Example: Solving for a variable Example of solving for a variable
Example: Solving for a variable
- 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.
- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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