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# Intro to equations with variables on both sides

Video transcript

Let's try to solve a
more involved equation. So, let's say that we have 2x
plus 3, 2x plus 3 is equal to is equal to 5x minus 2. So this might look a
little daunting at first. We have x's on both
sides of the equation. We're adding and
subtracting numbers. How do you solve it? And we'll do it in a
couple of different ways. The the important thing
to remember is, we just want to isolate an x. Once you've isolated an x,
you have x equals something. Or x equals something. You're done, you've
solved the equation. You can actually go back and
check whether that works, So what we're going to do is just
do a bunch of operations on both sides of the equation,
to eventually isolate the x. But while we do those, I
actually want to visualize what's occurring. Because I don't want you just
say, oh what are the rules or the steps of solving equations. And I forgot whether this is a
allowed or that isn't allowed. If you visualize what's
happening, it'll actually be common sense what's allowed. So let's visualize it. So we have 2x right here
on the left-hand side. So that's literally,
that's x plus x. And then you have plus 3. Plus 3, I'll do it like this. So that's equal to plus
1, plus 1, plus 1. That's the same thing as 3. I could've drawn 3
circles here as well. Let do the same color. Plus 3. And then that is
equal to 5 x's. Do that in blue. That is equal to 5 x's. So, 1, 2, 3, 4, 5. And I want to make it clear. You would never actually
have to do it this way when you're solving the problem. You would just have to
do the algebraic steps. But I'm doing this for you so
you can actually visualize what this equation is saying. the left-hand side is these
two orange x's plus 3. The right-hand side
is 5x minus 2. So minus 2, we could write
as -- so let me do this in a different color,
I'll do it in pink. So, minus 2, I'll do as
minus 1 and minus 1. Now, we want to isolate
the x's on the same side of the equation. So, how could we do that? Well, there's two
ways of doing it. We could subtract these
two x's from both sides of the equation. And that would be
pretty reasonable. Because then you'd have
5 x's minus the 2 x's. You'd have a positive number of
x's on the right-hand side. Or, you could actually
subtract 5x from both sides. And that's what's
neat about algebra. As long as you do legitimate
operations, you will eventually get the right answer. So let's just start off
subtracting 2x from both sides of the equation. And what I mean there, I mean
we're going to remove 2 x's from the left-hand side. And if we were to move 2 x's on
the left-hand side, we have to remove 2 x's the
right-hand side. Just like that. So what does that give us? We're subtracting 2 x's. 2 x's from the left. And we're also going to
subtract 2 x's from the right. Now, what does our left-hand
side simplify to? We have 2x plus 3 minus 2x. The 2 x's cancel out. So you're just left with --
you're just left with the 3. And you see that over here. We took 2 of these x's away. We're just left with the
plus 1, plus 1, plus 1. And then on the right-hand
side, 5x minus 2x. We have it over here. We have 5 x's minus 2 x's. You only have 1, 2,
3, x's left over. 3 is equal to 3x. And then you have
your minus 2 there. You have your minus 2. So, normally if you were to do
the problem, you would just have to write what we have
here on the left-hand side. So what can we do next? Remember, we want to
isolate the x's. Well, we have all of our x's on
the right-hand side right here. If we could get rid of this
negative 2, off of the right-hand side, then
the x's will be alone. They'll be isolated. So how can we get rid of
this negative 2, if we visualize it over here. This negative 1,
this negative 1. Well, we could add 2 to both
sides of this equation. Think about what happens there. So, if we add 2, so I'm
going to do it like this. Plus 1, plus 1. So you could literally see. We're adding 2. And then we're going to add
2 to the left-hand side. 1 plus, 1 plus. What happens? Let me do it over here as well. So we're going to add 2. We're going to add 2. So what happens to
the left-hand side? 3 plus 2 is going
to be equal to 5. And that is going to be
equal to 3x minus 2 plus 2. These guys cancel out. And you're just left with 3x. And we see it over here. We have the left-hand side is 1
plus 1 plus 1 plus 1 plus 1. We have 5 1's, or 5. And the right-hand side,
we have the 3 x's, right over there. And then we have the
negative 1, negative 1. Plus 1, plus 1, negative
1, these cancel out. They get us to 0. They cancel out. So we're just left with
5 is equal to 3x. So we have 1, 2, 3,
4, 5 is equal to 3x. Let me clear everything that
we've removed, so it looks a little bit cleaner. These are all of the things
that we've removed. Let me clear that out. And then let me clear
that out, like that. Edit. Clear. So now we are just left
with 1, 2, 3, 4, 5. Actually, let me
move this over. So I could just move this
over right over here. We now have 1, 2, 3, 4, 5. These are the two that we
added here, is equal to 3x. These guys canceled out. That's why we have
nothing there. Now, to solve this, we
just divide both sides of this equation by 3. And this is going to
be a little hard to visualize over here. But if we divide over here both
sides by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why we divided
by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy word, it literally
just means the number multiplying the variable. The number we're solving, the
variable we're solving for. So these 3's cancel out. The right-hand side of
the equation is just x. The left-hand side is 5/3. So 5/3, we could say
is is equal to 5/3. And this is different than
everything we've seen so far. I now have the x on the
right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as
saying 5/3 is is equal to x is the same thing as saying
x's equal to 5/3. Completely equivalent. Completely equivalent. We sometimes get more used
to this one, but this is completely the same thing. Now, if we wanted to write this
as a mixed number, if we want to write this as a mixed
number, 3 goes into 5 one time with remainder 2. So it's going to be 1 2/3. So it's going to be 1 2/3. So we could also write
that x is equal to 1 2/3. And I'll leave it up for you
to actually substitute back into this original equation. And see that it works out. Now, to visualize it over here,
you know, how did he get 1 2/3, let's think about it. Instead of doing 1, I'm
going to do circles. I am going to do circles. Actually, even better,
I'm going to do squares. So I'm going to have 5 squares
on the left-hand side. I'll do it in this same
yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be
equal to the 3 x's. x plus x plus x. Now, we're dividing both
sides of the equation by 3. We're dividing both sides
of the equation by 3. Actually, that's where
we did it up here, we divided both sides by 3. So how do you do that
right-hand side's pretty straightforward. You want to divide these
3 x's into 3 groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you
divide 5 into 3? And they have to be thought
through even groups. And the answer tells us. Each group is going
to be 1 2/3. So, 1 2/3. So it's going to be 2/3
of this, the next one. And then we're going
to have 1 2/3. So this is 1/3. We're going to need another. Another 1, so this is 1 1/3. We're going to need 1 more
1/3, so this is going to be right here. And then we're left
with 2/3 and 1. So we've broken it
up into 3 groups. This right here. Let me make it clear. Let me make it clear, this
right here is 1 2/3. 1 2/3. And then this right
here, this 1/3. That's another 1/3, so
that's 2/3, and then that's 1 right there. So that's 1 2/3. And then finally this
is 2/3 and this is 1, so this is 1 2/3. So when you divide both
sides by 3 you get 1 2/3. Each section, each bucket, is
1 2/3 on the left-hand side. On the left-hand side, or 5/3. And on the right-hand
side we just have an x. So it still works. A little bit harder to
visualize with fractions.