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## Manipulating formulas

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# Manipulating formulas: perimeter

CCSS Math: HSA.CED.A.4

## Video transcript

We are told that the formula for
finding the perimeter of a rectangle is P is equal to 2l
plus 2w, where P is the perimeter, l is the length,
and w is the width. And just to visualize what
they're saying, and you might already be familiar with this,
let me draw a rectangle. That looks like a rectangle. And if this side's length is l,
then this side's length is also going to be l. And if this width is w, then
this width up here is w. And the perimeter is just how,
what is the distance if you were to go around
this rectangle? And so, that distance is going
to be this w plus this l, plus this w-- or that width--
plus this length. And if you have 1 w and you add
it to another w, that's going to give you 2 w's. So that's 2 w's. And then if you have 1 l, and
then you have another l, that's going to give you, if you
add them together, that's going to give you 2 l's. So the perimeter is going
to be 2 l's plus 2 w's. They just wrote it in a
different order than the way I wrote it. But the same thing, so hopefully
that makes sense. Now, their question is, rewrite
the formula so that it solves for width. So the formula, the way it's
written now, it says P is equal to something. They want us to write it so
it's, this w, right here, they want it to be w is equal to a
bunch of stuff with l's and P's in it, and maybe
some numbers there. So let's think about
how we can do this. So they tell us that P is
equal to 2 times l, plus 2 times w. We want to solve for w. Well, a good starting point
might be to get rid of the l on this side of the equation. And to get rid of it on that
side of the equation, we could subtract the 2l from both
sides of the equation. So let's do it this way. So you subtract 2l over here. Minus 2l. You're also going to have to do
that on the left-hand side. So you're going to
have minus 2l. We're doing it on both sides
of the equation. And remember, an equation says
P is equal to that, so if you do anything to that, you
have to do it to P. So if you subtract 2l from this,
you're going to have to subtract 2l from P in order
for the equality to keep being true. So the left-hand side is going
to be P minus 2l, and then that is going to be equal to--
well, 2l minus 2l, the whole reason why we subtracted 2l is
because these are going to cancel out. So these cancel out,
and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have
to divide both sides of this equation by 2. And the whole reason why I'm
dividing both sides of this equation by 2 is to get rid of
this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of
this equation by 2, once again, if you do something to
one side of the equation, you do it to the other side. The whole reason why I divided
the right-hand side by 2 is 2 times anything divided by 2
is just going to be that anything, so this is
going to be a w. And then we have our
left-hand side. So we're done. If we flip these two sides, we
have our w will be equal to this thing over here-- equals
P minus 2l, all of that over 2. Now, this is the
correct answer. There's other ways to
write it, though. You might want to rewrite this,
so let me square this off, because this is completely
the correct answer. This is the correct answer, but
there's other ways that you might have been able to
get this answer, other expressions for this answer. You might have also, you know,
another completely legitimate way to do this problem-- let me
write it this way-- so our original problem is P is equal
to 2l plus 2w-- is on this right-hand side, what if
we factor out a 2? So let me make this clear. You have a 2 here, and
you have a 2 here. So you could imagine
undistributing the 2. So we would get P is equal
to 2 times l plus w. This is an equally legitimate
way to do this problem. Now, we can divide both sides of
this equation by 2, so that we get rid of this 2 on
the right-hand side. So if you divide both sides of
this equation by 2, these 2's are going to cancel out-- 2
times anything divided by 2 is just going to be the
anything-- is equal to P over 2. So let me just rewrite
this over here. Let me just rewrite this. So we will get P over 2 is going
to be equal to l plus w. And then if we want to solve for
w, we just subtract l from both sides. And sometimes, you know, you
could write it in a separate line like this. Sometimes you could just
write it like this. You could say, I'm going to
subtract an l on that side. If I do it on that side,
I have to do it on this side, too. That's the same thing as
adding a negative l. And so the right-hand side,
you're just left with a w. And then the left-hand side,
you're going to have, it could be a negative l plus P over 2,
or you could just change the order, and you can write this
as P over 2 minus l. And this is also an equally
legitimate answer. And you're probably saying,
hey Sal, wait. These things look different. P minus 2l over 2, that
looks different than P over 2 minus l. And they're not. Think about this. We could rewrite this as P-- let
me do this with the same colors-- this over here is the
same thing as P over 2 minus 2l over 2. Right? If I have a minus b and they're
both being divided by 2, I can just separate-- you can
imagine I'm distributing the division by 2
right over here. And over here, 2 times
l divided by 2, that's just an l. So this is going to be equal to
P over 2 minus l, which is the exact same thing as
this right there.