WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we will take a look at solving formulas and that is our only objective for our list of things to do.
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But one thing I want you to know while going through this one is how similar formulas to solve the equations.
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If you are curious what exactly is a formula and how they differ from those linear equations that we saw earlier.
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A formula is a type of the equation and usually conveys some sort of fundamental principle.
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Below, I have examples of all kinds of different formulas, D = RT and I = P × R × T.
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What these represents are usually something else.
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For example, D = RT we will look at that a little bit later on because it stands for distance = rate × time.
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I = PRT, that is an interesting formula and it stands for interest = principal, rate, time.
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It is not that these equations are too unusual but they actually have lots of good applications behind them.
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One thing that is a little bit scarier with these formulas is you will notice how they have a lot more variables than what we saw earlier.
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They have a lot of L, W and other could be multiple variables for just a single formula.
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You know we have those multiple variables present in there and usually we are only interested in solving for a single variable in the entire thing.
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The way we go about solving for that variable is we use exactly the same tools that we used for solving several linear equations.
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You have the addition property for equality and you have the multiplication property for equality.
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Both of these say exactly the same thing that they did before.
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One, you can add the same amount to both sides of an equation, keep it nice and balanced.
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And that you can multiply both sides of an equation by the same value and again keep it nice and balanced.
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The only difference here is that we will be usually adding and subtracting or multiplying, dividing by a variable of some sort.
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One thing that we usually thrown in as an assumption is that when we do multiply by something then we make sure that it is not 0.
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If I do multiply it by a variable, multiply by both sides by an X, then I will make the assumption that X is not 0.
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Just to make sure I do not violate the multiplication property of 0.
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You have lots of variables and the very first thing is probably identify what variable you are solving for.
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I usually like to underline it, highlight it in some sort of way just to keep my eye on it.
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If you have multiple copies of this variable, try to work on getting them together so that you can eventually isolate it.
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Some of these formulas do involve some fractions.
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Use the common denominator techniques so you can clear them out and not have to worry about them.
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Simplify each side of your equation as much as possible then isolate that variable that you are looking for.
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Try and get rid of all the rest of the things that are around you by using the addition and multiplication property.
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You can check to make sure that your solution works out by taking your solution for that variable and playing it back into the original.
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When you do this for your formula, what you often see is that a lot of stuff will cancel out and you will be left with a very simple statement.
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Usually we do not worry about that one too much unless we have some more numbers present in there but you can do it.
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Let us actually look at a formula and watch the solving process.
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You will see that it actually flows pretty easily.
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In this first one I have P=2L + 2W.
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This formula stands for the perimeter of a rectangle.
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It goes through and adds up both of the lengths and both of the widths.
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What we want to do with this one is we want to solve it for L.
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Let us identify where L is in our formula right there.
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What I’m going to do is I’m trying to get rid of everything else around it.
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The 2W is not part of what I'm interested in so I will subtract that from both sides.
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P and W are not like terms so even though it will end up on the other side of the equation, I will not be able to combine them any further.
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Now you have 2 - 2 or P - 2W = 2L, continuing on.
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I almost have that L completely isolated, let us get rid of that 2 by dividing.
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Notice how we are dividing the entire left side of this equation by 2 so that we can get that L isolated.
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What I have here is (P -2W)/2 = L.
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One weird part about solving a formula is sometimes it is tough to tell when you are done
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because usually with an equation when you are done you will have a number like L = 5 or L = 2.
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Since you have many variables in these formulas that when we get to the end what we developed is another formula right here.
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It is just in a different order or different way of looking at things.
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This formula that we have created could be useful if we were looking for L many times in a row and we had information about the perimeter and W.
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It is solved, it is a good the way that it is and the way we know it is solved is because L is completely isolated.
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Let us look at another one, in this one we want to solve for R.
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I have Q= 76R + 37, a lots of odd ball numbers but let us go ahead and underline what we would be looking for.
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We want to know about that R.
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I think we can get rid of a few fractions and we would have to multiply everything through by 6.
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Let me do that first, I will multiply the left side by 6, multiply the right side by 6.
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Q = 76R + 37, on the right side we better distribute that 6 and then we will see that it actually does take care of our fraction like it should.
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6Q = 7R, I know I have to take care of 37 × 6.
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I guess I better do some scratch work.
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37 × 6 I have 42, 3 × 6 is 18 + 4 = 22, a lot of 2.
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6Q = 7R + 222, I do not have to deal with any more fractions at this point so let us continue isolating the R and get it all by itself.
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I will subtract 222 from both sides, awesome.
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One final step to get R all by itself, we will divide it by 7.
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(6Q – 222)/7 = R and I will consider this one as solved.
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The reason why we can consider this one done is because we have isolated R completely.
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Even though we do have a Q still floating around in there, it is solved.
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Do not get too comfortable with that, it is not equal to just a single number, very nice.
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This one is a little different, we want to solve the following for V.
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The reason why this one is a little bit different is I have a V over here but I also have another V sitting over there.
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When you have more than one copy of the variable like this, you have to work on getting them together before you can get into the isolating process.
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Let us take care of our fractions and see if we can actually get those V’s together and work on isolating it.
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We only isolate one of them then it is not solved, we still have a V in there.
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To take care of our fraction I will multiply both sides of this one by W, it is my common denominator.
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(RV + Q) / W = W × 5V, on the left side those W’s would take care of each other.
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I will be left with RV + Q = W × 5V.
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In this point I do not have to deal with the fractions but notice we have not got those V’s any closer together so let us keep working on that.
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If I'm going to get them together, I at least better get them on the same side of the equation.
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I'm going to subtract say an RV from both sides right.
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Now comes the fun part, I still have two V’s, I’m going to make them into one.
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The way I’m going to do this is I’m going to think of my distributive property but I’m going to think of it in the other direction.
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If both of these have a V then I will pull it out front and I will be left with W5 – R.
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This step is a little bit tough when you see it at first but notice how it does work is that I'm taking both of these and moving them out front into a single V.
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I'm sure if it is valid, go ahead and take the V and put it back in using our distributive property and you will see that you be right back at this step.
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It is valid.
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The important part of why we are using it though is now we only have a single V and we can work further by isolating it.
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How do we get it all by itself, these V’s be multiplied by 5W – R, that entire thing inside the parentheses.
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We will divide both sides by that and then it should be all alone 5W – R = V.
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I literally just took this entire thing right here and divided it on both sides.
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To your left, now I can call this one done because V is completely isolated, it is all alone.
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There is no other V’s running around in there.
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I have worked hard to get them together and I know it is completely solved.