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have this fun and you can do it.
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In this video we're going to solve literal equations.
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What I mean by that is we're going to solve for one variable
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that's in a formula but there are other variables as well.
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So an example would be our common distance equals rate times time
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usually written d equals RT. Remember that means R times T.
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So let's say we are going to solve for T.
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Well, pretend like d and R are numbers.
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If you don't know where to start.
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So for instance, what if it had been something like 3 equals 5T for an example?
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How would you solve this? You would divide by 5, right?
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But there's not a 5, there's an R is in there.
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So what we want to do is divide by R to isolate the T.
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And so we get d over R equals T.
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And usually what you're solving for a variable you write it.
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The variable on the left that you're solving for.
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So this would be saying dT equals d over R.
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So what does this mean?
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It means that if you knew the distance that you traveled and the rate you were going
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you could figure out how long it took you the time.
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The original formula d equals RT.
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You would use that if you knew the rate and the time.
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You would multiply the rate times the time to get the distance.
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So for instance, take a practical example just so you can see how this would work.
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Let's say you knew that you went 30 miles per hour.
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And you went four hours.
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And most of us would know that you went 120 miles.
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All right?
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So let's say you only knew two of these things.
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Remember, this is the rate, that's the time, and that's the distance.
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I could see that R times T equals d.
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But let's say somebody told you only the miles and the rate.
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And you didn't know how many hours.
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So let's pretend you didn't know what the time was.
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So we have this formula.
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T equals d over R and you want to find T.
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And so if you plugged d over R, we would have 120 miles.
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Right?
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And this is a little bit tricky.
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The rate is 30 miles per hour.
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So miles per hour basically means the hour would be in the denominator here.
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And when you multiply by the reciprocal, you would get four hours, which is what we had.
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So there's an example how you could use this formula.
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T equals d over R to figure out the number of hours.
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All right, let's go on to another example.
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Another common formula is the interest formula.
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Interest equals principal times rate times time.
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And let's say you were asked to solve for R.
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Solve for R.
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All right, so it's a little bit tricky here because you see sort of the R in the middle of two other variables.
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But remember, it doesn't matter the order you write when you do multiplication.
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So it might be easier to think of this as P-T-R.
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So you sort of are thinking of this as the coefficient of R.
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Similar to if you had something like this, five equals seven R.
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Right?
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Except instead of a seven, it's a P-T.
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Now what I'm doing over on the right, like I did in the previous example, is if you're not sure how to get started,
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you could think about having numbers in for all those other variables.
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So you would keep the R and then see how would I figure out this problem.
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And you would do this by dividing by seven.
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But in this case, there isn't a seven.
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There's a P-T being multiplied by R.
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So we want to divide both sides by P-T.
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And that way the P-T's cancel.
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And I get I over P-T equals R.
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And again, it's easier to read it when you write the variable that you're solving for on the left-hand side.
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Okay, let's do another one.
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All right, here is another formula.
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It has to do with perimeter.
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2W would be the width plus twice the height plus the length equals P. So 2W plus 2H plus L equals P, and we're going to solve for W.
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So first you find the variable W, and it's over here.
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And that's the only W.
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So basically you want to isolate the term that has a W on one side of the equation just like if this was 2W plus 5 equals P.
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So we need to subtract 2H and subtract L from both sides.
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So 2W is on both sides.
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Just like if this is just plus 2, you would subtract 2.
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So let's do that.
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2W equals, all right. So we're going to put that on the other side by subtracting 2H from each side and subtracting L.
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Now I'm not showing the minus 2H minus L on both sides of the equation.
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Hopefully you know that if you subtracted on both sides, you're going to see that over there, right?
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On the right-hand side of the equation.
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Okay, we're close to being done.
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This was 2W equals like 8, for instance, you would just divide both sides by 2, and you're going to do the same thing here.
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You're going to divide both sides by 2.
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But you've got this whole side to divide by 2.
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So you have to put the whole thing over 2.
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So we have W equals P minus 2H minus L all over 2.
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Now some people might write this differently by dividing each term by 2.
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So it would also be correct, but just not all written as one fraction if you wrote P over 2 minus...
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Now if you had 2H over 2, the 2 is a cancel, so that would just be an H, right?
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And the minus you would have L over 2.
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That's also a legitimate way to write the answer.
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You have to decide which looks in simplest form.
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I think the one I've put a box around, but they're both really correct.
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All right, let's do another one.
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S R plus T R equals U, and we're going to solve for R.
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All right, this is a little bit different because when you're looking for the variable R,
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there are actually two terms that have the variable for R.
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So think about what you would do is if it was something like 3R plus 5R equals 11, for instance,
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you would combine the light terms on the left here to get 8R.
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And how do you do that?
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You've got to add the 3 and the 5 together.
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And by the way, that's also an application of the distributive property.
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And that's how I usually like to think of this over here.
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Either factor out R, using the distributive property, or you could think of just adding the coefficients,
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which is S plus R. I mean S plus T.
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So either way you look at it, if you factor out R, you would look like this, right?
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And this S plus T is the coefficient.
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Or somebody else might have think, oh, I'm just going to add the coefficients.
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And then write the R. Two students might write it either of these ways to be correct,
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depending on how you think about the problem.
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Alright, I'm going to look at this second way, and it doesn't really matter.
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The trick is how am I going to isolate R? I've got something times R.
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Okay, well let's see.
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How would I have done it over here if I had 8 times R?
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I would have divided by the coefficient, which is 8.
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Well the coefficient here is this whole quantity S plus T.
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So I have to divide by the whole thing S plus T.
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Now on the right, I'm just going to put S plus T.
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You don't need to put that S plus T in parentheses.
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I did here so you could see how those are going to cancel.
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So R equals U over S plus T.
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And I've already got the very one on the left side of the equal sign,
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so it's in a nice form.
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It's another problem.
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Mn equals P minus MR.
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And we're going to solve for M.
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Alright, so again notice we have the variable M in two different terms in the problem.
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So what I needed is to sort of do the combining of light terms,
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but they have to be on the same side of the equation first.
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So these two terms, the Mn and the MR, negative MR,
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need to be on the same side of the equation.
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So let's add MR to both sides.
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Alright, so I have Mn, I'm going to add MR to both sides.
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Well when I add it on the right hand side, I'll be left only with P.
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Okay, now how am I going to combine these light terms?
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I can think of this as the coefficients of each M.
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I've got an N for the coefficient of the first term,
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and R is the coefficient of the second term.
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So if I want to just add the coefficients, I could write N plus R.
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That's how many M's I have,
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or you could have gotten the same thing by factoring an M out and writing M times N plus R.
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Either way it works.
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And again, that's just like a 5M,
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it's an N plus R, so we need to divide both sides by the coefficient.
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So we've got M equals P over N plus R.
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Alright, I hope you learned something.
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Your mouth is fun.