INSTRUCTORS Carleen Eaton Grant Fraser

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• In a linear-quadratic system, use substitution to solve.
• When graphing inequalities, remember the conventions about graphing boundaries using either solid or dotted lines.
• If possible, check your solutions to systems of equations by graphing.

Solve:
x2 + y2 + 5x − y − 6 = 0
x − y = 2
• This is a linear\ quadratic system. Solve by substitution.
• Solve for x for the second equation
• x = y + 2
• Substitute x into the first equation
• x2 + y2 + 5x − y − 6 = 0
• (y + 2)2 + y2 + 5(y + 2) − y − 6 = 0
• y2 + 4y + 4 + y2 + 5y + 10 − y − 6 = 0
• Combine like terms
• 2y2 + 8y + 8 = 0
• Divide entire equation by 2
• y2 + 4y + 4 = 0
• Notice how this is a perfect square trinomial, solve for y
• (y + 2)2 = 0
• y = − 2
• Solve for x, using y = − 2
• x = y + 2
• x = − 2 + 2
• x = 0
Solution: (0, − 2)
Solve:
− 2y2 + 6x + 3y + 153 = 0
2x + y = 3
• This is a linear\ quadratic system. Solve by substitution.
• Solve for y for the second equation
• y = − 2x + 3
• Substitute y into the first equation
• − 2y2 + 6x + 3y + 153 = 0
• − 2( − 2x + 3)2 + 6x + 3( − 2x + 3) + 153 = 0
• − 2(4x2 − 12x + 9) + 6x − 6x + 9 + 153 = 0
• − 8x2 + 24x − 18 + 6x − 6x + 9 + 153 = 0
• Combine like terms
• − 8x2 + 24x + 144 = 0
• Divide entire equation by − 8
• x2 − 3x − 18 = 0
• Factor and solve using the zero - product property
• (x − 6)(x + 3) = 0
• x − 6 = 0 and x + 3 = 0
• x = 6 and x = − 3
• Solve for y, using x = 6;x = − 3
• y = − 2x + 3
• y = − 2(6) + 3 = − 9
• y = − 2( − 3) + 3 = 9
Solution: ( − 3,9) and (6, − 9)
Solve:
2x2 − x + 18y − 28 = 0
x − 2y = 4
• This is a linear\ quadratic system. Solve by substitution.
• Solve for x for the second equation
• x = 2y + 4
• Substitute x into the first equation
• 2x2 − x + 18y − 28 = 0
• 2(2y + 4)2 − (2y + 4) + 18y − 28 = 0
• 2(4y2 + 16y + 16) − 2y − 4 + 18y − 28 = 0
• 8y2 + 32y + 32 − 2y − 4 + 18y − 28 = 0
• Combine like terms
• 8y2 + 48y = 0
• Divide entire equation by 8
• y2 + 6y = 0
• Factor and solve using the zero - product property
• y = 0 and y + 6 = 0
• y = 0 and y = − 6
• Solve for x, using y = 0; y = − 6
• x = 2y + 4
• x = 2(0) + 4 = 4
• x = 2( − 6) + 4 = − 8
Solution: ( − 8, − 6) and (4,0)
Solve:
− 2x2 − 37x + 6y − 188 = 0
x + 2y + 4 = 0
• This is a linear\ quadratic system. Solve by substitution.
• Solve for x for the second equation
• x = − 2y − 4
• Substitute x into the first equation
• − 2x2 − 37x + 6y − 188 = 0
• − 2( − 2y − 4)2 − 37( − 2y − 4) + 6y − 188 = 0
• − 2(4y2 + 16y + 16) + 74y + 148 + 6y − 188 = 0
• − 8y2 − 32y − 32 + 74y + 148 + 6y − 188 = 0
• Combine like terms
• − 8y2 + 48y − 72 = 0
• Divide entire equation by − 8
• y2 − 6y + 9 = 0
• Factor and solve using the zero - product property
• (y − 3)2 = 0
• y − 3 = 0
• y = 3
• Solve for x, using y = 3;
• x = − 2y − 4
• x = − 2(3) − 4 = − 10
Solution: ( − 10,3)
Solve:
x2 + y2 − 6x + 6y − 19 = 0
x2 + y2 − 6x − 7y + 20 = 0
• Multiply the first equation by − 1 to create an equivalent system.
• -1* (x2 + y2 − 6x + 6y − 19 = 0)
• − x2 − y2 + 6x − 6y + 19 = 0
• Add the system of equation
• (− x2 − y2 + 6x − 6y + 19 = 0) + (x2 + y2 − 6x − 7y + 20 = 0)
• − 13y + 39 = 0
• Solve for y
• − 13y = − 39
• y = 3
• Solve for x using either the first or second equation form the system.
• x2 + y2 − 6x − 7y + 20 = 0
• x2 + (3)2 − 6x − 7(3) + 20 = 0
• x2 + 9 − 6x − 21 + 20 = 0
• x2 − 6x + 8 = 0
• Factor and solve using the zero - product property
• (x − 4)(x − 2) = 0
• x − 4 = 0 and x − 2 = 0
• x = 4 and x = 2
Solution: (4,3) and (2,3)
Solve:
− x2 + y2 + 14x − 13y − 8 = 0
x2 + y2 − 14x − 13y + 88 = 0
• Add the system of equatios
• (− x2 + y2 + 14x − 13y − 8 = 0) + (x2 + y2 − 14x − 13y + 88 = 0)
• 2y2 − 26y + 80 = 0
• Solve for y, factor but divide entire equation by 2 first
• y2 − 13y + 40 = 0
• (y − 5)(y − 8) = 0
• y = 5 and y = 8
• Solve for x using either the first or second equation form the system.
• y=5
x2 + y2 − 14x − 13y + 88 = 0
x2 + (5)2 − 14x − 13(5) + 88 = 0
x2 − 14x + 48 = 0
(x − 6)(x − 8) = 0
x = 6 and x = 8
• y=8
x2 + y2 − 14x − 13y + 88 = 0
x2 + 82 − 14x − 13(8) + 88 = 0
x2 − 14x + 48 = 0
(x − 6)(x − 8) = 0
x = 6 and x = 8
Solution: (6,5),(8,5),(6,8),(8,8)
Solve:
8x2 + 4y2 − 9x − 8y − 95 = 0
10x2 − 4y2 + 45x + 8y + 41 = 0
• Add the system of equatios
• (8x2 + 4y2 − 9x − 8y − 95 = 0) + (10x2 − 4y2 + 45x + 8y + 41 = 0)
• 18x2 + 36x − 54 = 0
• Solve for x, factor but divide entire equation by 18 first
• x2 + 2x − 3 = 0
• (x + 3)(x − 1) = 0
• x = − 3 and x = 1
• Solve for y using either the first or second equation form the system.
• x = − 3
8x2 + 4y2 − 9x − 8y − 95 = 0
8( − 3)2 + 4y2 − 9( − 3) − 8y − 95 = 0
4y2 − 8y + 4 = 0
y2 − 2y + 1 = 0
(y − 1)(y − 1) = 0
y = 1
• x=1 8x2 + 4y2 − 9x − 8y − 95 = 0
8(1)2 + 4y2 − 9(1) − 8y − 95 = 0
4y2 − 8y − 96 = 0
y2 − 2y − 24 = 0
(y − 6)(y + 4) = 0
y = 6 and y = − 4
Solution: ( − 3,1),(1,6),(1, − 4)
Solve:
2x2 − 24x − y + 69 = 0
− 2x2 + 5y2 + 24x − 9y − 144 = 0
• Add the system of equatios
• (2x2 − 24x − y + 69 = 0) + ( − 2x2 + 5y2 + 24x − 9y − 144 = 0)
• 5y2 − 10y − 75 = 0
• Solve for y, factor but divide entire equation by 5 first
• y2 − 2y − 15 = 0
• (y − 5)(y + 3) = 0
• y = 5 and y = − 3
• Solve for x using either the first or second equation form the system.
• y=−3 2x2 − 24x − y + 69 = 0
2x2 − 24x − ( − 3) + 69 = 0
2x2 − 24x + 72 = 0
x2 − 12x + 36 = 0
(x − 6)(x − 6) = 0
x = 6
• y=5
2x2 − 24x − y + 69 = 0
2x2 − 24x − (5) + 69 = 0
2x2 − 24x + 64 = 0
x2 − 12x + 32 = 0
(x − 4)(x − 8) = 0 x = 4 and x = 8
Solution: (6, − 3),(8,5),(4,5)
Solve:
x2 + y2< 9
(x − 4)2 + y2< 16
• Notice how both equations are circles. Draw both circles
• Circle 1 has a r = 3 and is centered at (0,0)
• Circle 2 has a r = 4 and is centered at (4,0)
• Circle 1 and 2 need to be dashed becuase both have the " < " sign, their circumference is not part of the solution set
• Check which way to shade using a test point.
• Circle One: Test (0,0)
x2 + y2< 9
02 + 02< 9
0 < 9
True
• Circle Two: Test (5,0)
(x − 4)2 + y2< 16
(5 − 4)2 + 02< 16
1 < 16
True
• For Circles 1 and 2, you must shade inside the boundary line of each circle
• The solution is whatever they have in common
Solve:
(x − 5)2 + y2 ≤ 36
(x − 5)2 + y2≥ 9
• Notice how both equations are circles. Draw both circles
• Circle 1 has a r = 6 and is centered at (5,0)
• Circle 2 has a r = 3 and is centered at (5,0)
• Circle 1 and 2 need be solid becuase both have the " ≤≥ " sign, their circumference is part of the solution set.
• Check which way to shade using a test point.
• Circle One: Test (5,0)
(x − 5)2 + y2 ≤ 36
02 + 02< 9
0 < 9
True
• Circle Two: Test (5,0)
(x − 5)2 + y2≥ 9
02 + 02≥ 16
0 ≥ 16
NotTrue
• For Circles 1, you must shade inside the circle.
• For circle 2, you must shade outside the circle.
• The solution is the shaded area they have in common. The solution set forms a doughnut shape.

*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.

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
• Solutions 2:49
• Graphs of Possible Solutions
• Example: Elimination
• Solutions 11:39
• Example: 0, 1, 2, 3, 4 Solutions
• Systems of Quadratic Inequalities 12:48
• Example 1: Solve Quadratic System 21:42
• Example 2: Solve Quadratic System 29:13
• Example 3: Solve Quadratic System 35:02
• Example 4: Solve Quadratic Inequality 40:29

Welcome to Educator.com.0000

Today, we are going to talk about solving quadratic systems of equations.0002

In earlier lectures, we discussed talking about linear systems of equations, and used various methods to solve those: for example, substitution and elimination.0006

And we are going to use some similar methods for quadratic systems, although these systems are more complex.0017

There are two types of quadratic systems that we are going to be working with today.0023

The first is a linear-quadratic system: linear-quadratic systems are a set of two equations in x and y,0027

in which one of the equations is linear, and the other is quadratic.0035

We will talk in a second about how to solve these; let's just stop at the definition right now.0041

x2 + 3x + y = 9; x = 2y: these two equations, considered together, would be a linear-quadratic system.0046

I have a linear equation here and a quadratic equation here.0057

And right here, this is a fairly simple linear equation to start out with.0063

And we can use that to apply the method discussed above.0068

So, given this type of system, these can be solved algebraically by isolating0072

one of the variables in the linear equation, and then substituting it into the quadratic equation.0077

OK, so conveniently, x is already isolated.0083

Often, you will be given a linear equation as part of a system where you have to do some manipulation to isolate either x or y.0088

But I already have x = 2y, so that is perfectly set up for me to substitute 2y in for the x variable in this quadratic equation.0096

So, I am rewriting the quadratic equation, and now substituting 2y in for x.0105

That gives me 2y2 + 3(2y) + y = 9.0113

What I am left is a quadratic equation, which I can solve by the usual methods that we have learned.0122

This is 22 is 4y2 + 6y + y = 9.0128

That gives me 4y2 + 7y = 9, or 4y2 + 7y - 9 = 0.0136

And then, you could go on to solve this, using the quadratic formula.0148

Once you find y, you can go ahead and substitute that value(s) in here, and then determine the corresponding value of x.0153

In a linear quadratic system, you may have either 0, 1, or 2 solutions.0170

And the easiest way to understand this is to think about it in terms of a graph.0176

A linear equation is going to give you a line: a quadratic equation might give you a parabola or an ellipse, a hyperbola, a circle...0180

So, let's look at some possibilities: maybe I have a system that ends up graphing out like this.0188

It gives me that line and this parabola.0196

Well, the solutions are going to be where these two intersect; but in this system, they will never intersect.0201

So, this system is going to have 0 solutions.0206

I may have another situation (again, I will use a parabola as my example) where I have a line that goes through right here at one point.0212

And this is going to give me one solution.0224

Perhaps I have an ellipse, and I have a line going through it like this.0228

It intersects at two spots, so I would have two solutions.0237

This helps to illustrate why you may have no solutions, one, or two solutions.0241

This would be a quadratic-quadratic system in which there is a set of two quadratic equations in x and y.0255

For example, x2 + y2 = 5 and 2x2 - y = 4:0262

you can see that you have two quadratic equations.0272

You could try to use substitution, but it could get a little bit messy.0276

So, often, elimination is the easiest way to use, so we solve them algebraically by elimination.0281

And you will recall, working with linear equations, that in elimination, what you try to do is get variables to have the same or opposite coefficients.0286

And then, either add or subtract the equations so that a variable drops out.0294

And here I want to get either x2 or y2 to drop out.0298

Well, I have an x2 term in each; so that, I could end up having drop out if I multiplied this top equation by 2.0304

So first, I am going to multiply the first equation by 2.0314

2 times x2 + y2 = 5: and this is going to give me 2x2 + 2y2 = 10.0317

So, I am going to then go ahead and write the other equation just below that: 2x2 - y = 4.0333

This is equation 1; this is equation 2; I multiplied equation 1 by 2, and then I ended up with this; and I have equation 2 right here.0347

Now, what I am going to do is subtract this second equation from the first.0359

Rewriting that: this gives me 2x2 - 2x2; these terms drop out.0366

The x2 terms drop out; a negative and a negative is a positive, so this is actually going to give me, let's see, 3...0375

actually, there are slightly different coefficients, so I can't add those; this is 2y2 - a negative0394

(that becomes a positive) y, and then 10 - 4 is equal to 6.0402

OK, so 2x2 - 2x2--those drop out; I have a 2y2 down here.0407

I have a negative and a negative, gives me a positive y; and then this becomes a -4; so 10 - 4 is 6.0414

So, I end up with 2y2 + y = 6 by using elimination.0422

Therefore, I have gotten rid of one of the variables.0427

From there, I have a regular quadratic equation that I can solve: that gives me 2y2 + y - 6 = 0.0431

I am going to try to solve that by factoring; and this is going to give me (2y + something) (y - something).0440

So, let's look at 6: if we take factors of 6, those are 1 and 6, and 2 and 3.0448

And I want them to be close together, so that I just end up with a coefficient of 1 here; so let's try 3 - 2 = 1.0455

Therefore, I am going to take + 3, - 2.0466

But I have to take this 2y into account; so let's go ahead and see if this works.0478

This becomes 2y2, and then this is minus 4y, plus 3y; so that gives me a -y; so that won't work--let's try it the other way.0482

Let's actually make this a negative and see what happens.0496

This gives me 2y2, and then this is 4y, minus 3y; that gives me a y; and then, -3 times 2 is -6.0500

So, this actually factored out, and it worked.0510

Using the zero product property, let's go back up here and get (2y - 3) (y + 2) = 0.0514

2y - 3 = 0, or y + 2 = 0; if either of these equals 0, the whole quantity equals 0.0522

Solving for y gives me 2y = 3 or y = 3/2; solving for y gives me y = -2.0530

Now, what I need to do is go back and find the corresponding x-value, so I have coordinate pairs as my answers.0541

All right, let's first work with y = -2.0552

And going back to this equation (because this one is simple to work with), I am going to substitute in that value for y2 and see what I get.0558

x2 + (-2)2 = 5: that is x2 + 4 = 5, or x2 = 1.0568

Therefore, x = 1 and -1; now, that is when y is -2.0580

So, my solutions are (1,-2) and (-1,-2)--I have two solutions there.0588

Now, I need to repeat that process for y = 3/2; let y equal 3/2, and then substitute in here.0603

This is going to give me x2 + (3/2)2 = 5, so x2 + 9/4 = 5.0610

So, x2 = 5 - 9/4; that gives me x2 =...a common denominator of 4; it would be 20/4 - 94, so x2 = 11/4.0620

Therefore, x = ±√11/4; now, looking over here, when y is 3/2, x could be √11/4, or it could be -√11/4.0637

So, I have another two members to my solutions: I have that x could be (√11/4,3/2), and (-√11/4,3/2)0656

Here, we have a situation where we actually have four solutions.0673

And I solved this by elimination; and then I had to go back and take each of the solutions I ended up with,0679

-2 and 3/2 for y, and find corresponding solutions for x.0686

And each of those yielded two values for x, so I ended up with 4 solutions altogether.0691

And I can also illustrate that point using graphing--of why you can end up with 0, 1, 2, 3, or 4 solutions.0700

Quadratic-quadratic systems: think about all of the different curves that you could come up with, and the possibilities.0708

You could maybe have a circle and a parabola as a system that never intersect: this is 0 solutions.0714

Perhaps you have an ellipse, and then another ellipse, like this: that gives you 1, 2, 3, 4 solutions.0722

You may have an ellipse, and then a circle, right here; and this gives you two solutions.0735

You could have, say, a parabola and an ellipse here that intersect at just one point.0747

So, you can see how that, with various combinations, you could get 0, 1, 2, 3, or 4 solutions.0754

And we just saw that demonstrated algebraically in the previous example--that we ended up with 4 solutions.0760

Systems of quadratic inequalities: these are systems of inequalities that have two inequalities, and at least one of these is a quadratic inequality.0769

So, for example, x2/9 + y2/4 <1; and then that is considered along with (x - 1)2 + (y - 3)2 < 9.0781

If you look at this, you have two quadratic equations; and you would recognize this0807

as an ellipse in standard form; and this gives the equation for a circle in standard form.0811

We have worked with some systems of inequalities before, and talked about how these can be solved by graphing.0818

And we are going to do the same thing here.0823

We are going to first graph the corresponding equation; and that will give us the boundary for the solution set.0825

Then, we will use a test point to determine on which side of the boundary the solution set lies.0832

We will do the same thing for the second equation that corresponds with the inequality:0840

find the boundary; use a test point; find where the solution set is.0845

And then, the overlap between those two solution sets gives you the solution for the system.0849

Illustrating this with this example: I am going to start out be graphing the boundary, and then finding the test point, for this first inequality.0855

So, the corresponding equation is going to be x2/9 + y2/4 = 1.0866

Since this is an ellipse, looking at it in this form, I know that it has a center at (0,0).0883

I know that A2 = 9, so a = 3; and since the larger term is associated0892

with the x2 term, I also know that this has a horizontal major axis.0901

The major axis is going to be oriented this way.0907

B2, right here, is equal to 4; therefore, B = 2.0911

So, that allows me to at least sketch this out: since A = 3, then I am going to have a vertex here, and the other vertex here, at (-3,0).0917

B is 2, so I am going to go up 2 and down 2; and this allows me to just sketch out the ellipse.0928

All right, the next step is to use a test point, because now I have a boundary.0941

And I actually need to be careful; I need to determine if the boundary should be dotted, or if it should be solid.0948

And looking here, I actually have a strict inequality.0956

What that tells me, recall, is that the boundary is not part of the solution set.0960

And the way we make that known is by using a dotted or dashed line.0966

Clarifying that, so we have the correct type of boundary...these breaks in the boundary indicate that this boundary is not part of the solution set.0976

OK, so I graph the boundary; I checked, and I had a strict inequality.0993

Now, I am going to take a test point; this is a convenient test point,0999

because the boundary has divided this into two regions: the region outside the ellipse, and the region inside it.1004

And I need to determine where the solution set is.1010

So, let's take the test point (0,0): I am going to put these values back into that original inequality.1013

0 is less than 1; this is true; therefore, this test point is part of the solution set.1033

So, the solution set must be inside this boundary.1040

So, I graphed my first boundary, and I determined where the solution set for this inequality is.1050

Looking at the second inequality: the corresponding equation will give me the boundary line for that.1055

Looking at this, this is written in standard form for a circle.1064

So, this is a circle with the center at (1,3), and r2 = 9, so the radius equals 3.1067

And this is a circle; and the center is at (1,3), so the center is right here.1078

And this is also a strict inequality, so I am going to use the dotted line when I draw this boundary line.1086

And the radius is 3: since the radius is 3, then that would be 4, 5, 6: that would go up to here.1099

And then, this is at 1, so then I would have the end right here, here, here.1106

OK, this is enough to get a rough sketch of the circle.1114

So, first I will just draw it as solid, and then go back and make it dashed, since this is a strict inequality.1133

All right, so this is a dashed line; it is a circle; it has a center at (1,3), and a radius of 3.1149

So again, this boundary line is not part of the solution set.1161

I am going to use another test point and insert it into that inequality to figure out where my solution set is.1165

So, for this, let's go ahead and use (1,1) as a test point, right there...the test point is going to equal (1,1).1173

So, that is going to give me 12/9 +...oops, I actually need to go back into that inequality...there is my equation1185

for the circle, (x - 1)2 + (y - 3)2 < 9; my test point is (1,1).1207

So, that is going to give me (1 - 1)2 + (1 - 3)2 < 9.1214

1 - 1 is 0, plus 1 - 3...that is -2, squared is less than 9; so 0...and this is -2 times -2 is 4...so 4 is less than 9.1222

Yes, this is true; therefore, this point is part of the solution set.1235

So, the solution set lies inside the circle.1240

The solution set for the system of inequalities is going to be this area here that is the overlap1251

between the solution set for the circle and the solution set for the inequality involving the ellipse.1260

And you can see, right here, the area where there is both red and black; it is that area of overlap.1266

So again, this is similar to methods we have used before, involving solving systems of inequalities,1270

where we graph the boundary line for one inequality; we graph the boundary line for the other inequality;1277

we use test points to find the solution sets for each inequality; and then, we determine the area of overlap1282

between those two solution sets, and that is the solution set for the system of inequalities.1289

And here, we are doing that with quadratic inequalities involving a circle and an ellipse.1295

Example 1: this is a linear-quadratic system: I have a linear equation here, and a quadratic equation here.1303

Recall that the easiest way to solve for these is by substitution.1310

Therefore, I am going to isolate x; I am going to rewrite this as x = y + 2.1314

Then, I go back to this first equation (this is from equation 2); I go back, and I make sure I substitute this into the other equation.1320

So, wherever there is an x, I am going to then put y + 2 instead; and I need to square that in this case.1331

This equals 36; I am going to write this out as y2 + 4x + 4 + y2 = 36.1339

OK, y2 + y2 gives me 2y2 + 4y (we are working with y here) + 4 = 36.1360

Now, it is a regular quadratic equation that I can just solve as I usually would.1372

And I see here that I have a common factor of 2; let's first go ahead and subtract 4 from both sides.1377

And this is going to give me 32; I am going to divide both sides by 2 to make this simpler, which is going to give me y2 + 2y = 16.1391

And now, I am going to solve it as I would any other quadratic equation: y2 + 2y - 16 = 0.1403

You could try this out, but it actually doesn't really factor out.1411

This is a situation where we have to go back to the quadratic formula, y = -b ±√(b2 - 4ac), divided by 2a.1415

It is a little more time-consuming, but it will get us the answer when factoring doesn't work.1431

So, let's rewrite this up here: y2 + 2y - 16 = 0.1436

y =...well, b is 2, so that is -2 ±√((-2)2 - 4 times a (is 1), and then c is negative 16), all divided by 2 times a, which is 1.1443

Simplifying: y = -2 ±√...-22 is 4; -4 times 1 is -4; -4 times -16 is + 641461

(those negatives, times each other, become positive) all over 2.1478

Therefore, y = -2±√...that is 64; -4 times 1 times -16 is 64, plus 4 is 68, divided by 2.1485

Now, you actually could determine that 68 is equal to 4 times 17; therefore, √68 equals the perfect square of 4, times 17.1507

So, I can then pull this 2 out by taking the square root of 4, which is 2; and it becomes 2√17.1523

So, I am going to do that over here, as well.1531

All right, you can factor out a 2, and those will cancel; I am just going to do that over here.1540

y = 2(-1) ± 1√17/2; that will cancel.1547

What you are left with -1 ± 1√17, divided by 2.1555

OK, so this is not an easy problem, because you ended up having to use the quadratic formula.1567

So, let's look at what we actually have: we have y = -1 + 1√17, and we also have y = -1 - √17.1576

So, we don't really need this 1 here.1591

The next step is to go back and substitute in: fortunately, we have an easy equation here1595

that we can substitute: x - y = 2, which is the same as x = y + 2.1601

I am going to work up here; and I am going to say, "OK, when y is -1 + √17, then x = this (that is y) + 2."1607

Therefore, x = 2 - 1 (is 1)...1 + √17.1624

That gives us an ordered pair: (1 + √17, -1 + √17).1632

That is an ordered pair; then I am going to take this second possibility, where y equals -1 - √17.1643

And I am going to substitute that into this, as well: x = -1 - √17 + 2.1656

This is going to give me x = -1 + 2...that is going to give me positive 1...so 1 - √17.1669

So, my second ordered pair is going to be (let's write it over here) x = 1 - √17, and then the y-value is -1 - √ 17.1677

So, it gets a little confusing with all of those signs; you want to make sure that you are careful1692

to check your work, and that you don't have any of the signs mixed up.1697

But what it came down to is using the quadratic equation to determine that y = -1 + √17, and y = -1 - √17.1701

Then, you take each possibility, starting with the first one; substitute for y in this equation,1711

which is the same as this (just rearranged); and substitute this to determine the corresponding value of x1719

(because remember, the solution is going to be an ordered pair, an x and a corresponding y value).1724

So, I did that for this first one; then I went and took the second one, repeated that process,1730

and got that, when y is -1 - √17, x is 1 - √17.1734

These two are an ordered pair, as well--that is where this came from; and these are the two solutions for this linear quadratic system of equations.1743

Example 2: we are going to solve...this is a quadratic-quadratic system, because I have a second-degree equation here and here.1755

I have two second-degree equations.1761

Recall that the best way to solve these is by elimination.1763

Looking at this, the first thing I can do to make this a little easier is actually to divide this first equation (equation 1 and equation 2) by 2,1768

because right now, it is kind of messy; it is bigger numbers than I need to be working with.1781

So, 2 goes into 8 four times; 2 goes into 2 once; and 2 goes into 40 twenty times.1786

All right, now I am going to go ahead and take equation 2, 4x2 + y2 = 100, and rewrite it down here.1797

And I see that, if I subtract this second equation, the y2 terms will drop out, because they already have the same coefficient.1805

I am going to rewrite this as 4x2 + y2 = 100; plus...I am going to make this a plus,1819

and then apply that as -4x; minus y2; and then I have a -20 here; I am just adding the opposite of each term.1828

So here, I end up with 4x2 - 4x (these are different, so I can't just combine them);1843

y2 - y2...the y2's drop out; 100 - 20 is 80.1857

All right, now I am going to divide both sides by a factor of 4 to simplify this.1863

This gives me x2 - 4 = 20; now it is just a quadratic equation that I need to solve.1867

x2 - 4x - 20 = 0: let's rewrite that up here.1875

x2 - 4x - 20 = 0: and let's hope that we can solve this by factoring,1882

x...the factors of 20 are 1 and 20, 2 and 10...actually, this is just x, because we got rid of that when we divided by 4...1893

this is just x; we divided by 4 to give us x2, x, and 20, which will make this even easier to factor.1905

OK, so since this is -1 for a coefficient, I want some factors that are close together, like 4 and 5.1912

And if I take 4 - 5, I am going to get -1; so I am going to use this combination.1920

This is going to give me (x - 5) (x + 4) = 0; just checking that, x times x is x2, and then outer terms is 4x,1926

minus the inner terms--that is 4x + -5x gives me -x; -5 times 4 is -20.1939

The zero product property tells me that, if x - 5 equals 0, or x + 4 = 0, this whole thing will equal 0.1948

So, solving for x will give me these two solutions: x = 5 and x = -4.1956

OK, I have the x-values: the next thing is to find the y-values.1962

So, I need to go back and substitute into one of these equations.1968

I am just going to select the top one; and what I need to do is determine what y will be when x is 5, and what y will be when x is -4.1973

So, starting out, I am letting x = -4, and then using this 8x + 2y2 = 40.1986

So, this is 8 times -4, plus 2y2, equals 40; that is -32 + 2y2 = 40.1997

Adding 32 to both sides gives me 2y2 = 72; dividing both sides by 2 gives me y2 = 36 (72/2 is 36).2008

Therefore, if I take the square root of 36, I get that y = ±6; so y = 6, and y = -6, when x = -4.2023

OK, let's write some ordered pairs up here as solutions.2035

When x = =4, you could equal 6; when x equals -4, another solution could be that y = -6.2039

I am going to repeat this process with x = 5, substituting into this equation.2047

This is going to give me 8(5) + 2y2 = 40.2053

That is 40 + 2y2 = 40; 40 - 40 is 0; 2y2 = 0; divide both sides by 2; I get y2 = 0.2058

The square root of 0 is 0, so I only get one solution for y here.2072

So, when x is 5, y is 0; I ended up with 3 solutions for this system of quadratic equations,2076

because it turned out that I had two values for x; one of these values of x, substituting in, yielded 2 values for y.2085

The second value for x yielded only one result for y; so I have three solutions here.2094

Another system: this is another quadratic-quadratic system, so I am going to use the approach of elimination.2104

Before I do anything, though, I can simplify these, because there are common factors.2110

I am going to take the first equation; and first, I will just rewrite it so that it is in more of a standard form.2115

So, I am going to subtract 4x2 from both sides; so it is -4x2 + 4y2 = -28.2124

OK, this is still equation 1; I am going to divide both sides by the common factor of 4.2134

That is going to give me -x2 + y2 = -7.2141

For the second equation, I have a common factor of 5; so I will divide both sides by 52147

to get x2 + y2... 125 divided by 5 is 25.2152

Now, all I have to do is add these two together, because I have a -1 for a coefficient here, and a 1 here; these cancel out.2159

y2 + y2 is 2y2; 25 - 7 is 18.2166

Just solve for y2: y2 = 18/2, so I divided both sides by 2; y2 is 9.2175

Therefore, y = ±3, by taking the square root of 9.2183

That means that y = 3, and y could also equal -3.2188

Now, I need to go back and substitute into one of these equations when y = 3, and figure out what x is.2196

Then, I need to see, when y is -3, what x is going to be.2204

Let's see, the easiest one to work with would be this: and I could go back in and use the top one,2211

but since I divided both sides by the same thing, I didn't really change this equation.2221

And it is a lot easier to work with this without these larger coefficients.2228

So, what I am going to do is say, "Let's let y equal 3."2232

And then, I am going to look at this: x2 + y2 =...actually, this first one; the first one is a smaller number, over here.2237

I am going to say -x2 + y2 = -7.2247

And let's rearrange this a bit, because we are looking for x.2251

So, let's move this y2 to the other side; and now I am stuck with a bunch of negatives.2254

And what I can do is just multiply both sides of the equation by -1, and that gives me x2 = y2 + 7.2261

Now, this is the one I am going to substitute back into: again, you could have taken either of these forms;2270

but I just took this and made it easier to work with, and solved for x2.2275

I am going to substitute 3 in wherever this is a y: so x2 = 32 + 7; therefore, x2 = 9 + 7.2281

So, x2 = 16, which means that x = ±4; so x could equal 4, and x could equal -4, when y is 3.2291

So, let's start our solutions up here.2307

When x is 4, y is 3; when x is -4, y is 3; that is two solutions so far.2311

That is when y is 3; but recall, y can also equal -3.2322

So, when y is -3, I am going to go back into that equation and substitute -3 and see what I get for x.2326

(-3)2 is 9; 9 + 7 is 16; so again, I get x = ±4; so x can equal 4, and x can equal -4.2340

But this time, y is -3; so this is two different solutions from what I had up here.2352

So, when x is 4, y is -3; that point is a solution for this system of equations.2358

When x is -4, y is -3; so there are four solutions.2366

If you graphed this out, you would find that these intersected at four points.2372

This is pretty complicated: the initial part actually wasn't that bad, but keeping track of all of the different solutions was a little bit challenging.2377

We started out with these two equations that I simplified by dividing the first by its common factor of 4, and the second by its common factor of 5.2385

Then, you added them together; the x2 terms dropped out, which allowed you to just solve for y.2393

I got two solutions for y: 3 and -3; I took each of those solutions, y = 3 and y = -3, and plugged them into this equation,2399

right here, to find corresponding values for x; that yielded these two solutions; and then y = -3,2411

when I found the x-values that corresponded to that...I got two more solutions, for a total of four solutions.2422

This time, we have a system of quadratic inequalities: I have x2 ≤ y, and then 4x2 + 4y2 < 36.2429

So, remember, we are going to solve these by graphing the corresponding equation to find the boundary line.2440

And then, we are going to use test points to find the solution sets for each, and then the solution set for the system.2446

Starting out with x2 ≤ y: this is x2 = y--that is the corresponding equation.2452

Finding a few points: x and y--recall that y = x2, so when x is 0, 02 gives you 0, so y is 0.2465

When x is 1, then we get 12; y is 1; when x is 2, we get...2 times 2 is 4; y is 4.2482

When x is -1, -1 times -1 is 1; when x is -2, -2 times -2 is 4; all right.2493

You recognize this as a parabola that opens upward; and it has its vertex, which is a minimum, right here at the center: (0,0).2503

I have a point here at (1,1), and a point at (-1,1); I have another point at (2,4), and then a point at (-2,4).2513

I could have also just graphed half, and used reflection symmetry to graph the other part of the parabola.2524

OK, now, this is less than or equal to; so I am actually using a solid line for my boundary,2531

because the boundary line is included as part of the solution set.2540

Now, I have the graph of the boundary; and I need to use a test point to determine where my solution set lies.2544

Does it lie inside the parabola, or outside?2552

And I am going to use the test point right here, (0,2); that would be convenient to work with.2555

x2 ≤ y; and I have a test point at (0,2); so when x is 0, that would give me 02 ≤ 2.2560

Is 0 less than or equal to 2? Yes, this is true; therefore, the test point is part of the solution set.2574

So, this solution set for this first inequality lies inside the parabola.2581

That is shaded in; and it is going to include the boundary.2594

This second equation that I have is 4x2 + 4y2 < 36.2600

So, let's rewrite this down here, and find the corresponding equation, 4x2 + 4y2 = 36.2608

And as you can see, there is a common factor of 4; so I am dividing both sides by that.2621

Looking at this, you can see that this describes a circle; and the circle has a center at (0,0).2630

r2 = 4; therefore...actually, that should be 9, because I divided both sides by 4; so 36/4 is 9: r2 = 9.2643

Therefore, r = 3; so this equation--all I did is divide both sides by 4, and I can see now that I have2658

an x2 and y2 term here on the same side of the equation, and they are equal to 9.2666

Therefore, it is a circle at center (0,0); they have the same coefficient; and r2 = 9, so the radius of the circle is 3.2674

So, the center of the circle is here; the radius is 3.2683

There would be a point there...the edge of the circle there...there...and there.2686

Then, you can fill in; now, this is a strict inequality, so I am just using a dashed, or dotted, line, as the border.2692

I am not going to make this a solid line, because this border is not part of the solution set.2706

The next step is to use a test point to determine, just for this inequality (this is the second inequality), where the solution set is.2711

And I am going to use the test point for the circle, right, test point (0,0), at the origin; that is easy to work with.2721

So, I go back up here to the original, 4x2 + 4y2 < 36.2730

So, 4 times 02 + 4 times 02 < 36.2736

This just gives me 0: 4 times 0 is 0; and this is 02...4 times 0 is also 0.2742

Is 0 less than 36? Yes, so again, I have a solution set that is on the inside of this boundary.2752

OK, so this is the solution set for the second inequality; this is the solution set for the first inequality.2766

And the solution set for the system is going to be right in here, where the blue and the black overlap.2771

This boundary is included in part of the solution set: the boundary of this circle is not.2780

So, we solved this by graphing the boundary for the first inequality, then the boundary for the second inequality,2786

and using test points to find the solution sets for the individual inequalities.2794

And then, the overlap between the two is the solution set for the entire system.2799

So today, we worked on systems involving quadratic equations--systems of equations where one was linear,2806