For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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

 Solve by substitution. For this problem, the equations are already in the form y=stuff, so we can just set those portions equal and solve for x:
x+5 = x^{2} + x −4  Solving for x, we see that we're working with a polynomial equation, so get everything on one side and a 0 on the other:
Factor to find the solutions for x:x+5 = x^{2} + x −4 ⇒ 0 = x^{2} − 9
Thus, we have two possible values: x=−3 and x=3.0 = x^{2} − 9 ⇒ 0 = (x+3)(x−3)  Before we're done, we need to find the yvalue that matches with each of those. x=−3:
x=3:y = x+5 ⇒ y = (−3) + 5 = 2 y = x+5 ⇒ y = (3) +5 = 8

 Solve by substitution. For this problem, the equations are already in the form y=stuff, so we can just set those portions equal and solve for x:
x^{2} − 3x + 5 = −x^{3} −x^{2} + 5x + 5  Solving for x, we see that we're working with a polynomial equation, so get everything on one side and a 0 on the other:
Factor to find the solutions for x:x^{2} − 3x + 5 = −x^{3} −x^{2} + 5x + 5 ⇒ 0 = −x^{3} − 2x^{2} + 8x
Thus, we have three possible values: x=−4, x=0, and x=2.0 = −x^{3} − 2x^{2} + 8x ⇒ 0 = −x(x^{2}+2x−8) ⇒ 0 = −x(x+4)(x−2)  Before we're done, we need to find the yvalue that matches with each of those. x=−4:
x=0:y = x^{2} − 3x + 5 ⇒ y = (−4)^{2} −3(−4)+5 = 33
x=2:y = x^{2} − 3x + 5 ⇒ y = (0)^{2} −3(0)+5 = 5 y = x^{2} − 3x + 5 ⇒ y = (2)^{2} −3(2)+5 = 3

 Solve by substitution. Solve for y in the first equation so we can plug it in to the second:
Now we can plug into the second equation:xy = 4 ⇒ y = 4 x3x − 2 ⎛
⎝4 x⎞
⎠= 5  It will be easier to not have a variable in the denominator of a fraction, so multiply both sides by x:
We now see we're working to solve a polynomial equation, so get everything on one side:3x − 8 x= 5 ⇒ 3x^{2} − 8 = 5x 3x^{2} − 8 = 5x ⇒ 3x^{2} − 5x − 8 = 0  This looks rather difficult to factor, so let's use the quadratic formula:
Simplify:x = −b ±
√b^{2} − 4ac2a⇒ −(−5) ±
√(−5)^{2} − 4(3)(−8)2(3)
Thus, we have two solutions:x = 5 ±
√25 +966= 5 ±
√1216= 5 ±11 6x = 5+11 6= 8 3and x = 5−11 6= −1  Before we're done, we need to find the yvalue that matches with each of those. x=[8/3]:
x = −1:xy = 4 ⇒ ⎛
⎝8 3⎞
⎠y = 4 ⇒ y = 4 · 3 8⇒ y = 3 2xy = 4 ⇒ (−1) y = 4 ⇒ y = −4

 Do not let yourself be worried that there are three variables instead of just two: we still approach it in the same mannersolve for a variable in terms of the others, plug in, repeat until you have an equation with just one variable.
 We can approach this by working through the variables in any order, but it looks easiest to plug the last equation (z=4y) into the middle equation:
We often tend to solve for y in terms of x, but for this problem, it's better to put everything in terms of y because we started with z in terms of y. [Although you could do it a different way, if you wanted.] We now have x = 3y and z=4y.z−x−y=0 ⇒ (4y) − x − y = 0 ⇒ 3y − x = 0 ⇒ 3y = x  We can now plug these in to our first equation:
We see we're working with a polynomial equation, so get everything on one side and factor:z = x^{2} + y^{2} ⇒ (4y) = (3y)^{2} + y^{2} ⇒ 4y = 9y^{2} + y^{2} ⇒ 4y = 10y^{2}
Which gives solutions of y=0 and y = [2/5].4y = 10y^{2} ⇒ 0 = 10y^{2} − 4y ⇒ 0 = y (10y−4)  Before we're done, we need to find the xvalue and zvalue that matches with each of those. y=0:
y=[2/5]:x=3y = 3(0) = 0 ⎢
⎢z=4y = 4(0) = 0 x = 3y = 3 ⎛
⎝2 5⎞
⎠= 6 5⎢
⎢z = 4y = 4 ⎛
⎝2 5⎞
⎠= 8 5

 If we tried to solve this using substitution and algebra, we would find that the problem would become extremely difficult. However, we can get a good approximation of the precise answer by using a graphing calculator. To do this, our first step is to get both of the equations into a format we can graph. That is, something in the form y=stuff.
 The second equation is easy, because it's already in the desired form. The first equation isn't too difficult, but it requires us to catch a little trick. Notice, if we worked to solve for y in the first equation (so we could graph it), we would do the following:
Next, we would need to take the square root of both sides. However!, we must remember that when we take the square root of both sides in algebra, we must put a ± symbol in. If we forget this, we will only get the positive, top half of the graph for the first equation. Thus, we get y = ±√{25−x^{2}}. So in the end, our first equation splits into two for graphing on a calculator:x^{2} + y^{2} = 25 ⇒ y^{2} = 25−x^{2} x^{2} + y^{2} = 25 ⇒ y =
√25−x^{2}y = −
√25−x^{2}  Plug in both of the "split" versions into your graphing calculator along with the second equation from the system and you should get a graph something along the lines of the below. [If your graph looks very different, make sure your graphing calculator is in radians mode and that you entered everything correctly.]
 From the graph, we see that we have a total of four intersections, so there are a total of four solutions to the system of equations. Using your graphing calculator, select the 'intersection' command (if you've never done this before, check out the Appendix to this course on graphing calculators), then select the two functions involved and the interval where the calculator should search for an intersection. Notice that the top half and the bottom half of the circle are treated as two different functions from the calculator's point of view. That means you have to make sure to choose the appropriate half of the circle (along with the trig function) when looking for an intersection point.

 It usually helps to begin by putting the inequalities in a format that is easy to graph, like one where y is alone on one side. Both of the inequalities are already in such a form, so we can immediately plot them.
 Plot out a line or curve based on each inequality. While plotting, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once every line/curve is plotted on the graph, (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's structured to have y alone on one side (like ours are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side.
 Once you have shaded each of the inequalities, look for where the shadings overlap. The overlap is the set of solutions.

 It usually helps to begin by putting the inequalities in a format that is easy to graph, like one where y is alone on one side. Both of the inequalities are already in such a form, so we can immediately plot them.
 Plot out a line or curve based on each inequality. While plotting, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once every line/curve is plotted on the graph, (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's structured to have y alone on one side (like ours are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side.
 Once you have shaded each of the inequalities, look for where the shadings overlap. The overlap is the set of solutions.

 It usually helps to begin by putting the inequalities in a format that is easy to graph, like one where y is alone on one side. However, it is not always the case that having y alone on one side is the easiest way to know how to graph something. You might already know from previous math classes (if you don't, and are curious to see more, check out the lessons on conic sections) that an equation of the form x^{2} + y^{2} = r^{2} gives a circle of radius r that is centered at the origin. Thus, we see that the first inequality describes a circle of radius 8 (since 64=8^{2}). For the other two inequalities though, it will be useful to put them in the normal form:
x+y ≥ −5 ⇒ y ≥ −x −5 ⎢
⎢10 > 2x−y ⇒ y > 2x −10  Plot out a line or curve based on each inequality. While plotting, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once every line/curve is plotted on the graph, (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's structured to have y alone on one side (like our two lines are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side. This testing method works well for the circle. If we test (0, 0) in the circle's inequality, we get that 0^{2} + 0^{2}< 64, so the test point satisfies the inequality, meaning we should shade the inside of the circle.
 Once you have shaded each of the inequalities, look for where the shadings overlap. The overlap is the set of solutions.
 If you have difficulty understanding the problem, begin by sketching out a picture of it. Let's call the two unknown legs a and b. We can connect the two legs to the hypotenuse through the Pythagorean Theorem: the sum of each leg squared then added together equals the hypotenuse squared.
We can also connect the length of all the sides through the perimeter. By definition, the perimeter is the sum of the lengths of all the sides, so we havea^{2} + b^{2} = 53^{2} a+b+ 53 = 126  From here, we have a system of equalities to solve. Let's solve for b in the perimeter equation:
We can now plug this in to the Pythagorean Theorem equation:a+b+ 53 = 126 ⇒ b = 73 − a a^{2} + b^{2} = 53^{2} ⇒ a^{2} + (73−a)^{2} = 53^{2}  We see that we will be working with a polynomial equation, so expand, simplify, and get everything on one side:
Since the numbers are so large, it would be a little difficult to solve this by factoring (but not impossible), so let's use the quadratic formula:a^{2} + (73−a)^{2} = 53^{2} ⇒ a^{2} +5329 − 146a + a^{2} = 2809 ⇒ 2a^{2} −146a + 2520 = 0
Thus, there are two possibilities for a:a = −(−146) ±
√(−146)^{2} − 4(2)(2520)2(2)= 146 ±34 4
[Eeek! Shouldn't we be worried about the fact that we just got two answers for a? It is a word problem after all, and it doesn't make sense for a length to be two things simultaneously. However, don't worry. This will clear itself up shortly.]a = 146+34 4= 45 ⎢
⎢a = 146−34 4= 28  Let's now try using one of the possible values we have: a=45. Plugging that in, we can find b:
Thus, if a=45, we have b=28. Similarly, if we try using a=28, we getb = 73 − a ⇒ b = 73−(45) = 28
So if a=28, then we have b=45. Therefore, the two sides are length 28 m and 45 m. [This clears up why we got two values for a from the quadratic formula. Because the two legs must come out to be 28 and 45. We don't know which one a represents, so it could be either. But whatever a winds up being, b must be the other value.]b = 73−a ⇒ b = 73 − (28) = 45


 While the problem tells us how to connect the distance the car has traveled to the time elapsed, it does not directly tell us what the location of each car is at a given time. We must first set that up. At t=0, arbitrarily say that car A is at the starting location of L=0. Thus, after t seconds, car A's location will be
For car B, we do a similar thing, but we want it to be based on the same location system that car A is. Thus, we need to know where car B is relative to car A at t=0. According to the problem, car B starts 100 m behind car A, so B must have a starting location of L=−100. Thus, after t seconds, car B's location will beL_{A} = 20t. L_{B} = 1 2t^{2} + 15 t − 100.  At the moment when car B passes car A, they must have the same location. Thus, we want to consider when L_{A} = L_{B}. We can now set the right side of each equation equal:
Now we solve for the time t.20t = 1 2t^{2} + 15 t − 100  It's a polynomial equation, so get everything on one side:
20t = 1 2t^{2} + 15 t − 100 ⇒ 0 = 1 2t^{2} −5t − 100 ⇒ 0 = t^{2} − 10t −200  If you can't see how to factor it, just use the quadratic formula. But we might see that we can factor it as
Thus the two possible times (from the equation) are t=−10 and t=20. However, it makes no sense to talk about negative time, since we don't know what was happening before the start of the problem. We only know what happens in positive time, so that is our answer.0 = t^{2} − 10t −200 ⇒ 0 = (t+10)(t−20)
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Nonlinear Systems
Lecture Slides are screencaptured 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.
Precalculus with Limits Online Course
Transcription: Nonlinear Systems
Hiwelcome back to Educator.com.0000
Today, we are going to talk about nonlinear 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 = y^{2} 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 nonlinear 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 indepth.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 = y^{2}, 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, y^{2}, plus 6y, equals 9.0087
We have 3y^{2} + 6y = 9; it is nice not to have as many numbers.0097
We see that we can divide by the number 3: we have y^{2} + 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 y^{2} + 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 y^{2}; 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 = y^{2}; so now, we plug it in: what is x going to be when y is 1?0171
We plug that in; x = 1^{2}, 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 = +9that 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 = y^{2} 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 = 3y^{2}.0240
So now, we can bring this over; we can add it over; and we have + 3x = 3y^{2}.0247
We can add that on either side: 3x and 3x cancel out; we have 6y = 9  3y^{2}.0255
We move this over; we have 3y^{2} + 6y  9 = 0.0265
This is starting to look familiar; we divide everything by 3; y^{2} + 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 nonlinear 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 y^{2} over here, but y^{2} doesn't show up anywhere over here.0307
There is no y^{2} over here, so we can't try to cancel out y^{2}.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 = y^{2} and 3x + 6y = 9, we could look at this and say,0381
"x = y^{2} 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 x0418
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 = y^{2}, 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 = y^{2} is a little bit different; we have x = y^{2}, 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 nonlinear 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 thereany 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 nonlinear 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 nonlinear inequalities: when we are working with nonlinear 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 dashed...you use dashed when it is a strict inequalitystrictly less than or strictly greater than.0621
Notice: in this case, we have that y is strictly greater than 2x^{2}  5.0629
So, this red one is the graph of 2x^{2}  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 lookit 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 lineit is not a solution."0656
And then, after that, you shade it appropriatelyyou 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 > 2x^{2}  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 1/5x^{2} + 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 shading0790
each one of them individually, and then where all of the shadings agree0793
where all of the shadings overlapthat 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 + x^{2} = 2; this looks like a polynomial, so let's solve it like a polynomial.0833
x^{2}  x...subtract the 2 over...we have x^{2}  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 x^{2}great; x + 1...so 1x  2x gives us x; 2 times 1 is 2; greatthat 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, yvalue 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 x^{2} + 2 in for y over here.0955
We plug that in; we have x^{2} + 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
x^{2} + 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 x^{4} + 4x^{2} + 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 xvalue that would make the lefthand side the same thing as the righthand 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
x^{4} + 4x^{2} + 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, x^{4} + 4x^{2} + 5...well, what does that look likewhat would we end up seeing there?1035
Well, that is going to be a really, really fastgrowing graph that starts at some height of 5,1041
and then shoots up really quickly: x^{4} + 4x^{2}...it never becomes negative.1048
That is what the lefthand side is equal to.1052
But the righthand side, xwhat 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 lefthand side1063
is going to always be putting out much larger numbers, no matter what x we put in,1067
than the righthand side is ever going to be able to put out.1070
We have x^{4} + 4x^{2} + 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/some...in this case, it is not a parabola, because it is a fourthdegree...1096
but some polynomials don't have solutions; they never touch the xaxis,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, x^{2} + 2 = y and y^{2} = x  1.1130
We will make a tickmark 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, x^{2} + 2 = y.1152
That one is pretty easy for us to graph; we will graph this one in blue.1157
x^{2} + 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 confusedthe graph is not actually connecting to that circled part.1182
The blue graph goes to x^{2} + 2 = y; what about y^{2} = 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 y^{2} + 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 1^{2} + 1, or 2.1234
At a height of 2, when y is at the height of 2, we plug that in; we have y^{2} + 1, so 2^{2} + 1; 4 + 1;1242
5 is going to be our xvalue, somewhere out here.1252
The same thing happens if we plug in a negative height: 1 will end up getting us an xvalue of 2.1255
2 will end up giving us an xvalue of +5, as well...1260
I'm sorry: did I say 2 for 1? A height of 1 will give us an xvalue 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 ≥ x^{2}; 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 xaxis 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 pointwhere 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 ≥ x^{2} in blue.1589
This one is much easierwe 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 parabolamuch 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 1it 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 ≥ x^{2}...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 0^{2}...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 xvalue has to be below it.1710
We could also use a test point, like (0,0); we don't care about the yvalue, 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 edgescount 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 herethe very front faceshow 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 sideboth 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 volumewhat 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 be2009
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 2w...plus l...l is (6  3w), times h...is 2w; equals 11.2097
6  3w times w...w distributes, so we have 6w  3w^{2}, plus...w distributes onto 2w...2118
well, not "distributes," but 2w^{2} + (6  3w)2w...6 times 2w becomes 12w,2126
minus 3w times 2w, becomes 6w^{2}; equals 11.2134
Great; let's simplify things a bit; we have 3w here; plus 2w^{2} here; and 6w^{2} here.2139
So, 3w^{2} + 2w^{2} gets us 1w^{2};  6w^{2} brings us to a total of 7w^{2}.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 easytosolve polynomial format.2166
Add 7w^{2} 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 7w^{2}; 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 = 1if 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 feetthat 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 thisif you wanted to make sure that everything was great2420
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 Educator.com latergoodbye!2459
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