WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about non-linear systems.
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Previously, we have only worked with linear equations or inequalities.
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But a system is not required to be linear; a system of equations or inequalities is just multiple relations that are true at the same time.
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It is just different things that we know are all true at the same time; that is a system.
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For example, we could look for the solutions to the system x = y² and 3x + 6y = 9.
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For it to be a solution to the system, it has to be some (x,y) pair that makes both equations true at the same time.
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It has to satisfy each part of our system.
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Solving the system will end up being very similar to solving a system of linear equations.
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We will look at how we can apply the three methods of substitution, elimination, and graphing,
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just like we talked about when we were solving systems of linear equations.
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After that, we will also look at finding solutions to a system of non-linear inequalities.
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It would be helpful to watch the previous two lessons, if you are just jumping to this one as the first one on these things.
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We will be reviewing methods, but we won't be really teaching any of them in-depth.
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So, if you really want to get the chance to learn substitution/elimination/graphing, it would be great to watch the previous ones.
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But if you have already seen them, you are good to go.
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All right, substitution is the most fundamental way to solve any system of equations.
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You just put one variable in terms of the other, or the others; then you substitute it in, and you work through a solution.
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For example, we have x = y², 3x + 6y = 9; well, we notice that we have x here; we have x here.
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So, we can plug in, and we will have 3 times what it was over here, y², plus 6y, equals 9.
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We have 3y² + 6y = 9; it is nice not to have as many numbers.
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We see that we can divide by the number 3: we have y² + 2y = 3.
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At this point, we look, and we say, "Oh, that looks just like solving a polynomial; let's do it like we are solving a polynomial."
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So, we move everything over, so that we have a 0 on one side: we have y² + 2y - 3 = 0.
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We see...can we factor this? Yes, it is not too difficult for us to factor.
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We find (y + 3)(y - 1) = 0; we check that; y times y is y²; good; -y + 3y gets us +2y; good; 3 times -1 is -3; good.
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At this point, we can solve for what our y's are going to be.
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We have y + 3 = 0 as one possibility; y - 1 = 0 is another possibility; we have y = -3 and y = +1.
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Those are our two possible worlds; so, in one world (let's make it the red world), we have y = 1.
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Now, we can solve for what our x is going to be by just plugging it in.
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x = y²; so now, we plug it in: what is x going to be when y is 1?
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We plug that in; x = 1², so x = 1.
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So, in the red world, when we plug in y = 1, we get x = 1; so we have the point (1,1) as an answer to both of these equations.
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However, we also have another world that we can check out; let's make this one the green world.
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We have y = -3; so what happens when we plug that one in?
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We will end up having x = (-3)²; so we have x = +9--that is the other one.
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So, in the green world, we have x = 9; y = -3; (9,-3) is the other point that would solve this.
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And that is how you do it by substitution; you just get one variable on its own, plug it in, swap out the other ones, and then work your way to a solution.
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In elimination, we add a multiple of one equation to the other to eliminate variables.
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In this case, if we have x = y² and 3x + 6y = 9, well, we notice that we have x's here; we have x's here.
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We can make the x's on our left be the same number as the other one, but opposite (negative).
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So, let's multiply everything by -3; we have -3x = -3y².
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So now, we can bring this over; we can add it over; and we have + -3x = -3y².
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We can add that on either side: 3x and -3x cancel out; we have 6y = 9 - 3y².
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We move this over; we have 3y² + 6y - 9 = 0.
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This is starting to look familiar; we divide everything by 3; y² + 2y - 3 = 0.
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And now, we are just where we were with substitution, at this point.
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We solve this like a polynomial to figure out what our values of y are.
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And then, we use that to figure out what our values of x are going to be.
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What are our possible values of y?--we figure out our possible values of x from that.
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That is how we would use elimination.
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Now, in general, elimination, I would say, is less useful for non-linear systems.
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It still works, but it can be difficult to eliminate variables, because the equations aren't linear, so they don't match up as easily for cancellation.
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In this one we have here, we have y² over here, but y² doesn't show up anywhere over here.
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There is no y² over here, so we can't try to cancel out y².
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We were lucky enough that we had an x in one of them and an x in the other one, so we could cancel out x terms.
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But we can't necessarily cancel out everything, because there are various ways.
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Since we are no longer stuck with just being limited to using linear things, we can have all sorts of weird things.
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We could have sine of x, exponential function of t...all sorts of things for our variables that make them not really fit together.
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So, elimination doesn't really work that well.
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It is still useful and useable when you are in the right situation, so you can keep a lookout for it.
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But don't rely on it as much as you do when you are working with linear systems.
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Graphing: we can also graph each equation in the system.
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Wherever they intersect is a solution to the system.
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Remember: this is because a graph shows us all of the points that are true; all of the solutions to a single equation are its graph.
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So, if we find a solution to both of our equations at the same time, that would have to be on both of the graphs at the same time.
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So, wherever they intersect is a location that is a solution to both systems, because it is on both graphs.
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Cool; so if we had x = y² and 3x + 6y = 9, we could look at this and say,
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"x = y² ends up making the red curve; 3x + 6y = 9 makes the blue curve."
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They intersect at those locations, and so those are our solutions; great.
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Now, if you have a graphing calculator, you can use that to find the points of intersection for the system.
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Remember: graphing calculators are this really great tool for being able to quickly find it.
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If you can graph the two of them, wherever the two graphs intersect on your graphing calculator,
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you can tell it to calculate what is the value of that intersection point, and it will just put out the numbers to get that intersection point.
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Now, if you are going to do this, you have to first put equations into the form y = stuff involving x--
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y = stuff involving some other variable, because that is how the graphing calculator takes things in.
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So, you would have to get this x = y², 3x + 6y = 9, into the forms that will be y = things.
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So, for example, 3x + 6y = 9: we would have 6y = -3x + 9.
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We divide everything by 6, and we would have y = -3/2x + 3.
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And so, that is our blue curve, right there.
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We could plug that into a graphing calculator, and that would appear.
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x = y² is a little bit different; we have x = y², but we want to get y on its own.
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To do that, we take the square root of both sides.
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But remember: if you take the square root of both sides, you have to have a plus or minus show up.
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We have ±x = y; but that is two equations at the same time.
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That is positive and negative; so we have to take this, and we split it into two different things.
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We split it into y equals the positive √x, and y equals the negative √x.
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And so, that gives us...the positive √x is the top half of our sideways parabola, and -√x is the bottom half of our sideways parabola.
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So, if we plugged in all three of those into our graphing calculator, we could then use the intersection ability to figure out where they are going to be.
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If we just solved for y = +√x and put in just the top half, we would end up getting only one of the answers, and miss the other one.
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So, it is important, when you are working through with a graphing calculator,
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to really pay attention to how you are getting this into a form that you can plug into your graphing calculator.
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Is this really the same as the equation I started with?
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OK, the number of solutions isn't going to be as fixed as it was when we were dealing with linear systems.
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When we worked with linear systems, there were only three possibilities for the number of solutions.
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There was going to be one solution, no solutions whatsoever, or infinitely many solutions (when we just ended up having the same line on top of itself).
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But with a non-linear system, all guesses are futile; there can be any number of solutions at all.
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The only way to figure out how many solutions there are is by solving the system or by looking at a really good graph of it.
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For example, with this one on the left, we end up seeing that it has three solutions, because it intersects here, here, and here.
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And this one just has a crazy, huge number of solutions, because we have the first intersections.
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But then here, we have one here and here and here, and then it just starts to pack in and pack in and pack in
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as we get closer, because that red graph is going up and down really, really quickly.
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So, you end up seeing lots of solutions in some cases.
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You won't end up having to work with any in there--any like this; it would probably be a little too difficult at this point.
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But just understand that the number of solutions that you are going to get out of a non-linear system isn't any fixed value.
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It is not like linear systems, where you can rely on knowing that it is going to just be one.
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There is no known number that it is going to be, and the only way to work it out is by working it out.
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Systems of non-linear inequalities: when we are working with non-linear inequalities, it is basically the same as when we were working with linear inequalities.
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The first step is to graph each of the inequalities.
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Graph it; and the way you graph it is as if it were an equation.
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Oops, there was a mistake in the graph here; we will fix it in just a second.
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So, you graph them as if they were equations with lines.
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However, you don't necessarily just use straight lines all the time.
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If it is dashed...you use dashed when it is a strict inequality--strictly less than or strictly greater than.
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Notice: in this case, we have that y is strictly greater than 2x² - 5.
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So, this red one is the graph of 2x² - 5; it needs to be dashed, because it is a strictly greater than.
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Let's go through and dash that, really quickly; that is how it should look--it should be dashed.
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OK, a dashed line is for the greater than, because it is saying, "If you are actually on the line-- if the point is on the line--it is not a solution."
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And then, after that, you shade it appropriately--you use test points to help you figure it out.
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So, for this one, let's use (0,0), since neither of the lines falls straight on (0,0); it makes a great test point.
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If we plug that in to y > 2x² - 5, we have 0 (for y) > 2(0)² - 5.
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0 is, indeed, greater than -5, so that checks out.
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We know that the side that we are going to be shading in is the side facing this purple dot for our red equation.
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So, we shade in this stuff here; OK.
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Now, the blue curve is less than or equal to 1/5x² + 2; that is our blue curve, right here.
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Let's use that same test point: we plug in (0,0); let's check what happens with that.
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Plug in 0 for our y; 0 ≤ 1/5(0)² + 2; and indeed, 0 is less than or equal to 2.
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So, that checks out; that tells us that we are going to have the blue side shade towards our purple test point.
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We shade towards our purple test point, so we are shading everything underneath and including that blue curve.
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So, everything underneath this blue curve is included in this inequality.
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And everything above the red parabola is included in this one.
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The part where they overlap is the space between the two parabolas.
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So, we can color that out; now it is getting a little confusing to see things, but see, we color that out with the purple;
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and we can see that this is the space that satisfies our system of inequalities,
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because if you are inside of this space, you end up being true for both of the inequalities at the same time.
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And that is how we figure it out; we shade it, and we will be able to figure it out by shading--
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each one of them individually, and then where all of the shadings agree--
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where all of the shadings overlap--that is our set of solutions; that is the set of points that satisfies our system of inequalities.
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All right, we are ready for some examples.
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The first example: xy = 2; we want to get something where we can plug it into the other one.
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Well, we see y = -1 + x, so let's take this; we will swap out y for -1 + x.
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We plug that in over here; we have x times the -1 + x for y; -1 + x = 2.
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Multiply our x over; we have -x + x² = 2; this looks like a polynomial, so let's solve it like a polynomial.
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x² - x...subtract the 2 over...we have x² - x - 2 = 0.
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At this point, we factor; x looks like it is going to have minus 2, and x + 1.
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Check it really quickly in our head: x times x is x²--great; x + 1...so 1x - 2x gives us -x; -2 times 1 is -2; great--that checks out.
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So, at this point, we can solve for it: x - 2 = 0, or x + 1 = 0.
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Those are the two possible worlds: we have x = 2 and x = -1; great, those are our two possibilities.
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I will arbitrarily choose two different colors for them; let's make the green world be when x = 2;
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we can plug it in over here; x = 2...actually, we can plug it into either one.
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It looks to me like it is probably a little easier to plug it into this one.
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But we could plug it into either of the equations, if we wanted.
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We have y = -1 +...1x is 2, so y = -1 + 2; that gets us +1.
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Our first point that we figured out is (2,1)--our first answer.
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Then, we will arbitrarily choose another color; in the purple world, we are going to have x equal to -1.
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When x = -1, we plug that one in; so we have y = -1 plus the x value that we are plugging in, -1.
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So, y = -1 + -1; that gets us -2, so that gives us the point...x value of -1, y-value of -2; and that is our two solutions.
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Our purple solution and our green solution are all of the solutions to this system.
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All right, the next example: Find the solutions to this system.
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In this case, we see that y is already over here, just by itself; so it looks like an easy candidate for substitution.
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Let's plug x² + 2 in for y over here.
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We plug that in; we have x² + 2, since that was what y used to be; that equals...
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oops, squared: we have to have everything continue to be the same...x - 1.
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x² + 2, squared, equals x - 1; at this point, that might raise our suspicions a little bit.
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But let's keep working it out and see what happens.
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So, that gets x⁴ + 4x² + 4 = x - 1.
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This might start to raise our suspicions about what we are looking at.
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Is it possible for this equation to ever be true?
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Is there some x-value that would make the left-hand side the same thing as the right-hand side here?
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Well, let's keep working it out and see if we can get something that will make it obvious what we are looking at.
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x⁴ + 4x² + 5 = x; OK, if we look at this, we might realize that we could try to solve it from this point;
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but we aren't certain that there are solutions: it said "If possible."
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We want to be just a little bit suspicious, because it can be a real pain to try to solve something for a long time,
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if it turns out that it is impossible to solve; so you want to be able to figure out if this is possible to solve,
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before you get too deep into the process of trying to solve it.
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So, x⁴ + 4x² + 5...well, what does that look like--what would we end up seeing there?
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Well, that is going to be a really, really fast-growing graph that starts at some height of 5,
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and then shoots up really quickly: x⁴ + 4x²...it never becomes negative.
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That is what the left-hand side is equal to.
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But the right-hand side, x--what would that end up being?
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That is going to be here; and if we graphed just x, that would go like this.
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Now, we are appealing to a graph to understand this, but we see that the left-hand side
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is going to always be putting out much larger numbers, no matter what x we put in,
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than the right-hand side is ever going to be able to put out.
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We have x⁴ + 4x² + 5; that is going to make really big numbers, always positive, really quickly.
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Now, x can end up getting large positive numbers, but it has to put in a very large x to do that.
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And if we put in a very large x on the right side, the left side will be enormous.
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So, we see that these two sides can never match up; so our suspicion is that there are actually no solutions here.
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We could try to move that x over and have it equal 0.
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But remember: some parabolas/some...in this case, it is not a parabola, because it is a fourth-degree...
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but some polynomials don't have solutions; they never touch the x-axis,
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if we are searching for when it is equal to 0; we are searching for roots.
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So, we end up being able to figure out that this won't work.
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But we want something that really makes it more obvious than just having to turn this into an equation and see this.
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It just seems a little bit uncertain; so the best way for this is actually going to be to graph it.
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If we graph it, we will see that this very clearly can never work out.
00:18:50.600 --> 00:18:58.600
We can graph both of our original equations, x² + 2 = y and y² = x - 1.
00:18:58.600 --> 00:19:11.900
We will make a tick-mark length of 1, 3, 4...go up to 4 on each...1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4.
00:19:11.900 --> 00:19:17.300
Great; so let's first graph the easy one, x² + 2 = y.
00:19:17.300 --> 00:19:22.500
That one is pretty easy for us to graph; we will graph this one in blue.
00:19:22.500 --> 00:19:30.300
x² + 2 = y; well, we start at a height of 2; we plug in x = 0, and we get a height of 2.
00:19:30.300 --> 00:19:36.800
Plug in x = 1; now we are at a height of 3; with -1, the same thing; at 2, we will be way out at a height of 6.
00:19:36.800 --> 00:19:42.600
So, we see that we are going to curve out very quickly; we are going to shoot out like this.
00:19:42.600 --> 00:19:49.300
That is what the curve ends up being; don't get that part confused--the graph is not actually connecting to that circled part.
00:19:49.300 --> 00:19:54.100
The blue graph goes to x² + 2 = y; what about y² = x - 1?
00:19:54.100 --> 00:20:02.500
Now, we are used to solving things in terms of y = stuff; but that is actually not going to be the easiest way to do this,
00:20:02.500 --> 00:20:06.700
because then you have to take square root, and you have a positive and a negative side, because it is plus or minus square root.
00:20:06.700 --> 00:20:11.800
What we can do is y² + 1 (we add 1 to both sides) = x.
00:20:11.800 --> 00:20:16.500
And so, we can solve this from y's point of view, as being the input, and x being the output.
00:20:16.500 --> 00:20:19.900
For example, if y is 0, what does x end up being?
00:20:19.900 --> 00:20:30.600
If y is 0, then x ends up being 1; so at y = 0, a height of 0 on the horizontal axis, x will end up being positive 1.
00:20:30.600 --> 00:20:34.000
0 squared, plus one, equals positive 1.
00:20:34.000 --> 00:20:41.900
At a height of 1, when we plug in y = 1, x will end up being 1² + 1, or 2.
00:20:41.900 --> 00:20:51.900
At a height of 2, when y is at the height of 2, we plug that in; we have y² + 1, so 2² + 1; 4 + 1;
00:20:51.900 --> 00:20:55.100
5 is going to be our x-value, somewhere out here.
00:20:55.100 --> 00:20:59.900
The same thing happens if we plug in a negative height: -1 will end up getting us an x-value of 2.
00:20:59.900 --> 00:21:03.900
-2 will end up giving us an x-value of +5, as well...
00:21:03.900 --> 00:21:11.600
I'm sorry: did I say -2 for -1? A height of -1 will give us an x-value of +2, just in case I said the wrong thing back there.
00:21:11.600 --> 00:21:19.600
And then, we curve this out like this, because it is going to be a parabola, as well, because it is of that degree.
00:21:19.600 --> 00:21:23.900
Curve that just a little bit better, so we can see it more accurately.
00:21:23.900 --> 00:21:29.700
Curve that like this; it goes out like this.
00:21:29.700 --> 00:21:35.400
So notice: the red one is going to keep going out to the right; the blue one is going to keep going up.
00:21:35.400 --> 00:21:37.100
They are never going to end up touching each other.
00:21:37.100 --> 00:21:41.400
This cup goes like this; this cup goes like this; they go out like that.
00:21:41.400 --> 00:21:44.000
They are never going to touch each other; they are never going to intersect.
00:21:44.000 --> 00:21:48.100
There is no way for these things to ever connect to each other.
00:21:48.100 --> 00:21:53.000
They can't ever connect to each other, because we see that when graphing them, they fail to ever touch.
00:21:53.000 --> 00:22:00.900
With that in mind, we see that there are no solutions to this system.
00:22:00.900 --> 00:22:10.500
All right, next: Graph the solutions to the system of inequalities: y < 3cos(x); y ≥ x²; and x < 1.
00:22:10.500 --> 00:22:14.200
Now, we haven't explicitly said that we are going to be using trigonometric things.
00:22:14.200 --> 00:22:19.200
But we are basically at that point, where we are going to assume that you have gone through the trigonometry lessons.
00:22:19.200 --> 00:22:25.400
So, you are probably used to it; if you are not used to it, just trust that I am giving you an accurate depiction of how 3cos(x) works.
00:22:25.400 --> 00:22:35.000
y < 3cos(x); the first thing we do, let's set up a nice, large graph, because we are trying to graph the solutions to the system of inequalities.
00:22:35.000 --> 00:22:40.300
Since we are working with cosine, we know that we are going to want it at least out to 2π on either side.
00:22:40.300 --> 00:22:51.300
So, π is 3.14; 2π is 6.28, approximately; we want to go out to probably at least 7 marks horizontally in either direction.
00:22:51.300 --> 00:23:02.500
A tick mark will be a length of 1; so 1, 2, 3, 4, 5, 6, 7...I actually drew a little bit more than was necessary;
00:23:02.500 --> 00:23:19.900
1, 2, 3, 4, 5, 6, 7; up 1, 2, 3, 4, 5, 1, 2, 3, 4, 5; OK.
00:23:19.900 --> 00:23:26.000
Let's graph each one of these: first, we will graph y < 3cos(x) in green.
00:23:26.000 --> 00:23:28.400
Now, notice: it is less than, so it is going to be dashed.
00:23:28.400 --> 00:23:32.500
Let's get some points, so that we can draw in this line properly.
00:23:32.500 --> 00:23:39.000
If we plug in x = 0, remember: we are in radians; we are not using degrees here; otherwise our x-axis would have to be huge.
00:23:39.000 --> 00:23:45.500
So, with radians, if we plug in 0 (x is 0), then we are going to have a cosine of 0 put out 1;
00:23:45.500 --> 00:23:51.600
so we will have 3 times 1, so we will be at a height of 3 when we are at a horizontal location of 0.
00:23:51.600 --> 00:24:05.100
Halfway, at π/2, which is a little bit over 1 and 1/2 (1.5 and a little bit), we will be at 0; cos(π/2) is 0; 3 times 0 is 0.
00:24:05.100 --> 00:24:11.900
The same thing if we go over to the negative π/2, around halfway between 1 and 2.
00:24:11.900 --> 00:24:22.800
Then, at π, we are going to end up having -3, because cos(π) is -1, so 3cos(π) will be -3.
00:24:22.800 --> 00:24:29.900
So, 3.14 is a little bit past the 3 marker over here.
00:24:29.900 --> 00:24:38.400
And then, the same thing happens over here at -π; so -3...a little bit past the -3 marker...
00:24:38.400 --> 00:24:53.400
We can curve out what cosine looks like, and what we knew before, in this portion right here.
00:24:53.400 --> 00:24:58.500
All right, OK, we can do the same thing with continuing this graph out.
00:24:58.500 --> 00:25:09.000
We know that at 2π, 1, 2, 3, 4, 5, 6...around 6.28 we will be back up to a height of 3.
00:25:09.000 --> 00:25:18.800
We know that, halfway between the two, at around 4.6 or 4.7 or so, we will be at 3π/2; 1, 2, 3, 4...
00:25:18.800 --> 00:25:25.900
And a little bit over halfway, we will be back at a height of 0; the same thing going the other way: 1, 2, 3, 4...
00:25:25.900 --> 00:25:36.100
a little bit over -4 1/2...and then 5, 6...a little bit over 6; and there we go...a little bit over -6.
00:25:36.100 --> 00:25:45.400
We can curve in a little bit more of this, like that.
00:25:45.400 --> 00:25:53.400
Now, notice: at this point, I technically made a mistake, because was it a strict inequality?
00:25:53.400 --> 00:25:56.400
Yes, it was a strict inequality; it was strictly less than.
00:25:56.400 --> 00:26:00.800
So, if that is the case, we can't be using a solid line; we have to be using a dashed line.
00:26:00.800 --> 00:26:06.000
It is a little bit easier for me to draw it; I am just going to come in with my eraser, and I am going to erase this into being dashed,
00:26:06.000 --> 00:26:10.700
which I am sure we have all done at some point--where we accidentally draw something as solid,
00:26:10.700 --> 00:26:12.500
and realize, "Oh, I have to make this dashed now."
00:26:12.500 --> 00:26:20.900
We came through with an eraser and just erase it into a dashed form.
00:26:20.900 --> 00:26:27.200
At this point, we have y < 3cos(x); at least, the line, the curve generated by it, graphed in.
00:26:27.200 --> 00:26:29.500
We will shade it in, in a little bit.
00:26:29.500 --> 00:26:33.500
Next, we will do y ≥ x² in blue.
00:26:33.500 --> 00:26:39.900
This one is much easier--we know it: x is 0; we are at a height of 0; at 1, 1; at 2, +4;
00:26:39.900 --> 00:26:51.600
the same thing on the other side: -2, +4; and we curve out, just in a nice, handy parabola--much faster.
00:26:51.600 --> 00:26:59.100
And since it is greater than or equal to, it actually makes a solid line, because it is not a strict inequality; it is not strict.
00:26:59.100 --> 00:27:02.600
And then finally, x < 1 we will do in red.
00:27:02.600 --> 00:27:10.400
That means that x has to be something less than 1; so that means we have fixed x at 1 for drawing the actual graph, for drawing the curve.
00:27:10.400 --> 00:27:16.400
We are going to fix x at 1, and then we will see that y is allowed to go to anything it wants,
00:27:16.400 --> 00:27:21.300
because for x < 1, y can be anything it wants; x < 1 doesn't care what y is.
00:27:21.300 --> 00:27:27.000
So, we fix x at 1, and we draw a dashed line, because it is a strict inequality.
00:27:27.000 --> 00:27:32.900
It says that x has to be less than 1--it is not allowed to actually be equal to 1; we have a dashed line for that.
00:27:32.900 --> 00:27:37.600
So now, we do some shading and figure out what is allowed for each one of these.
00:27:37.600 --> 00:27:47.100
y < 3cos(x): we could do a test point at (0,0); 0 < 3cos(0), so 0 < 3.
00:27:47.100 --> 00:27:54.200
That is perfectly true, which makes sense, because y < ...also means that we are going to be looking at the part that would be below the curve.
00:27:54.200 --> 00:28:01.600
So, what is below the curve? Stuff like this.
00:28:01.600 --> 00:28:05.500
I am not doing a very extreme job of shading, just because we want to have some idea of where we are looking.
00:28:05.500 --> 00:28:09.400
So, we don't have to shade too much right now, until we figure out where they all agree.
00:28:09.400 --> 00:28:16.000
y ≥ x²...we can't use the test point (0,0) because it is actually on our line; so we have to choose a new one.
00:28:16.000 --> 00:28:21.100
Let's try (0,1): 1 is greater than or equal to 0²...indeed, that is true.
00:28:21.100 --> 00:28:25.300
Also, since it is y ≥, we know that it has to be above the curve.
00:28:25.300 --> 00:28:30.200
So, what is above the curve is this stuff here; great.
00:28:30.200 --> 00:28:36.700
And then finally, if x < 1, we have to be to the left side of it, because our x-value has to be below it.
00:28:36.700 --> 00:28:44.300
We could also use a test point, like (0,0); we don't care about the y-value, but 0 is less than 1, so we shade towards that test point of (0,0).
00:28:44.300 --> 00:28:47.900
So, we would go in to the left; great.
00:28:47.900 --> 00:29:00.800
At this point, we have seen that the only thing that they can agree on is this little part right in the middle that we are now shading with green.
00:29:00.800 --> 00:29:07.000
And there we go; and that is how we figure out where these things go; cool.
00:29:07.000 --> 00:29:10.500
All right, Example 4: A rectangular box has the following properties.
00:29:10.500 --> 00:29:15.800
The sum of its edges is 24 feet; adding together the area of each of its faces gives a total of 22 square feet;
00:29:15.800 --> 00:29:19.900
its height is twice its width; and then we are asked to find out what its volume is.
00:29:19.900 --> 00:29:24.100
The first thing to do is: we want to get a sense of what we are looking at, so let's draw a quick picture.
00:29:24.100 --> 00:29:39.200
If we have a box, some box, it has three things to it: length, width, and height.
00:29:39.200 --> 00:29:51.100
We can see that; we have...what is its length? What is its width? And what is its height?
00:29:51.100 --> 00:29:56.700
Great; with that in mind, let's start trying to figure out how these properties turn into math.
00:29:56.700 --> 00:30:05.000
The sum of its edges is 24 feet; now technically, we don't really know: are they saying just one of its edges each time?
00:30:05.000 --> 00:30:07.300
Or are they saying all of its edges?
00:30:07.300 --> 00:30:10.300
So sometimes, you end up seeing things that are a little bit confusing in math.
00:30:10.300 --> 00:30:13.900
And if you saw this on a test, it would probably be a good idea to ask your teacher, because you are not sure.
00:30:13.900 --> 00:30:19.700
Does that mean h + w + l, or does that mean each edge of the box, all added together?
00:30:19.700 --> 00:30:26.100
Let's go with the sum of every edge; let's say that is what it means.
00:30:26.100 --> 00:30:31.100
But notice how each of its edges...the sum of its edges...well, that could be considered to just be the three edges,
00:30:31.100 --> 00:30:36.800
the fundamental edges (height, width, and length); but it could also be all of the times that they show up on the box.
00:30:36.800 --> 00:30:40.200
If you have a rectangular object, it shows up here and here and here.
00:30:40.200 --> 00:30:47.200
Height actually shows up four times, because we have each of the columns that make up our box.
00:30:47.200 --> 00:30:52.400
We have a height here, a height here, a height here, and a height here.
00:30:52.400 --> 00:30:57.800
So, we can think of it, if we are looking at every edge, as 4(height) will be part of what is going in that.
00:30:57.800 --> 00:31:03.000
So, the sum of not just its edges, but every edge, is how we will do this problem.
00:31:03.000 --> 00:31:09.000
All right, so if that is the case, this first thing is going to end up coming out to be...
00:31:09.000 --> 00:31:21.400
The first idea will come out to be 4h + 4w + 4l = 24 feet.
00:31:21.400 --> 00:31:26.400
So, height, width, and length, all combined together, comes out to a total of 24 feet.
00:31:26.400 --> 00:31:30.600
because it is the sum of every edge, every one of the edges, and each edge...
00:31:30.600 --> 00:31:35.800
height will show up 4 times; width will show up 4 times on the box; length will show up 4 times on the box.
00:31:35.800 --> 00:31:39.100
If 4you find this confusing, try just finding some rectangular object that you can look at;
00:31:39.100 --> 00:31:43.600
pick up an actual box and count the edges--count how many times its height shows up.
00:31:43.600 --> 00:31:46.600
Any rectangular box will be able to show you this idea, if it is a little confusing.
00:31:46.600 --> 00:31:50.100
Physical things are a great way to explore things in math.
00:31:50.100 --> 00:31:53.000
Next is adding together the area of each of its faces.
00:31:53.000 --> 00:32:00.200
This one is a little bit tougher: how many times does, let's say, this face right here--the very front face--show up?
00:32:00.200 --> 00:32:04.300
Well, we see that height times width would be the area of that face.
00:32:04.300 --> 00:32:08.800
So, height times width is going to get multiplied together.
00:32:08.800 --> 00:32:13.700
But it doesn't show up just on the front; it also shows up on the back side.
00:32:13.700 --> 00:32:16.300
It is the front side, but also the back side--both sides.
00:32:16.300 --> 00:32:22.600
So, it is not going to be just h times w, but 2 times h times w.
00:32:22.600 --> 00:32:30.400
By that same logic, each of the side faces...we are going to have l times h, because it is h over here...
00:32:30.400 --> 00:32:36.600
so the side faces will show up twice, as well; so we have 2lh.
00:32:36.600 --> 00:32:49.900
And then finally, the top face is going to be the length times the width, so 2 length times width.
00:32:49.900 --> 00:32:53.800
And we were told that that came out to be 22 square feet total.
00:32:53.800 --> 00:32:59.200
So, we can find the area of each one of the faces; hw will be one of the faces, and then it doubles up each time.
00:32:59.200 --> 00:33:02.800
lh will be the area of one face; it doubles up, and so on, and so forth.
00:33:02.800 --> 00:33:10.100
So, we have 2hw + 2lh + 2lw = 22; and then finally, we were told that the height is twice its width.
00:33:10.100 --> 00:33:16.800
That is probably the easiest one of all: if the height is equal to twice the width, it is 2 times the width; great.
00:33:16.800 --> 00:33:20.200
At this point, how do we find volume--what is volume based on?
00:33:20.200 --> 00:33:29.200
Well, if it is a nice rectangular box, volume is just equal to all three of these variables, multiplied together: h times l times w; great.
00:33:29.200 --> 00:33:33.100
That is all of the steps that we need together to be able to figure out what this is going to be--
00:33:33.100 --> 00:33:35.200
to be able to get this in a position where we can solve it.
00:33:35.200 --> 00:33:37.400
So now, let's start working it out.
00:33:37.400 --> 00:33:44.600
We have 4l + 4w + 4h = 24, 2lw + 2wh + 2lh = 22, and h = 2w.
00:33:44.600 --> 00:33:47.700
And we are looking for volume, which is going to be l times w times h.
00:33:47.700 --> 00:33:53.100
Great; so I would say that the very first thing to do...let's take 4l + 4w + 4h, and let's make it a little bit easier.
00:33:53.100 --> 00:34:00.500
Let's divide everything by 4; we have l + w + h = 24/4, which gets us 6.
00:34:00.500 --> 00:34:10.700
The same thing over here: let's divide everything by 2, so that gets us lw + wh + lh = 11; great.
00:34:10.700 --> 00:34:12.800
So, at this point, we can actually start figuring things out.
00:34:12.800 --> 00:34:21.000
Let's try to solve in terms of w; we have w over here; h is already ready to be substituted in somewhere.
00:34:21.000 --> 00:34:26.200
So, since it is connected to w, we can probably figure out w easiest from what we have set up here.
00:34:26.200 --> 00:34:40.600
We can plug that in over here: if h = 2w, then we have l + w +...what was h equal to? h was equal to 2w, so + 2w = 6.
00:34:40.600 --> 00:34:47.500
l + w + 2w...3w = 6, so l = 6 - 3w.
00:34:47.500 --> 00:34:50.200
Now, at this point, we are ready to substitute l in, as well.
00:34:50.200 --> 00:34:57.500
We have h ready to substitute and l ready to substitute; so we can now go into our big equation, lw + wh + lh = 11.
00:34:57.500 --> 00:35:18.500
So, we plug in here: l is 6 - 3w, so (6 - 3w) times w, plus w times...h is 2w...plus l...l is (6 - 3w), times h...is 2w; equals 11.
00:35:18.500 --> 00:35:26.600
6 - 3w times w...w distributes, so we have 6w - 3w², plus...w distributes onto 2w...
00:35:26.600 --> 00:35:34.100
well, not "distributes," but 2w² + (6 - 3w)2w...6 times 2w becomes 12w,
00:35:34.100 --> 00:35:39.400
minus 3w times 2w, becomes -6w²; equals 11.
00:35:39.400 --> 00:35:45.900
Great; let's simplify things a bit; we have -3w here; plus 2w² here; and -6w² here.
00:35:45.900 --> 00:35:54.000
So, -3w² + 2w² gets us -1w²; - 6w² brings us to a total of -7w².
00:35:54.000 --> 00:36:00.400
Our 6w and 12w combine to +18w, equals 11.
00:36:00.400 --> 00:36:05.700
We have squared; we have a single degree of 1; and we have a constant; this looks like a polynomial.
00:36:05.700 --> 00:36:08.200
Let's get it into an easy-to-solve polynomial format.
00:36:08.200 --> 00:36:12.900
Add 7w² to both sides; add -18w to both sides; +11.
00:36:12.900 --> 00:36:17.900
At this point, we could toss this into the quadratic formula and solve out the answers.
00:36:17.900 --> 00:36:23.100
But we might be able to get lucky; and even though it looks a little complex, we might realize that we can factor this.
00:36:23.100 --> 00:36:24.500
It is not too difficult to factor.
00:36:24.500 --> 00:36:34.000
We get lucky; we notice that it turns out pretty easy to factor: 7w and w here...
00:36:34.000 --> 00:36:38.000
We need to have minus on both of them, because it comes out to be positive 11.
00:36:38.000 --> 00:36:43.500
Minus 18w...we have a 7w here, so this will be -1, and this will be -11.
00:36:43.500 --> 00:36:56.100
7w times w is 7w²; 7w - 1 - 7w...minus 11 times w...minus 11w; so -7w - 11w is 18w; it checks out; -11 times -1 checks out as a positive number.
00:36:56.100 --> 00:36:58.800
Great; it is always a good idea to check when you are factoring.
00:36:58.800 --> 00:37:05.800
So, we can now solve each one of these: 7w - 11 = 0, or w - 1 = 0.
00:37:05.800 --> 00:37:08.600
We actually have two different possible worlds.
00:37:08.600 --> 00:37:15.400
It didn't say that in the problem, but there are two different possible worlds for what the width of the box can be.
00:37:15.400 --> 00:37:21.400
It can be 11/7 or 1; so 11/7 and 1 are our two possible things.
00:37:21.400 --> 00:37:25.700
So, let's call these, arbitrarily...we will make colors for these.
00:37:25.700 --> 00:37:30.200
w = 1 looks easiest to solve and deal with first, so we will make that the purple world.
00:37:30.200 --> 00:37:41.400
w = 1--if that is the case, we can figure out that h = 2w; we plug that in; h = 2(1), so we have that h is 2.
00:37:41.400 --> 00:37:55.900
And then, we also have, if it is w = 1...we plug that into l = 6 - 3w; so l = 6 - 3(1); so 6 - 3 = 3.
00:37:55.900 --> 00:38:04.000
We have w = 1, h = 2, l = 3 in our first world, in this purple world,
00:38:04.000 --> 00:38:06.800
because remember: there were two possibilities for what our width could be.
00:38:06.800 --> 00:38:14.100
But if this is the case, then our volume is going to be the three of these multiplied together: 1 times 2 times 3, which equals 6.
00:38:14.100 --> 00:38:24.800
So, one possible value for our volume is going to be 6 cubic feet--that is one possible answer.
00:38:24.800 --> 00:38:28.300
It turns out that there are two different worlds here, so we want to check out both of them.
00:38:28.300 --> 00:38:32.600
The other one: we will make this the green world, where the width is equal to 11/7...
00:38:32.600 --> 00:38:43.600
Well, if width equals 11/7, then we can plug that into h = 2w, so h = 2(11/7), which equals 22/7.
00:38:43.600 --> 00:38:50.900
So, if our width is 11/7, then our height is 22/7.
00:38:50.900 --> 00:38:54.900
And then, we can also figure out what our length is going to be.
00:38:54.900 --> 00:39:09.900
Length equals 6 - 3w; w, in this case, is 11/7; so we have 6 - 33/7, which ends up simplifying out to 9/7; great.
00:39:09.900 --> 00:39:23.200
At this point, we have width = 11/7; height = 22/7; and length = 9/7, all of these in feet as the units.
00:39:23.200 --> 00:39:26.600
So, at this point, we know that the volume is equal to each one of these, multiplied together:
00:39:26.600 --> 00:39:40.200
so 11/7 times 22/7 times 9/7; we multiply these all out together, and it becomes 2178/343,
00:39:40.200 --> 00:39:43.300
which is really not that easy to see what that means.
00:39:43.300 --> 00:39:48.900
So, let's approximate that using a calculator; and that comes out to be 6.35.
00:39:48.900 --> 00:39:56.100
So, our other possibility is that the volume comes out to be 6.35 cubic feet.
00:39:56.100 --> 00:40:02.500
So, it turns out that there is actually a larger possible box if we are not going with these nice, friendly integer things.
00:40:02.500 --> 00:40:07.000
But we can still follow the three requirements; there are three conditions that were given to us.
00:40:07.000 --> 00:40:11.700
It turns out that there are two possible boxes that actually fit those conditions; and those are the volumes.
00:40:11.700 --> 00:40:15.300
In our purple world, where our width was 1, we got 6 cubic feet.
00:40:15.300 --> 00:40:20.400
And in our green world, where our width was 11/7, we got 6.35 cubic feet.
00:40:20.400 --> 00:40:23.200
And if you wanted to check this--if you wanted to make sure that everything was great--
00:40:23.200 --> 00:40:29.200
a good thing to do would be to see that you have width = 1, height = 2, length = 3,
00:40:29.200 --> 00:40:33.400
and then just try plugging that into each of these three equations that we started with,
00:40:33.400 --> 00:40:36.800
and making sure..."Yes, that checks out; yes, that checks out; yes, that checks out."
00:40:36.800 --> 00:40:47.500
The same thing over here for the width = 11/7, height = 22/7, length = 9/7:
00:40:47.500 --> 00:40:52.200
you can just plug that into each of these three equations, and make sure that it checks out in each one,
00:40:52.200 --> 00:40:55.400
because if it checks out in each one, you know that that is a workable answer.
00:40:55.400 --> 00:40:59.300
All right, that finishes our work with systems of equations of all types.
00:40:59.300 --> 00:41:01.000
And we will see you at Educator.com later--goodbye!