08.549 --> 00:03:11.186
can we break up each of these terms

00:03:11.186 --> 00:03:13.809
so that they have a common factor?

00:03:13.809 --> 00:03:17.353
Well, this one over here,
six X literally represents

00:03:17.353 --> 00:03:22.126
six times X, and then 30,
if I want to break out

00:03:22.126 --> 00:03:24.346
a six, 30 is divisible
by six, so I could write

00:03:24.346 --> 00:03:27.549
this as six times five,

00:03:27.549 --> 00:03:30.017
30 is the same thing as six times five.

00:03:30.017 --> 00:03:31.330
And when you write it this way,

00:03:31.330 --> 00:03:34.237
you see, "Hey, I can factor out a six!"

00:03:34.237 --> 00:03:35.703
Essentially, this is the reverse

00:03:35.703 --> 00:03:37.657
of the distributive property!

00:03:37.657 --> 00:03:40.521
So I'm essentially undoing
the distributive property,

00:03:40.521 --> 00:03:43.699
taking out the six, and you
are going to end up with,

00:03:43.699 --> 00:03:48.699
so if you take out the six,
you end up with six times,

00:03:49.823 --> 00:03:51.316
so if you take out the six here,

00:03:51.316 --> 00:03:52.805
you have an X, and you take out

00:03:52.805 --> 00:03:55.239
the six here, you have plus five.

00:03:55.239 --> 00:03:57.628
So six X plus 30, if you factor it,

00:03:57.628 --> 00:04:00.815
we could write it as
six times X plus five.

00:04:00.815 --> 00:04:02.985
And you can verify with
the distributive property.

00:04:02.985 --> 00:04:05.491
If you distribute this six, you get

00:04:05.491 --> 00:04:09.005
six X + five times six or six X + 30.

00:04:09.005 --> 00:04:09.927
Let's do something that's a little bit

00:04:09.927 --> 00:04:11.419
more interesting where we might

00:04:11.419 --> 00:04:13.770
want to factor out a fraction.

00:04:13.770 --> 00:04:16.290
So let's say we had the situation ...

00:04:16.290 --> 00:04:18.652
Let me get a new color here.

00:04:18.652 --> 00:04:20.404
So let's say we had

00:04:20.404 --> 00:04:25.187
1/2 minus 3/2,

00:04:25.187 --> 00:04:27.293
minus 3/2 X.

00:04:27.293 --> 00:04:29.541
How could we write this in a, I guess

00:04:29.541 --> 00:04:31.245
you could say, in a factored form,

00:04:31.245 --> 00:04:33.573
or if we wanted to factor out something?

00:04:33.573 --> 00:04:34.913
I encourage you to pause the video

00:04:34.913 --> 00:04:37.324
and try to figure it out,
and I'll give you a hint.

00:04:37.324 --> 00:04:39.730
See if you can factor out 1/2.

00:04:39.960 --> 00:04:41.559
Let's write it that way.

00:04:41.559 --> 00:04:42.845
If we're trying to factor out 1/2,

00:04:42.845 --> 00:04:44.816
we can write this first term as

00:04:44.816 --> 00:04:47.851
1/2 times one and this second one

00:04:47.851 --> 00:04:52.851
we could write as minus 1/2 times three X.

00:04:53.578 --> 00:04:55.956
That's what this is,
3/2 X is the same thing

00:04:55.956 --> 00:04:59.360
as three X divided by
two or 1/2 times three X.

00:04:59.360 --> 00:05:01.345
And then here we can see
that we can just factor

00:05:01.345 --> 00:05:03.520
out the 1/2 and you're going to get

00:05:03.520 --> 00:05:08.233
1/2 times one minus three X.

00:05:08.233 --> 00:05:10.560
Another way you could
have thought about it is,

00:05:10.560 --> 00:05:13.017
"Hey, look, both of these are products

00:05:13.017 --> 00:05:15.376
"involving 1/2," and
that's a little bit more

00:05:15.376 --> 00:05:18.428
confusing when you're
dealing with a fraction here.

00:05:18.428 --> 00:05:20.119
But one way to think about it is,

00:05:20.119 --> 00:05:23.252
I can divide out a 1/2
from each of these terms.

00:05:23.252 --> 00:05:25.787
So if I divide out a 1/2 from this,

00:05:25.787 --> 00:05:28.253
1/2 divided by 1/2 is one.

00:05:28.253 --> 00:05:31.267
And if I take 3/2 and divide it by 1/2,

00:05:31.267 --> 00:05:33.751
that's going to be
three, and so I took out

00:05:33.751 --> 00:05:35.798
a 1/2, that's another
way to think about it.

00:05:35.798 --> 00:05:37.287
I don't know if that confuses you more

00:05:37.287 --> 00:05:38.893
or it confuses you less, but hopefully

00:05:38.893 --> 00:05:42.808
this gives you the sense of
what factoring an expression is.

00:05:42.808 --> 00:05:44.911
I'll do another example,
where we're even using

00:05:44.911 --> 00:05:49.911
more abstract things, so
I could say, "AX plus AY."

00:05:50.705 --> 00:05:52.929
How could we write this in factored form?

00:05:52.929 --> 00:05:55.309
Well, both of these terms have products

00:05:55.309 --> 00:06:00.309
of A in it, so I could write
this as A times X plus Y.

00:06:00.554 --> 00:06:02.625
And sometimes you'll hear people say,

00:06:02.625 --> 00:06:04.981
"You have factored out the A,"

00:06:04.981 --> 00:06:07.532
and you can verify it if
you multiply this out again.

00:06:07.532 --> 00:06:11.903
If you distribute the A,
you'd be left with AX plus AY.


Algebraic expressions with fraction division

WEBVTT
Kind: captions
Language: en

00:00:00.007 --> 00:00:02.006
- Let's deal with some
algebraic expressions

00:00:02.006 --> 00:00:04.673
that involve multiplying fractions.

00:00:04.673 --> 00:00:09.673
So let's say that I had a over b,

00:00:09.746 --> 00:00:14.746
a over b times c over d.

00:00:17.014 --> 00:00:18.443
What is this going to be?

00:00:18.443 --> 00:00:19.701
I encourage you to pause the video

00:00:19.701 --> 00:00:22.534
and try to figure it out on your own.

00:00:22.534 --> 00:00:24.216
Well, when you multiply fractions,

00:00:24.216 --> 00:00:25.582
you just multiply the numerators

00:00:25.582 --> 00:00:27.352
and multiply the denominators.

00:00:27.352 --> 00:00:29.201
So the numerators here, a, c,

00:00:29.201 --> 00:00:30.698
you're just going to multiply those out.

00:00:30.698 --> 00:00:32.373
It's going to a times c,

00:00:32.373 --> 00:00:35.902
which we can just write as ac,

00:00:35.902 --> 00:00:38.890
that just a times c, all of that over

00:00:38.890 --> 00:00:42.381
the denominator, b times d,

00:00:42.381 --> 00:00:47.339
b times d.

00:00:48.691 --> 00:00:50.150
Instead of multiplying,
what would of happened

00:00:50.150 --> 00:00:51.463
if we were dividing?

00:00:51.463 --> 00:00:56.265
So if we had a over b,

00:00:56.265 --> 00:01:00.400
a over b divided by,

00:01:00.400 --> 00:01:04.573
divided by c over d,

00:01:04.573 --> 00:01:05.475
what would this be?

00:01:05.475 --> 00:01:07.719
And once again, encourage you,

00:01:07.719 --> 00:01:09.328
encourage you to pause the video

00:01:09.328 --> 00:01:11.800
and figure it out on your own.

00:01:11.800 --> 00:01:13.868
Well, when you divide by a fraction,

00:01:13.868 --> 00:01:15.695
it is equivalent to multiplying

00:01:15.695 --> 00:01:18.081
by it's, by it's reciprocal.

00:01:18.081 --> 00:01:20.345
So this is going to be the same thing

00:01:20.345 --> 00:01:25.328
as a over b, a over b times,

00:01:27.617 --> 00:01:30.517
times the reciprocal of this.

00:01:30.517 --> 00:01:34.477
So times d over, I'm going to use

00:01:34.477 --> 00:01:36.806
the same color just so
I don't confuse you,

00:01:36.806 --> 00:01:41.481
that d was purple, times d over c,

00:01:41.481 --> 00:01:44.644
times d over c and then it
reduces to a problem like this.

00:01:44.644 --> 00:01:45.827
You know, and I shouldn't even use

00:01:45.827 --> 00:01:49.114
this multiplication symbol
now that we're in algebra

00:01:49.114 --> 00:01:50.660
because you might confuse that with an x,

00:01:50.660 --> 00:01:55.347
so let me write that as times, times d,

00:01:56.560 --> 00:02:01.500
d over c, times d over c,

00:02:01.500 --> 00:02:02.634
and what are you left with?

00:02:02.634 --> 00:02:05.475
Well the numerator you're
going to have a times d,

00:02:05.475 --> 00:02:10.285
so it's ad, a, d, over,

00:02:12.484 --> 00:02:17.393
over bc, over b times c.

00:02:19.946 --> 00:02:24.114
Now let's do one that's maybe
a little bit more involved

00:02:24.114 --> 00:02:25.574
and see if you can pull it off.

00:02:25.574 --> 00:02:29.270
So let's say that I had,
let's say that I had,

00:02:29.270 --> 00:02:34.012
I don't know, let me write it as 1 over a,

00:02:34.012 --> 00:02:39.012
minus 1 over b, all of that over,

00:02:40.397 --> 00:02:44.441
all of that over c, and let's say,

00:02:44.441 --> 00:02:49.029
let's also divide that by 1 over d.

00:02:50.122 --> 00:02:52.161
So this is a more involved expression

00:02:52.161 --> 00:02:53.839
then what we've seen so
far but I think we have

00:02:53.839 --> 00:02:55.331
all the tools to tackle
it so I encourage you

00:02:55.331 --> 00:02:58.194
to pause the video and see
if you can simplify this,

00:02:58.194 --> 00:02:59.916
if you can actually carry
out these operations

00:02:59.916 --> 00:03:04.011
and come up with a one
fraction that represents this.

00:03:04.011 --> 00:03:05.813
All right so let's work
through it step by step,

00:03:05.813 --> 00:03:07.493
so 1 over a minus 1 over b,

00:03:07.493 --> 00:03:10.108
let me work through just
that part by itself.

00:03:10.108 --> 00:03:14.005
So 1 over a minus 1 over b,

00:03:14.005 --> 00:03:15.301
we know how to tackle that,

00:03:15.301 --> 00:03:16.716
we can find a common denominator,

00:03:16.716 --> 00:03:17.884
let me write it up here.

00:03:17.884 --> 00:03:22.283
So 1 over a minus 1 over b,

00:03:22.283 --> 00:03:23.536
is going to be equal to,

00:03:23.536 --> 00:03:26.447
we can multiply 1 over a times b over b,

00:03:26.447 --> 00:03:28.807
so it's going to be b over ba,

00:03:28.807 --> 00:03:30.190
notice I haven't changed it's value,

00:03:30.190 --> 00:03:32.697
I just multiplied it times 1, b over b,

00:03:32.697 --> 00:03:34.352
minus, well I'm going to multiply

00:03:34.352 --> 00:03:36.599
the numerator and denominator here by a,

00:03:36.599 --> 00:03:41.428
- a over ab, or I could write that as ba.

00:03:41.428 --> 00:03:42.487
And the whole reason why I did that

00:03:42.487 --> 00:03:44.794
is to have the same denominator.

00:03:44.794 --> 00:03:48.055
So this is going to be equal to b-a,

00:03:48.055 --> 00:03:52.781
over, I could write is a ba or ab.

00:03:52.781 --> 00:03:54.998
So this is going to be equal to,

00:03:54.998 --> 00:03:59.666
this is going to be equal to
this numerator right over here,

00:03:59.666 --> 00:04:03.877
b-a, over ab,

00:04:03.877 --> 00:04:07.505
and then, if I'm dividing it by c,

00:04:07.505 --> 00:04:09.096
that's the same thing as multiplying

00:04:09.096 --> 00:04:11.130
by the reciprocal of c.

00:04:11.130 --> 00:04:12.923
So if I'm dividing it by c,

00:04:12.923 --> 00:04:14.823
that's the same thing as multiplying,

00:04:14.823 --> 00:04:19.476
that's the same thing as
multiplying times 1 over c.

00:04:19.476 --> 00:04:22.794
And if I am, and I'll
just keep going here,

00:04:22.794 --> 00:04:25.690
if I'm dividing by 1 over d,

00:04:25.690 --> 00:04:27.428
if I'm dividing, notice
this is the same thing

00:04:27.428 --> 00:04:29.060
as division right over here.

00:04:29.060 --> 00:04:30.988
If I'm dividing by c,
that's the same thing

00:04:30.988 --> 00:04:33.429
as multiplying by the reciprocal of c.

00:04:33.429 --> 00:04:35.708
And then finally, I'm
dividing by 1 over d,

00:04:35.708 --> 00:04:36.908
that's the same thing as multiplying

00:04:36.908 --> 00:04:39.140
by the reciprocal of 1 over d.

00:04:39.140 --> 00:04:40.613
So the reciprocal of 1 over d

00:04:40.613 --> 00:04:44.236
is d over, d over 1.

00:04:44.236 --> 00:04:46.020
And so what does this result with?

00:04:46.020 --> 00:04:49.687
Well in the numerator I
have b-a times 1 times d.

00:04:49.687 --> 00:04:54.687
So we can write this as d times (b-a),

00:04:54.717 --> 00:04:58.821
times (b-a) and then in the denominator

00:04:58.821 --> 00:05:03.821
I have abc, ab and c.

00:05:05.620 --> 00:05:08.671
And then finally, we can use
the distributive property here,

00:05:08.671 --> 00:05:11.953
we can distribute this d,

00:05:11.953 --> 00:05:14.358
and we're going to be left with,

00:05:14.358 --> 00:05:17.151
we deserve a minor drum
roll at this point,

00:05:17.151 --> 00:05:21.473
we can write this as d times b,

00:05:21.473 --> 00:05:26.317
d times b minus d,
woops, I want to do that

00:05:26.317 --> 00:05:27.838
in the same green color so you really see

00:05:27.838 --> 00:05:32.090
how it got distributed, minus d times a,

00:05:32.090 --> 00:05:37.090
all of that over ab, abc.

00:05:40.308 --> 00:05:43.586
And we are done.


Algebraic expression adding fractions

WEBVTT
Kind: captions
Language: en

00:00:00.236 --> 00:00:02.168
- What I want to do in this video is

00:00:02.168 --> 00:00:04.781
really make sure that we feel comfortable

00:00:04.781 --> 00:00:09.125
manipulating algebraic expressions
that involve fractions.

00:00:09.125 --> 00:00:11.610
So we'll start with some
fairly straightforward ones.

00:00:11.610 --> 00:00:14.024
So let's say that I had, let's say I had

00:00:14.024 --> 00:00:19.024
A over B plus C over D,

00:00:20.698 --> 00:00:21.849
and if I actually wanted to

00:00:21.849 --> 00:00:24.589
add these things, so it
is just one fraction,

00:00:24.589 --> 00:00:26.434
how would I do that?

00:00:26.434 --> 00:00:28.129
Well, what we could do is,

00:00:28.129 --> 00:00:30.010
we could find a common denominator.

00:00:30.010 --> 00:00:32.309
Well, over here, we don't know what B is,

00:00:32.309 --> 00:00:33.911
we don't know what D is,

00:00:33.911 --> 00:00:35.164
but we know a common denominator

00:00:35.164 --> 00:00:37.208
is just going to be B times D.

00:00:37.208 --> 00:00:41.039
That is going to be a
common multiple of B and D.

00:00:41.039 --> 00:00:45.265
So we could rewrite this as two fractions,

00:00:45.265 --> 00:00:47.564
with the common denominator BD,

00:00:47.564 --> 00:00:52.395
so, plus, BD, actually, let
me color code it a little bit.

00:00:52.395 --> 00:00:55.430
So A over B is going to be the

00:00:55.430 --> 00:00:58.472
same thing as what over BD?

00:00:58.472 --> 00:01:02.210
Well, to get BD, I multiplied
the denominator by D,

00:01:02.210 --> 00:01:04.439
so let me multiply the
numerator by D as well,

00:01:04.439 --> 00:01:06.296
then I haven't changed
the value of the fraction,

00:01:06.296 --> 00:01:09.060
I'm just multiplying by D over D.

00:01:09.060 --> 00:01:13.285
So this is going to be A
times D over B times D.

00:01:13.285 --> 00:01:14.470
Notice I could divide the numerator

00:01:14.470 --> 00:01:16.257
and the denominator by D, and

00:01:16.257 --> 00:01:18.974
I'm going to get back to A over B.

00:01:18.974 --> 00:01:21.946
And then we can look at the
second fraction, C over D,

00:01:21.946 --> 00:01:26.946
to go from D to BD, we multiplied by B

00:01:27.729 --> 00:01:30.352
and so, if I multiply
the denominator by B,

00:01:30.352 --> 00:01:32.070
if I don't want to change
the value of the fraction,

00:01:32.070 --> 00:01:34.949
I have to multiply the
numerator by B as well.

00:01:34.949 --> 00:01:38.479
So let's multiply the
numerator by B as well,

00:01:38.479 --> 00:01:43.105
and it's going to be BC, BC.

00:01:43.105 --> 00:01:45.899
BC over BD.

00:01:45.899 --> 00:01:48.333
This is C over D.

00:01:48.333 --> 00:01:50.432
So what I have here in magenta,

00:01:50.432 --> 00:01:52.255
this fraction is equivalent
to this fraction.

00:01:52.255 --> 00:01:54.493
I just multiplied it by D over D,

00:01:54.493 --> 00:01:55.792
which we can assume is one,

00:01:55.792 --> 00:01:57.944
if we assume that D is not equal to zero,

00:01:57.944 --> 00:02:01.427
and then, if we just
multiply C over D times one,

00:02:01.427 --> 00:02:03.140
which is the same thing as B over B,

00:02:03.140 --> 00:02:05.184
if we assume that B is not equal to zero,

00:02:05.184 --> 00:02:07.856
then this fraction and this
fraction are equivalent.

00:02:07.856 --> 00:02:10.153
Now, why did I go through
all of this trouble?

00:02:10.153 --> 00:02:12.614
Well, now, I have a common denominator,

00:02:12.614 --> 00:02:15.168
so I can add these two fractions.

00:02:15.168 --> 00:02:16.770
So what's this going to be?

00:02:16.770 --> 00:02:19.115
Well, common denominator is BD,

00:02:19.115 --> 00:02:22.550
so let me just, so the
common denominator is BD,

00:02:22.550 --> 00:02:24.041
and I could just add the numerators,

00:02:24.041 --> 00:02:26.099
just like you would've
done if these were numbers,

00:02:26.099 --> 00:02:28.867
if this wasn't an algebraic expression.

00:02:28.867 --> 00:02:32.693
So this is going to
be, this is going to be

00:02:32.693 --> 00:02:36.693
AD plus BC, all of that over BD.