### Solving Quadratic Systems

- In a linear-quadratic system, use substitution to solve.
- In a quadratic-quadratic system, use elimination 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.

### Solving Quadratic Systems

x

^{2}+ y

^{2}+ 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
- x
^{2}+ y^{2}+ 5x − y − 6 = 0 - (y + 2)
^{2}+ y^{2}+ 5(y + 2) − y − 6 = 0 - y
^{2}+ 4y + 4 + y^{2}+ 5y + 10 − y − 6 = 0 - Combine like terms
- 2y
^{2}+ 8y + 8 = 0 - Divide entire equation by 2
- y
^{2}+ 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

− 2y

^{2}+ 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
- − 2y
^{2}+ 6x + 3y + 153 = 0 - − 2( − 2x + 3)
^{2}+ 6x + 3( − 2x + 3) + 153 = 0 - − 2(4x
^{2}− 12x + 9) + 6x − 6x + 9 + 153 = 0 - − 8x
^{2}+ 24x − 18 + 6x − 6x + 9 + 153 = 0 - Combine like terms
- − 8x
^{2}+ 24x + 144 = 0 - Divide entire equation by − 8
- x
^{2}− 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

2x

^{2}− 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
- 2x
^{2}− x + 18y − 28 = 0 - 2(2y + 4)
^{2}− (2y + 4) + 18y − 28 = 0 - 2(4y
^{2}+ 16y + 16) − 2y − 4 + 18y − 28 = 0 - 8y
^{2}+ 32y + 32 − 2y − 4 + 18y − 28 = 0 - Combine like terms
- 8y
^{2}+ 48y = 0 - Divide entire equation by 8
- y
^{2}+ 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

− 2x

^{2}− 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
- − 2x
^{2}− 37x + 6y − 188 = 0 - − 2( − 2y − 4)
^{2}− 37( − 2y − 4) + 6y − 188 = 0 - − 2(4y
^{2}+ 16y + 16) + 74y + 148 + 6y − 188 = 0 - − 8y
^{2}− 32y − 32 + 74y + 148 + 6y − 188 = 0 - Combine like terms
- − 8y
^{2}+ 48y − 72 = 0 - Divide entire equation by − 8
- y
^{2}− 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

x

^{2}+ y

^{2}− 6x + 6y − 19 = 0

x

^{2}+ y

^{2}− 6x − 7y + 20 = 0

- This is a quadratic \ quadratic system. Solve by elimination.
- Multiply the first equation by − 1 to create an equivalent system.
- -1* (x
^{2}+ y^{2}− 6x + 6y − 19 = 0) - − x
^{2}− y^{2}+ 6x − 6y + 19 = 0 - Add the system of equation
- (− x
^{2}− y^{2}+ 6x − 6y + 19 = 0) + (x^{2}+ y^{2}− 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.
- x
^{2}+ y^{2}− 6x − 7y + 20 = 0 - x
^{2}+ (3)^{2}− 6x − 7(3) + 20 = 0 - x
^{2}+ 9 − 6x − 21 + 20 = 0 - x
^{2}− 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

− x

^{2}+ y

^{2}+ 14x − 13y − 8 = 0

x

^{2}+ y

^{2}− 14x − 13y + 88 = 0

- This is a quadratic \ quadratic system. Solve by elimination.
- Add the system of equatios
- (− x
^{2}+ y^{2}+ 14x − 13y − 8 = 0) + (x^{2}+ y^{2}− 14x − 13y + 88 = 0) - 2y
^{2}− 26y + 80 = 0 - Solve for y, factor but divide entire equation by 2 first
- y
^{2}− 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

x^{2}+ y^{2}− 14x − 13y + 88 = 0

x^{2}+ (5)^{2}− 14x − 13(5) + 88 = 0

x^{2}− 14x + 48 = 0

(x − 6)(x − 8) = 0

x = 6 and x = 8 - y=8

x^{2}+ y^{2}− 14x − 13y + 88 = 0

x^{2}+ 8^{2}− 14x − 13(8) + 88 = 0

x^{2}− 14x + 48 = 0

(x − 6)(x − 8) = 0

x = 6 and x = 8

8x

^{2}+ 4y

^{2}− 9x − 8y − 95 = 0

10x

^{2}− 4y

^{2}+ 45x + 8y + 41 = 0

- This is a quadratic \ quadratic system. Solve by elimination.
- Add the system of equatios
- (8x
^{2}+ 4y^{2}− 9x − 8y − 95 = 0) + (10x^{2}− 4y^{2}+ 45x + 8y + 41 = 0) - 18x
^{2}+ 36x − 54 = 0 - Solve for x, factor but divide entire equation by 18 first
- x
^{2}+ 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

8x^{2}+ 4y^{2}− 9x − 8y − 95 = 0

8( − 3)^{2}+ 4y^{2}− 9( − 3) − 8y − 95 = 0

4y^{2}− 8y + 4 = 0

y^{2}− 2y + 1 = 0

(y − 1)(y − 1) = 0

y = 1 - x=1 8x
^{2}+ 4y^{2}− 9x − 8y − 95 = 0

8(1)^{2}+ 4y^{2}− 9(1) − 8y − 95 = 0

4y^{2}− 8y − 96 = 0

y^{2}− 2y − 24 = 0

(y − 6)(y + 4) = 0

y = 6 and y = − 4

2x

^{2}− 24x − y + 69 = 0

− 2x

^{2}+ 5y

^{2}+ 24x − 9y − 144 = 0

- This is a quadratic\ quadratic system. Solve by elimination.
- Add the system of equatios
- (2x
^{2}− 24x − y + 69 = 0) + ( − 2x^{2}+ 5y^{2}+ 24x − 9y − 144 = 0) - 5y
^{2}− 10y − 75 = 0 - Solve for y, factor but divide entire equation by 5 first
- y
^{2}− 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 2x
^{2}− 24x − y + 69 = 0

2x^{2}− 24x − ( − 3) + 69 = 0

2x^{2}− 24x + 72 = 0

x^{2}− 12x + 36 = 0

(x − 6)(x − 6) = 0

x = 6 - y=5

2x^{2}− 24x − y + 69 = 0

2x^{2}− 24x − (5) + 69 = 0

2x^{2}− 24x + 64 = 0

x^{2}− 12x + 32 = 0

(x − 4)(x − 8) = 0 x = 4 and x = 8

x

^{2}+ y

^{2}< 9

(x − 4)

^{2}+ y

^{2}< 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)

x^{2}+ y^{2}< 9

0^{2}+ 0^{2}< 9

0 < 9

True - Circle Two: Test (5,0)

(x − 4)^{2}+ y^{2}< 16

(5 − 4)^{2}+ 0^{2}< 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

(x − 5)

^{2}+ y

^{2}≤ 36

(x − 5)

^{2}+ y

^{2}≥ 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}+ y^{2}≤ 36

0^{2}+ 0^{2}< 9

0 < 9

True - Circle Two: Test (5,0)

(x − 5)^{2}+ y^{2}≥ 9

0^{2}+ 0^{2}≥ 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.

Answer

### Solving Quadratic 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
- Linear Quadratic Systems 0:22
- Example: Linear Quadratic System
- Solutions 2:49
- Graphs of Possible Solutions
- Quadratic Quadratic System 4:10
- Example: Elimination
- Solutions 11:39
- Example: 0, 1, 2, 3, 4 Solutions
- Systems of Quadratic Inequalities 12:48
- Example: Quadratic Inequality
- 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

### Algebra 2

### Transcription: Solving Quadratic Systems

*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

*x ^{2} + 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 isolating*0072

*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 2y ^{2} + 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 2 ^{2} is 4y^{2} + 6y + y = 9.*0128

*That gives me 4y ^{2} + 7y = 9, or 4y^{2} + 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

*More complicated systems involve quadratic-quadratic equations.*0251

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

*For example, x ^{2} + y^{2} = 5 and 2x^{2} - 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 x ^{2} or y^{2} to drop out.*0298

*Well, I have an x ^{2} 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 x ^{2} + y^{2} = 5: and this is going to give me 2x^{2} + 2y^{2} = 10.*0317

*So, I am going to then go ahead and write the other equation just below that: 2x ^{2} - 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 2x ^{2} - 2x^{2}; these terms drop out.*0366

*The x ^{2} 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 2y ^{2} - a negative*0394

*(that becomes a positive) y, and then 10 - 4 is equal to 6.*0402

*OK, so 2x ^{2} - 2x^{2}--those drop out; I have a 2y^{2} 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 2y ^{2} + 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 2y ^{2} + 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 2y ^{2}, 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 2y ^{2}, 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 y ^{2} and see what I get.*0558

*x ^{2} + (-2)^{2} = 5: that is x^{2} + 4 = 5, or x^{2} = 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 x ^{2} + (3/2)^{2} = 5, so x^{2} + 9/4 = 5.*0610

*So, x ^{2} = 5 - 9/4; that gives me x^{2} =...a common denominator of 4; it would be 20/4 - 94, so x^{2} = 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, x ^{2}/9 + y^{2}/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 this*0807

*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 x ^{2}/9 + y^{2}/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 A ^{2} = 9, so a = 3; and since the larger term is associated*0892

*with the x ^{2} term, I also know that this has a horizontal major axis.*0901

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

*B ^{2}, 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 r ^{2} = 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 1 ^{2}/9 +...oops, I actually need to go back into that inequality...there is my equation*1185

*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 overlap*1251

*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 overlap*1282

*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 y ^{2} + 4x + 4 + y^{2} = 36.*1339

*OK, y ^{2} + y^{2} gives me 2y^{2} + 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 y ^{2} + 2y = 16.*1391

*And now, I am going to solve it as I would any other quadratic equation: y ^{2} + 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 ±√(b ^{2} - 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: y ^{2} + 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 ±√...-2 ^{2} is 4; -4 times 1 is -4; -4 times -16 is + 64*1461

*(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 here*1595

*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 careful*1692

*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 x*1719

*(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, 4x ^{2} + y^{2} = 100, and rewrite it down here.*1797

*And I see that, if I subtract this second equation, the y ^{2} terms will drop out, because they already have the same coefficient.*1805

*I am going to rewrite this as 4x ^{2} + y^{2} = 100; plus...I am going to make this a plus,*1819

*and then apply that as -4x; minus y ^{2}; and then I have a -20 here; I am just adding the opposite of each term.*1828

*So here, I end up with 4x ^{2} - 4x (these are different, so I can't just combine them);*1843

*y ^{2} - y^{2}...the y^{2}'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 x ^{2} - 4 = 20; now it is just a quadratic equation that I need to solve.*1867

*x ^{2} - 4x - 20 = 0: let's rewrite that up here.*1875

*x ^{2} - 4x - 20 = 0: and let's hope that we can solve this by factoring,*1882

*instead of having to go back to the messy quadratic formula.*1888

*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 x ^{2}, 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 x ^{2}, 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 + 2y ^{2} = 40.*1986

*So, this is 8 times -4, plus 2y ^{2}, equals 40; that is -32 + 2y^{2} = 40.*1997

*Adding 32 to both sides gives me 2y ^{2} = 72; dividing both sides by 2 gives me y^{2} = 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) + 2y ^{2} = 40.*2053

*That is 40 + 2y ^{2} = 40; 40 - 40 is 0; 2y^{2} = 0; divide both sides by 2; I get y^{2} = 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 4x ^{2} from both sides; so it is -4x^{2} + 4y^{2} = -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 -x ^{2} + y^{2} = -7.*2141

*For the second equation, I have a common factor of 5; so I will divide both sides by 5*2147

*to get x ^{2} + y^{2}... 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

*y ^{2} + y^{2} is 2y^{2}; 25 - 7 is 18.*2166

*Just solve for y ^{2}: y^{2} = 18/2, so I divided both sides by 2; y^{2} 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: x ^{2} + y^{2} =...actually, this first one; the first one is a smaller number, over here.*2237

*I am going to say -x ^{2} + y^{2} = -7.*2247

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

*So, let's move this y ^{2} 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 x ^{2} = y^{2} + 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 x ^{2}.*2275

*I am going to substitute 3 in wherever this is a y: so x ^{2} = 3^{2} + 7; therefore, x^{2} = 9 + 7.*2281

*So, x ^{2} = 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 x ^{2} 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 x ^{2} ≤ y, and then 4x^{2} + 4y^{2} < 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 x ^{2} ≤ y: this is x^{2} = y--that is the corresponding equation.*2452

*Finding a few points: x and y--recall that y = x ^{2}, so when x is 0, 0^{2} gives you 0, so y is 0.*2465

*When x is 1, then we get 1 ^{2}; 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

*x ^{2} ≤ y; and I have a test point at (0,2); so when x is 0, that would give me 0^{2} ≤ 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 4x ^{2} + 4y^{2} < 36.*2600

*So, let's rewrite this down here, and find the corresponding equation, 4x ^{2} + 4y^{2} = 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

*r ^{2} = 4; therefore...actually, that should be 9, because I divided both sides by 4; so 36/4 is 9: r^{2} = 9.*2643

*Therefore, r = 3; so this equation--all I did is divide both sides by 4, and I can see now that I have*2658

*an x ^{2} and y^{2} 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 r ^{2} = 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, 4x ^{2} + 4y^{2} < 36.*2730

*So, 4 times 0 ^{2} + 4 times 0^{2} < 36.*2736

*This just gives me 0: 4 times 0 is 0; and this is 0 ^{2}...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

*one was quadratic, or both were quadratic, as well as some systems of quadratic inequalities.*2814

*And that concludes today's lesson; thanks for visiting Educator.com.*2822

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