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Nonlinear Systems

  • A system of equations does not necessarily have to be linear (made up of straight lines only). Nonetheless, we can still use the methods we learned about in previous lessons.
  • Substitution is the most fundamental way to solve any system of equations: put one variable in terms of the other(s), then substitute and work to a solution.
  • In elimination, we add a multiple of one equation to the other to eliminate variables. In general, elimination is less useful for nonlinear systems. While it still works, it can be difficult to eliminate variables because the equations aren't linear, so they don't match up as easily for cancellation.
  • We can also graph each equation in the system: wherever they intersect is a solution to the system. If you have a graphing calculator, you can use that to find the points of intersection for the system. [However, you will have to first put the equations into a form that you can enter into your graphing calculator: y = .]
  • Unlike a linear system of equations, there can be any number of solutions. The only way to figure out how many solutions there are is by solving the system.
  • Working with nonlinear inequalities is very similar to working with linear inequalities:
    • Graph each inequality (as if they were equations);
    • Dashed   < ,   > ;        Solid:   ≤ ,   ≥ ;
    • Shade appropriately (use test points to help).

Nonlinear Systems

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:06
  • Substitution 1:12
    • Example
  • Elimination 3:46
    • Example
    • Elimination is Less Useful for Nonlinear Systems
  • Graphing 5:56
    • Using a Graphing Calculator
  • Number of Solutions 8:44
  • Systems of Nonlinear Inequalities 10:02
    • Graph Each Inequality
    • Dashed and/or Solid
    • Shade Appropriately
  • Example 1 13:24
  • Example 2 15:50
  • Example 3 22:02
  • Example 4 29:06
    • Example 4, cont.

Transcription: Nonlinear Systems

Hi--welcome back to

Today, we are going to talk about non-linear systems.0002

Previously, we have only worked with linear equations or inequalities.0005

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.0009

It is just different things that we know are all true at the same time; that is a system.0018

For example, we could look for the solutions to the system x = y2 and 3x + 6y = 9.0022

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.0029

It has to satisfy each part of our system.0035

Solving the system will end up being very similar to solving a system of linear equations.0038

We will look at how we can apply the three methods of substitution, elimination, and graphing,0042

just like we talked about when we were solving systems of linear equations.0047

After that, we will also look at finding solutions to a system of non-linear inequalities.0050

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.0054

We will be reviewing methods, but we won't be really teaching any of them in-depth.0059

So, if you really want to get the chance to learn substitution/elimination/graphing, it would be great to watch the previous ones.0062

But if you have already seen them, you are good to go.0068

All right, substitution is the most fundamental way to solve any system of equations.0070

You just put one variable in terms of the other, or the others; then you substitute it in, and you work through a solution.0074

For example, we have x = y2, 3x + 6y = 9; well, we notice that we have x here; we have x here.0080

So, we can plug in, and we will have 3 times what it was over here, y2, plus 6y, equals 9.0087

We have 3y2 + 6y = 9; it is nice not to have as many numbers.0097

We see that we can divide by the number 3: we have y2 + 2y = 3.0105

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."0111

So, we move everything over, so that we have a 0 on one side: we have y2 + 2y - 3 = 0.0117

We see...can we factor this? Yes, it is not too difficult for us to factor.0123

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.0128

At this point, we can solve for what our y's are going to be.0146

We have y + 3 = 0 as one possibility; y - 1 = 0 is another possibility; we have y = -3 and y = +1.0148

Those are our two possible worlds; so, in one world (let's make it the red world), we have y = 1.0158

Now, we can solve for what our x is going to be by just plugging it in.0166

x = y2; so now, we plug it in: what is x going to be when y is 1?0171

We plug that in; x = 12, so x = 1.0176

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.0181

However, we also have another world that we can check out; let's make this one the green world.0191

We have y = -3; so what happens when we plug that one in?0195

We will end up having x = (-3)2; so we have x = +9--that is the other one.0200

So, in the green world, we have x = 9; y = -3; (9,-3) is the other point that would solve this.0209

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.0217

In elimination, we add a multiple of one equation to the other to eliminate variables.0225

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.0229

We can make the x's on our left be the same number as the other one, but opposite (negative).0236

So, let's multiply everything by -3; we have -3x = -3y2.0240

So now, we can bring this over; we can add it over; and we have + -3x = -3y2.0247

We can add that on either side: 3x and -3x cancel out; we have 6y = 9 - 3y2.0255

We move this over; we have 3y2 + 6y - 9 = 0.0265

This is starting to look familiar; we divide everything by 3; y2 + 2y - 3 = 0.0271

And now, we are just where we were with substitution, at this point.0278

We solve this like a polynomial to figure out what our values of y are.0281

And then, we use that to figure out what our values of x are going to be.0284

What are our possible values of y?--we figure out our possible values of x from that.0287

That is how we would use elimination.0292

Now, in general, elimination, I would say, is less useful for non-linear systems.0294

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.0299

In this one we have here, we have y2 over here, but y2 doesn't show up anywhere over here.0307

There is no y2 over here, so we can't try to cancel out y2.0314

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.0318

But we can't necessarily cancel out everything, because there are various ways.0323

Since we are no longer stuck with just being limited to using linear things, we can have all sorts of weird things.0327

We could have sine of x, exponential function of t...all sorts of things for our variables that make them not really fit together.0332

So, elimination doesn't really work that well.0341

It is still useful and useable when you are in the right situation, so you can keep a lookout for it.0343

But don't rely on it as much as you do when you are working with linear systems.0348

Graphing: we can also graph each equation in the system.0353

Wherever they intersect is a solution to the system.0356

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.0358

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.0365

So, wherever they intersect is a location that is a solution to both systems, because it is on both graphs.0371

Cool; so if we had x = y2 and 3x + 6y = 9, we could look at this and say,0381

"x = y2 ends up making the red curve; 3x + 6y = 9 makes the blue curve."0387

They intersect at those locations, and so those are our solutions; great.0394

Now, if you have a graphing calculator, you can use that to find the points of intersection for the system.0399

Remember: graphing calculators are this really great tool for being able to quickly find it.0403

If you can graph the two of them, wherever the two graphs intersect on your graphing calculator,0407

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.0411

Now, if you are going to do this, you have to first put equations into the form y = stuff involving x--0418

y = stuff involving some other variable, because that is how the graphing calculator takes things in.0427

So, you would have to get this x = y2, 3x + 6y = 9, into the forms that will be y = things.0432

So, for example, 3x + 6y = 9: we would have 6y = -3x + 9.0438

We divide everything by 6, and we would have y = -3/2x + 3.0443

And so, that is our blue curve, right there.0450

We could plug that into a graphing calculator, and that would appear.0452

x = y2 is a little bit different; we have x = y2, but we want to get y on its own.0456

To do that, we take the square root of both sides.0461

But remember: if you take the square root of both sides, you have to have a plus or minus show up.0463

We have ±x = y; but that is two equations at the same time.0469

That is positive and negative; so we have to take this, and we split it into two different things.0474

We split it into y equals the positive √x, and y equals the negative √x.0478

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.0488

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.0498

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.0504

So, it is important, when you are working through with a graphing calculator,0511

to really pay attention to how you are getting this into a form that you can plug into your graphing calculator.0513

Is this really the same as the equation I started with?0518

OK, the number of solutions isn't going to be as fixed as it was when we were dealing with linear systems.0522

When we worked with linear systems, there were only three possibilities for the number of solutions.0529

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).0534

But with a non-linear system, all guesses are futile; there can be any number of solutions at all.0541

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.0546

For example, with this one on the left, we end up seeing that it has three solutions, because it intersects here, here, and here.0551

And this one just has a crazy, huge number of solutions, because we have the first intersections.0559

But then here, we have one here and here and here, and then it just starts to pack in and pack in and pack in0564

as we get closer, because that red graph is going up and down really, really quickly.0569

So, you end up seeing lots of solutions in some cases.0573

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.0576

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.0581

It is not like linear systems, where you can rely on knowing that it is going to just be one.0588

There is no known number that it is going to be, and the only way to work it out is by working it out.0591

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.0597

The first step is to graph each of the inequalities.0605

Graph it; and the way you graph it is as if it were an equation.0608

Oops, there was a mistake in the graph here; we will fix it in just a second.0612

So, you graph them as if they were equations with lines.0616

However, you don't necessarily just use straight lines all the time.0619

If it is use dashed when it is a strict inequality--strictly less than or strictly greater than.0621

Notice: in this case, we have that y is strictly greater than 2x2 - 5.0629

So, this red one is the graph of 2x2 - 5; it needs to be dashed, because it is a strictly greater than.0633

Let's go through and dash that, really quickly; that is how it should look--it should be dashed.0641

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."0656

And then, after that, you shade it appropriately--you use test points to help you figure it out.0665

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.0669

If we plug that in to y > 2x2 - 5, we have 0 (for y) > 2(0)2 - 5.0676

0 is, indeed, greater than -5, so that checks out.0686

We know that the side that we are going to be shading in is the side facing this purple dot for our red equation.0690

So, we shade in this stuff here; OK.0701

Now, the blue curve is less than or equal to 1/5x2 + 2; that is our blue curve, right here.0707

Let's use that same test point: we plug in (0,0); let's check what happens with that.0715

Plug in 0 for our y; 0 ≤ 1/5(0)2 + 2; and indeed, 0 is less than or equal to 2.0721

So, that checks out; that tells us that we are going to have the blue side shade towards our purple test point.0730

We shade towards our purple test point, so we are shading everything underneath and including that blue curve.0737

So, everything underneath this blue curve is included in this inequality.0749

And everything above the red parabola is included in this one.0757

The part where they overlap is the space between the two parabolas.0762

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;0768

and we can see that this is the space that satisfies our system of inequalities,0774

because if you are inside of this space, you end up being true for both of the inequalities at the same time.0782

And that is how we figure it out; we shade it, and we will be able to figure it out by shading--0790

each one of them individually, and then where all of the shadings agree--0793

where all of the shadings overlap--that is our set of solutions; that is the set of points that satisfies our system of inequalities.0796

All right, we are ready for some examples.0803

The first example: xy = 2; we want to get something where we can plug it into the other one.0805

Well, we see y = -1 + x, so let's take this; we will swap out y for -1 + x.0811

We plug that in over here; we have x times the -1 + x for y; -1 + x = 2.0823

Multiply our x over; we have -x + x2 = 2; this looks like a polynomial, so let's solve it like a polynomial.0833

x2 - x...subtract the 2 over...we have x2 - x - 2 = 0.0842

At this point, we factor; x looks like it is going to have minus 2, and x + 1.0848

Check it really quickly in our head: x times x is x2--great; x + 1x - 2x gives us -x; -2 times 1 is -2; great--that checks out.0856

So, at this point, we can solve for it: x - 2 = 0, or x + 1 = 0.0865

Those are the two possible worlds: we have x = 2 and x = -1; great, those are our two possibilities.0870

I will arbitrarily choose two different colors for them; let's make the green world be when x = 2;0878

we can plug it in over here; x = 2...actually, we can plug it into either one.0884

It looks to me like it is probably a little easier to plug it into this one.0889

But we could plug it into either of the equations, if we wanted.0893

We have y = -1 +...1x is 2, so y = -1 + 2; that gets us +1.0895

Our first point that we figured out is (2,1)--our first answer.0904

Then, we will arbitrarily choose another color; in the purple world, we are going to have x equal to -1.0912

When x = -1, we plug that one in; so we have y = -1 plus the x value that we are plugging in, -1.0918

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.0928

Our purple solution and our green solution are all of the solutions to this system.0940

All right, the next example: Find the solutions to this system.0945

In this case, we see that y is already over here, just by itself; so it looks like an easy candidate for substitution.0948

Let's plug x2 + 2 in for y over here.0955

We plug that in; we have x2 + 2, since that was what y used to be; that equals...0958

oops, squared: we have to have everything continue to be the same...x - 1.0965

x2 + 2, squared, equals x - 1; at this point, that might raise our suspicions a little bit.0970

But let's keep working it out and see what happens.0977

So, that gets x4 + 4x2 + 4 = x - 1.0980

This might start to raise our suspicions about what we are looking at.0989

Is it possible for this equation to ever be true?0993

Is there some x-value that would make the left-hand side the same thing as the right-hand side here?0996

Well, let's keep working it out and see if we can get something that will make it obvious what we are looking at.1002

x4 + 4x2 + 5 = x; OK, if we look at this, we might realize that we could try to solve it from this point;1006

but we aren't certain that there are solutions: it said "If possible."1018

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,1023

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,1028

before you get too deep into the process of trying to solve it.1032

So, x4 + 4x2 + 5...well, what does that look like--what would we end up seeing there?1035

Well, that is going to be a really, really fast-growing graph that starts at some height of 5,1041

and then shoots up really quickly: x4 + never becomes negative.1048

That is what the left-hand side is equal to.1052

But the right-hand side, x--what would that end up being?1054

That is going to be here; and if we graphed just x, that would go like this.1058

Now, we are appealing to a graph to understand this, but we see that the left-hand side1063

is going to always be putting out much larger numbers, no matter what x we put in,1067

than the right-hand side is ever going to be able to put out.1070

We have x4 + 4x2 + 5; that is going to make really big numbers, always positive, really quickly.1073

Now, x can end up getting large positive numbers, but it has to put in a very large x to do that.1078

And if we put in a very large x on the right side, the left side will be enormous.1083

So, we see that these two sides can never match up; so our suspicion is that there are actually no solutions here.1087

We could try to move that x over and have it equal 0.1094

But remember: some parabolas/ this case, it is not a parabola, because it is a fourth-degree...1096

but some polynomials don't have solutions; they never touch the x-axis,1102

if we are searching for when it is equal to 0; we are searching for roots.1107

So, we end up being able to figure out that this won't work.1110

But we want something that really makes it more obvious than just having to turn this into an equation and see this.1114

It just seems a little bit uncertain; so the best way for this is actually going to be to graph it.1120

If we graph it, we will see that this very clearly can never work out.1125

We can graph both of our original equations, x2 + 2 = y and y2 = x - 1.1130

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, 4.1138

Great; so let's first graph the easy one, x2 + 2 = y.1152

That one is pretty easy for us to graph; we will graph this one in blue.1157

x2 + 2 = y; well, we start at a height of 2; we plug in x = 0, and we get a height of 2.1162

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 6.1170

So, we see that we are going to curve out very quickly; we are going to shoot out like this.1177

That is what the curve ends up being; don't get that part confused--the graph is not actually connecting to that circled part.1182

The blue graph goes to x2 + 2 = y; what about y2 = x - 1?1189

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,1194

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.1202

What we can do is y2 + 1 (we add 1 to both sides) = x.1207

And so, we can solve this from y's point of view, as being the input, and x being the output.1212

For example, if y is 0, what does x end up being?1216

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 1.1220

0 squared, plus one, equals positive 1.1230

At a height of 1, when we plug in y = 1, x will end up being 12 + 1, or 2.1234

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;1242

5 is going to be our x-value, somewhere out here.1252

The same thing happens if we plug in a negative height: -1 will end up getting us an x-value of 2.1255

-2 will end up giving us an x-value of +5, as well...1260

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.1264

And then, we curve this out like this, because it is going to be a parabola, as well, because it is of that degree.1271

Curve that just a little bit better, so we can see it more accurately.1279

Curve that like this; it goes out like this.1284

So notice: the red one is going to keep going out to the right; the blue one is going to keep going up.1290

They are never going to end up touching each other.1295

This cup goes like this; this cup goes like this; they go out like that.1297

They are never going to touch each other; they are never going to intersect.1301

There is no way for these things to ever connect to each other.1304

They can't ever connect to each other, because we see that when graphing them, they fail to ever touch.1308

With that in mind, we see that there are no solutions to this system.1313

All right, next: Graph the solutions to the system of inequalities: y < 3cos(x); y ≥ x2; and x < 1.1321

Now, we haven't explicitly said that we are going to be using trigonometric things.1330

But we are basically at that point, where we are going to assume that you have gone through the trigonometry lessons.1334

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.1339

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.1345

Since we are working with cosine, we know that we are going to want it at least out to 2π on either side.1355

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.1360

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;1371

1, 2, 3, 4, 5, 6, 7; up 1, 2, 3, 4, 5, 1, 2, 3, 4, 5; OK.1382

Let's graph each one of these: first, we will graph y < 3cos(x) in green.1400

Now, notice: it is less than, so it is going to be dashed.1406

Let's get some points, so that we can draw in this line properly.1408

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.1412

So, with radians, if we plug in 0 (x is 0), then we are going to have a cosine of 0 put out 1;1419

so we will have 3 times 1, so we will be at a height of 3 when we are at a horizontal location of 0.1425

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 0.1431

The same thing if we go over to the negative π/2, around halfway between 1 and 2.1445

Then, at π, we are going to end up having -3, because cos(π) is -1, so 3cos(π) will be -3.1452

So, 3.14 is a little bit past the 3 marker over here.1463

And then, the same thing happens over here at -π; so -3...a little bit past the -3 marker...1470

We can curve out what cosine looks like, and what we knew before, in this portion right here.1478

All right, OK, we can do the same thing with continuing this graph out.1493

We know that at 2π, 1, 2, 3, 4, 5, 6...around 6.28 we will be back up to a height of 3.1498

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...1509

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...1519

a little bit over -4 1/2...and then 5, 6...a little bit over 6; and there we go...a little bit over -6.1526

We can curve in a little bit more of this, like that.1536

Now, notice: at this point, I technically made a mistake, because was it a strict inequality?1545

Yes, it was a strict inequality; it was strictly less than.1553

So, if that is the case, we can't be using a solid line; we have to be using a dashed line.1556

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,1561

which I am sure we have all done at some point--where we accidentally draw something as solid,1566

and realize, "Oh, I have to make this dashed now."1571

We came through with an eraser and just erase it into a dashed form.1572

At this point, we have y < 3cos(x); at least, the line, the curve generated by it, graphed in.1581

We will shade it in, in a little bit.1587

Next, we will do y ≥ x2 in blue.1589

This one is much easier--we know it: x is 0; we are at a height of 0; at 1, 1; at 2, +4;1593

the same thing on the other side: -2, +4; and we curve out, just in a nice, handy parabola--much faster.1600

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.1611

And then finally, x < 1 we will do in red.1619

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.1622

We are going to fix x at 1, and then we will see that y is allowed to go to anything it wants,1630

because for x < 1, y can be anything it wants; x < 1 doesn't care what y is.1636

So, we fix x at 1, and we draw a dashed line, because it is a strict inequality.1641

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.1647

So now, we do some shading and figure out what is allowed for each one of these.1653

y < 3cos(x): we could do a test point at (0,0); 0 < 3cos(0), so 0 < 3.1657

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.1667

So, what is below the curve? Stuff like this.1674

I am not doing a very extreme job of shading, just because we want to have some idea of where we are looking.1681

So, we don't have to shade too much right now, until we figure out where they all agree.1685

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.1689

Let's try (0,1): 1 is greater than or equal to 02...indeed, that is true.1696

Also, since it is y ≥, we know that it has to be above the curve.1701

So, what is above the curve is this stuff here; great.1705

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.1710

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).1717

So, we would go in to the left; great.1724

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.1728

And there we go; and that is how we figure out where these things go; cool.1741

All right, Example 4: A rectangular box has the following properties.1747

The sum of its edges is 24 feet; adding together the area of each of its faces gives a total of 22 square feet;1750

its height is twice its width; and then we are asked to find out what its volume is.1756

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.1760

If we have a box, some box, it has three things to it: length, width, and height.1764

We can see that; we have...what is its length? What is its width? And what is its height?1779

Great; with that in mind, let's start trying to figure out how these properties turn into math.1791

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?1797

Or are they saying all of its edges?1805

So sometimes, you end up seeing things that are a little bit confusing in math.1807

And if you saw this on a test, it would probably be a good idea to ask your teacher, because you are not sure.1810

Does that mean h + w + l, or does that mean each edge of the box, all added together?1814

Let's go with the sum of every edge; let's say that is what it means.1820

But notice how each of its edges...the sum of its edges...well, that could be considered to just be the three edges,1826

the fundamental edges (height, width, and length); but it could also be all of the times that they show up on the box.1831

If you have a rectangular object, it shows up here and here and here.1837

Height actually shows up four times, because we have each of the columns that make up our box.1840

We have a height here, a height here, a height here, and a height here.1847

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.1852

So, the sum of not just its edges, but every edge, is how we will do this problem.1858

All right, so if that is the case, this first thing is going to end up coming out to be...1863

The first idea will come out to be 4h + 4w + 4l = 24 feet.1869

So, height, width, and length, all combined together, comes out to a total of 24 feet.1881

because it is the sum of every edge, every one of the edges, and each edge...1886

height will show up 4 times; width will show up 4 times on the box; length will show up 4 times on the box.1890

If 4you find this confusing, try just finding some rectangular object that you can look at;1896

pick up an actual box and count the edges--count how many times its height shows up.1899

Any rectangular box will be able to show you this idea, if it is a little confusing.1903

Physical things are a great way to explore things in math.1906

Next is adding together the area of each of its faces.1910

This one is a little bit tougher: how many times does, let's say, this face right here--the very front face--show up?1913

Well, we see that height times width would be the area of that face.1920

So, height times width is going to get multiplied together.1924

But it doesn't show up just on the front; it also shows up on the back side.1929

It is the front side, but also the back side--both sides.1934

So, it is not going to be just h times w, but 2 times h times w.1936

By that same logic, each of the side faces...we are going to have l times h, because it is h over here...1942

so the side faces will show up twice, as well; so we have 2lh.1950

And then finally, the top face is going to be the length times the width, so 2 length times width.1956

And we were told that that came out to be 22 square feet total.1970

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.1974

lh will be the area of one face; it doubles up, and so on, and so forth.1979

So, we have 2hw + 2lh + 2lw = 22; and then finally, we were told that the height is twice its width.1983

That is probably the easiest one of all: if the height is equal to twice the width, it is 2 times the width; great.1990

At this point, how do we find volume--what is volume based on?1997

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.2000

That is all of the steps that we need together to be able to figure out what this is going to be--2009

to be able to get this in a position where we can solve it.2013

So now, let's start working it out.2015

We have 4l + 4w + 4h = 24, 2lw + 2wh + 2lh = 22, and h = 2w.2017

And we are looking for volume, which is going to be l times w times h.2024

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.2028

Let's divide everything by 4; we have l + w + h = 24/4, which gets us 6.2033

The same thing over here: let's divide everything by 2, so that gets us lw + wh + lh = 11; great.2040

So, at this point, we can actually start figuring things out.2051

Let's try to solve in terms of w; we have w over here; h is already ready to be substituted in somewhere.2053

So, since it is connected to w, we can probably figure out w easiest from what we have set up here.2061

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 = 6.2066

l + w + 2w...3w = 6, so l = 6 - 3w.2080

Now, at this point, we are ready to substitute l in, as well.2087

We have h ready to substitute and l ready to substitute; so we can now go into our big equation, lw + wh + lh = 11.2090

So, we plug in here: l is 6 - 3w, so (6 - 3w) times w, plus w times...h is l...l is (6 - 3w), times 2w; equals 11.2097

6 - 3w times w...w distributes, so we have 6w - 3w2, plus...w distributes onto 2w...2118

well, not "distributes," but 2w2 + (6 - 3w)2w...6 times 2w becomes 12w,2126

minus 3w times 2w, becomes -6w2; equals 11.2134

Great; let's simplify things a bit; we have -3w here; plus 2w2 here; and -6w2 here.2139

So, -3w2 + 2w2 gets us -1w2; - 6w2 brings us to a total of -7w2.2146

Our 6w and 12w combine to +18w, equals 11.2154

We have squared; we have a single degree of 1; and we have a constant; this looks like a polynomial.2160

Let's get it into an easy-to-solve polynomial format.2166

Add 7w2 to both sides; add -18w to both sides; +11.2168

At this point, we could toss this into the quadratic formula and solve out the answers.2173

But we might be able to get lucky; and even though it looks a little complex, we might realize that we can factor this.2178

It is not too difficult to factor.2183

We get lucky; we notice that it turns out pretty easy to factor: 7w and w here...2184

We need to have minus on both of them, because it comes out to be positive 11.2194

Minus 18w...we have a 7w here, so this will be -1, and this will be -11.2198

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.2203

Great; it is always a good idea to check when you are factoring.2216

So, we can now solve each one of these: 7w - 11 = 0, or w - 1 = 0.2219

We actually have two different possible worlds.2226

It didn't say that in the problem, but there are two different possible worlds for what the width of the box can be.2228

It can be 11/7 or 1; so 11/7 and 1 are our two possible things.2235

So, let's call these, arbitrarily...we will make colors for these.2241

w = 1 looks easiest to solve and deal with first, so we will make that the purple world.2246

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 2.2250

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 = 3.2261

We have w = 1, h = 2, l = 3 in our first world, in this purple world,2276

because remember: there were two possibilities for what our width could be.2284

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 6.2287

So, one possible value for our volume is going to be 6 cubic feet--that is one possible answer.2294

It turns out that there are two different worlds here, so we want to check out both of them.2305

The other one: we will make this the green world, where the width is equal to 11/7...2308

Well, if width equals 11/7, then we can plug that into h = 2w, so h = 2(11/7), which equals 22/7.2312

So, if our width is 11/7, then our height is 22/7.2323

And then, we can also figure out what our length is going to be.2331

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.2335

At this point, we have width = 11/7; height = 22/7; and length = 9/7, all of these in feet as the units.2350

So, at this point, we know that the volume is equal to each one of these, multiplied together:2363

so 11/7 times 22/7 times 9/7; we multiply these all out together, and it becomes 2178/343,2366

which is really not that easy to see what that means.2380

So, let's approximate that using a calculator; and that comes out to be 6.35.2383

So, our other possibility is that the volume comes out to be 6.35 cubic feet.2389

So, it turns out that there is actually a larger possible box if we are not going with these nice, friendly integer things.2396

But we can still follow the three requirements; there are three conditions that were given to us.2402

It turns out that there are two possible boxes that actually fit those conditions; and those are the volumes.2407

In our purple world, where our width was 1, we got 6 cubic feet.2412

And in our green world, where our width was 11/7, we got 6.35 cubic feet.2415

And if you wanted to check this--if you wanted to make sure that everything was great--2420

a good thing to do would be to see that you have width = 1, height = 2, length = 3,2423

and then just try plugging that into each of these three equations that we started with,2429

and making sure..."Yes, that checks out; yes, that checks out; yes, that checks out."2433

The same thing over here for the width = 11/7, height = 22/7, length = 9/7:2437

you can just plug that into each of these three equations, and make sure that it checks out in each one,2447

because if it checks out in each one, you know that that is a workable answer.2452

All right, that finishes our work with systems of equations of all types.2455

And we will see you at later--goodbye!2459