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### Course: Algebra (all content)>Unit 15

Lesson 5: Manipulating formulas

# Manipulating formulas: perimeter

Sal rewrites the formula for the perimeter of a rectangle so it is solved for width. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• This is not a question but hopefully some useful info for people who get stuck on the set of questions afterwards.
I have spent the last few months brushing up on algebra and I read these posts and understand your frustrations so I will sum up what I think the key things to remember are which may help some of you (someone correct me if I am wrong because there is a reason why I am relearning algebra!)

1) Change sides, change signs means if you have a negative on one side of the equals sign, when you move it to the other side, it becomes positive i.e. x +2 = y becomes x = y -2. Similarly x-2=y +2
Usually you don't write out what Sal shows in the video of him adding the same thing to both sides of the equation as it is a given that if you add -2 to +2 it will cancel out (because it adds to zero)

2) You don't change signs for the variable when the number is a coefficient and is moving into the denominator.
eg -4y =12 becomes y = 12/-4 (you divide by -4 rather than subtract -4 because you need to break the 'bond' between -4 and y (i.e. -4 times y which = -4y)
However the general rule seems to be multiply by -1 so you don't have a negative sign in the denominator so you could choose to make it y= -12/4 (both would give you a result of -3)

3) Learn how to factorise - this is how you end up isolating the unknown. I can't summarise this succinctly but do a search on this and get your head around what factorisation is. As algebra increases in complexity this gets more important.

4) When you are presented with a problem with denominators first priority is to get rid of the denominator by multiplying both sides of the equation by the denominator. From the exercises it appears preference is given to dealing with the negative denominaotr first. You won't end up actually multiplying the part of the equation which has the denominator you are dealing with as it will be cancelled out
so
x + y / 10 + 30 = 50
you would do 10 (x +y) / 10 + (10) 30 = (10) 50
but you cancel the first 10 in the numerator because it is cancelled by the denominator so
(x + y) + 300 = 500
x+ y = 500 - 300
x + y = 200

I hope this helps someone in some small fashion. Do the exercises, and click 'show me a hint' and it will step you through so you can uncover what mistakes you tend to make.
(18 votes)
• This is GREAT! Thank you so much jacinta!
(2 votes)
• Why or when do you have to multiple the answer of an equation that is in fraction form by -1??? I know that isn't in this video, but it is in some of the problems in "practice this concept" for this video. And I don't see it explained in either video and it's driving me nuts!

For example,
-g-4h-4i+2 = -h+10i+4

The last step you have -1 as the denominator, and it say you have to make the denominator positive. Well, I swear I've seen answers with negatives in the denominator. ???

I found one!

8uv-6uw-9u+1 = 2v-7

The last step says to multiple top and bottom by -1. WHY!? You still are left with a negative in the denominator. The denominator goes from 8v-6w-9 to -8v+6w+9

Then ya got this!

4np+3nq-n-8 = 5p-6

Denominator is 4p+3q-1 but it doesn't say to change the denominator even though there is a negative number?? The only thing I can think is that the largest number in the denominator must be positive??

AND THIS!?!? What's going on?

5pq-5pr+p+6 = -q-5

You end up with the denominator being 5q-5r+1 but it say to change that to -5q+5r-1 WHY?
(11 votes)
• In the denominator and the numerator, they want to have the least amount of negative numbers.
Here is my work on the last problem:
Task, solve for p:
5pq-5pr+p+6=-q-5
-6 -6 =
5pq-5pr+p=-q-11 =
p(5q-5r+1)=-q-11 =
p(5q-5r+1)/(5q-5r+1)=-q-11/(5q-5r+1) =
p=-q-11/5q-5r+1* =
p=-q-11/-1/5q-5r+1/-1
p=q+11/5r-5q-1
*(Right now, they're 3 negatives. But if we multiply both sides by -1, they are now 2 negatives in the entire fraction(as opposed to the alternative of not dividing by -1 and having 3 negatives in the entire fraction))
Hope this helps.
(5 votes)
• I don't get the part where Sal made 2l -2l
(7 votes)
• It would seem to be confusing unless you realize that half of the work he is doing is off the left side of the screen. He is subtracting 2l from both sides of the equation. Alas, we can only see the right side of the equation. This video should be re-shot.
(3 votes)
• i was literally dying mentally with the sinsuoidal unit (the last 2 lessons) that made absolutely no sense and I barley know it and this unit is literally common sense that i'm confident a 7th grader would have no problem with this
(7 votes)
• Yes, I TOTALLY AGREE!
also, I am a 7th grader so...
(1 vote)
• Why are they called "variables"?
(5 votes)
• Because they can vary; it is in the name: VARIable. That is an over-simplification, but I think it gets to the root of what you are after. Take x = y + 5. Each time you are given a different y it will change the value of x.
(5 votes)
• The "Practice this concept" for this topic is way too hard compared to this video.
(5 votes)
• A lot of people have this problem including me...
Someone needs to ask him for more videos!
(1 vote)
• So is factoring basically "un-distributing" a number??
(3 votes)
• Yep! factoring and simplifying using the distributive property are basically the opposites..
Example:
Distributive property
3*(2x+3y) => 6x+9y
Factoring
6x+9y => 3*(x+y)
It works almost like multiplication and division, when they work together..
as in.. when you multiply 1*2.. the answer is 2..
If you divide the answer by the same number you multiplied by.. 2/2 makes 1... In simpler terms they reverse each other.
Hope this helps!
(2 votes)
• When solving for w in P=2w+2l, the final product is P-2l/2. My question is: can you divide the 2s somehow to further simplify it or can it not even be simplified? Thanks!
(3 votes)
• You can divide both terms in the numerator by 2, but it's not any more correct (or incorrect) than the current form.
W = (P-2L)/2 = P/2 - L
(2 votes)
• I don't really get it? Help.
(2 votes)
• To do this you need to use the whatever formula you're given and then use algebraic manipulation.
(4 votes)
• Quick question regarding the exercise; I noticed that once the problem has been simplified, there is a tendency to multiply both the numerator & denominator by -1 to change the signs all around, especially to convert the denominator into positives. Is this an algebraic law? What is the reason? And will the answer remain correct prior to making any changes to the signs?
(3 votes)

## 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.