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This is the first part in the unit on functions.
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So in this first lesson, we're going to talk about equations as relations
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and talk about what it means for something to be an equation,
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graphing equations, and using equations with two variables.
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So this is something you might have already seen in a previous lesson or in a previous unit.
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But it's something a good way to start off functions.
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So here's examples of equations in two variables.
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Y equals 3x plus 12. That has two variables x and y.
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Or d equals 3t. D and t are the two variables.
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So it just is an equation and it has two variables in it and that's all that means.
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So what is a solution to an equation?
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If a statement is true after an ordered pair is substituted into an equation,
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then the ordered pair is a solution.
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So it just means if you put an ordered pair in to an equation and it makes sense,
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it's a valid statement, then that is a solution to the equation.
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So what are some solutions to the above equations up here?
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So for this first one,
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if I'm putting ordered pairs like xy,
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then if x were 1, then y would be 15.
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So that's a solution to that equation.
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If I put 1 in for x and 15 in for y,
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I would get 15 equals 3 plus 12,
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which is a true statement, so that would be a solution.
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Okay, for this d comma t equation, if I had db6,
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then t would have to be 2.
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And so I would have 6 equals 3 times 2,
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which is a correct statement or a true statement,
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so that is also a solution.
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So these are solutions to the equations above.
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All right, here's another expert equation with two variables.
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F equals, if you can't see this,
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F equals c times 9 fifths plus 32.
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So that's the conversion from Celsius to Fahrenheit.
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And these were high temperatures for the week and degrees Celsius.
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Actually, they were actual data for the week of October 30th
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to November 3rd in Okamus, Michigan for the year of 2006.
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So if I take these high temperatures and degrees Celsius,
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that would be called my domain.
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It's sort of like the dependent variable.
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It's the thing that I put in to get something else out.
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So my domain or the independent variable are these data right here.
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15.5, 12.8, 6.7, 4.4, and 1.7.
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And now I want to know my range or my Fahrenheit values.
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So these would be degrees Celsius.
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So my range would be degrees Fahrenheit.
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So remember the domain is the thing you put into the equation.
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The thing you put in to get something else out
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and that thing you get out is the range.
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So I need to calculate it for some of these.
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So I need to do 15.5 times 9 divided by 5.
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And then I want to add 32 to that.
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So 59.9 degrees Fahrenheit.
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That's my first one.
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For 12.8.
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Whoa!
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59.9.
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Okay, for 12.8, I multiply by 9 into by by 5.
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And then I'll add 32.
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55.04.
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6.7.
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As you can see, as this week went along,
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the high temperatures got lower and lower.
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We add 32.
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44.06.
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4.4 degrees Celsius.
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Times 9 divided by 5 plus 32.
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39.92.
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And then finally, 1.7 times 9 divided by 5 plus 32.
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35.06.
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Let me just do that first one again just to make sure that's right.
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Okay, that was 59.9.
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So those are my ordered pairs.
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Well, those are my domain and range values.
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And the ordered pairs are just the domain comma the range.
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And the domain comma the range.
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And I'm not going to do it for all of them.
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All right, so then if I were to graph these,
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and probably want to see the values,
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so I would want my independent variable or my domain on the x-axis.
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That's degrees Celsius.
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And my range is going to go on the y-axis.
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So the independent variable is always the x-axis here.
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And the dependent variable is always the y-axis.
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Okay, so I probably want my x-axis to go from 0 to 20.
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So this would be 20.
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And 10 would be right in the middle.
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And then my y-axis could go from 0 up to 60.
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So this would be 60.
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30.
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And then 15 would be right in the middle of that.
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And 45 would be right there.
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Okay, so let's plot a few of these.
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So 15.5.
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And almost 60.
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And we have 12.8 and 55.
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12.8 and 55.
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Then 6.7 and 44.
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4.4 and 40.
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And 1.7 and 35.
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Okay, if I had been a little more accurate, those should form a straight line.
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That was a little bit off though, but yes, those should form a straight line because this equation is a straight line.
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It looks like y equals mx plus b. It looks like that form.
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So it should form a straight line.
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All right, here's another relation we could look at.
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If there were only cows and people on a ranch and there are 26 feet,
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what are the possible combinations of cows and people?
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So our equation could be 4 times the number of cows because cows have 4 feet.
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Plus 2 times the number of people because people have 2 feet, those should equal 26.
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So there's a few questions I would want to ask.
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First of all, what's the independent variable?
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Well, in this case, it doesn't matter. Independent could be either one because cows don't depend on people and people don't depend on cows for how many feet there are.
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So I'm going to just call it C, P just to go an alphabetical order.
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So cows then would be the independent variable and people would be the dependent variable.
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For this case, as I said, though, it really does not make a difference.
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So the dependent variable would be people.
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And then we're going to pick some values and put them in and I ask, is there an easier way to do this?
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Well, let's pick some values first and then we'll get to that question.
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So this is going to be cows. The domain is the independent variable and the range is people, the dependent variable.
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So here's the easy way I was thinking of.
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If there were zero cows, how many people would there be?
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Well, if there were zero cows, this term would be gone and I would have 2 times the number of people equals 26.
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I would divide by 2 and the people would be 13. So our ordered pair would be 0, 13.
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Now, if there were zero people, how many cows would there be? Well, this would go to 0.
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And I would do 26 divided by 4, which would be 8 and a half.
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So, of course, we could not have 8 and a half cows and we also probably could not have zero people on a ranch to take care of those cows.
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But, nonetheless, that is an ordered pair that would fit this situation.
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Oh, I'm sorry, it should be 6 and a half.
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My math's a little off today. 6 times 4 is 24. So, 6 and a half would be 26.
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So, then we can just pick some values in between. So, if there were one person,
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actually, I'll put that down here.
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If there were one person, then we would have 4c plus 2 equals 26.
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So, 4c would equal 24 and then c would equal exactly 6.
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If there were 5 people, then that would be 4c plus 10 equals 26 and 4c would equal 16.
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So, c would equal 4.
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And then, what if there were 2 cows? So, 2 times 4 is 8.
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8 plus 2p is 26. So, 2p would equal 18.
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And so, 9 people.
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And you could write these all as ordered pairs, which I'm not going to do for the moment.
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And then, if I graphed these, I'm just going to use a scale of 1.
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So, I'm not going to put my scale down.
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And 0, 13 would be about right there. It's a little off, I think.
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29, 45, 61,
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and then 6.50.
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So, let's see how good my points were. And that wasn't that bad.
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And they again should form a straight line because this is a linear equation.
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So, this was cows. And this was people.
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As the number of people go up, the cows are going to go down.
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And as the number of cows go up, the number of people are going to go down.
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And that makes sense.
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So, the reason that I picked these two values to start with is because they're the intercepts.
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When there's zero people, I know how many cows there are. When there's zero cows, I know how many people there are.
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And then I can just connect those two points. And then that gives me my line.
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Alright, what if I know the perimeter of a rectangle is 24 centimeters?
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What are the possible lengths and widths?
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So, I ask what steps should you take?
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So, you should find an equation to model the situation first.
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So, we know what the perimeter is two times the width plus two times the length equals the perimeter.
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And the perimeter we know is 24. So, I'm going to put equals 24.
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So, there's my equation.
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What's the next step?
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Solve for one of the variables. So, we don't have to do this.
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It just makes things a little easier. So, then you can just keep plugging in one value and to find the other value.
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So, let's do that. So, we'll subtract two w. I'll solve for l.
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So, two l equals 24 minus two w. And then divide by two.
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So, I'll have l equals 12 minus w.
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So, now, as you see, if I plug in values for w, I'll get values for l out right away.
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So, it's a little quicker if you solve for one of the variables.
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Okay, let's determine values for domain and range that makes sense.
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So, well, now my domain is probably going to be w because it's easier to put things in for w and get values for l out.
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So, I'm going to do w, l.
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Alright, so, what values would make sense for w? Well, w can be between 0 and 12, right?
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Because a width can't be negative. And if the width were any longer than 12, I would do 12 minus a bigger number to get a negative number.
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And l can't be negative either. Neither the length or the width can be negative and make sense.
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So, the width can only be between 0 and 12.
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And we're going to choose five values.
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Okay, so we're going to choose values for w and then find what l would be.
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So, the first thing I'm going to do is do a 0.
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If the width is 0, the length is going to be 12, which would give me 0, 10 or 12.
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If the length were 2, sorry, if the width were 2, the length would be 10, 4 and 8, 6 and 6, and then I'm going to do a 12 and the 0.
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Alright, and let's plot these. So, 0, 12 is going to be right up here.
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And now the other one's 2, 10, 4, 8, 6, 6.
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And you can see the slope in this one already.
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Whenever you go down to, you go to the right too.
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So, the slope is just a negative one. And it looks like some of my points are a little bit off.
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That's okay. So, this was the width and this was the length.
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Alright, so that's how you would give in a situation you could find an equation, solve for a variable, and then find values that make sense in order to graph it.
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Alright, so that's it for that lesson. Thanks for watching.
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Part 2 will be talking more about graphing linear equations and what it means for something to be a linear equation.