WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we are going to talk about solving quadratic systems of equations.
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In earlier lectures, we discussed talking about linear systems of equations, and used various methods to solve those: for example, substitution and elimination.
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And we are going to use some similar methods for quadratic systems, although these systems are more complex.
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There are two types of quadratic systems that we are going to be working with today.
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The first is a linear-quadratic system: **linear-quadratic systems** are a set of two equations in x and y,
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in which one of the equations is linear, and the other is quadratic.
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We will talk in a second about how to solve these; let's just stop at the definition right now.
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x² + 3x + y = 9; x = 2y: these two equations, considered together, would be a linear-quadratic system.
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I have a linear equation here and a quadratic equation here.
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And right here, this is a fairly simple linear equation to start out with.
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And we can use that to apply the method discussed above.
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So, given this type of system, these can be solved algebraically by isolating
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one of the variables in the linear equation, and then substituting it into the quadratic equation.
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OK, so conveniently, x is already isolated.
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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.
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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.
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So, I am rewriting the quadratic equation, and now substituting 2y in for x.
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That gives me 2y² + 3(2y) + y = 9.
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What I am left is a quadratic equation, which I can solve by the usual methods that we have learned.
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This is 2² is 4y² + 6y + y = 9.
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That gives me 4y² + 7y = 9, or 4y² + 7y - 9 = 0.
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And then, you could go on to solve this, using the quadratic formula.
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Once you find y, you can go ahead and substitute that value(s) in here, and then determine the corresponding value of x.
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In a linear quadratic system, you may have either 0, 1, or 2 solutions.
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And the easiest way to understand this is to think about it in terms of a graph.
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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...
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So, let's look at some possibilities: maybe I have a system that ends up graphing out like this.
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It gives me that line and this parabola.
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Well, the solutions are going to be where these two intersect; but in this system, they will never intersect.
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So, this system is going to have 0 solutions.
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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.
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And this is going to give me one solution.
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Perhaps I have an ellipse, and I have a line going through it like this.
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It intersects at two spots, so I would have two solutions.
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This helps to illustrate why you may have no solutions, one, or two solutions.
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More complicated systems involve quadratic-quadratic equations.
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This would be a quadratic-quadratic system in which there is a set of two quadratic equations in x and y.
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For example, x² + y² = 5 and 2x² - y = 4:
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you can see that you have two quadratic equations.
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You could try to use substitution, but it could get a little bit messy.
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So, often, elimination is the easiest way to use, so we solve them algebraically by elimination.
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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.
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And then, either add or subtract the equations so that a variable drops out.
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And here I want to get either x² or y² to drop out.
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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.
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So first, I am going to multiply the first equation by 2.
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2 times x² + y² = 5: and this is going to give me 2x² + 2y² = 10.
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So, I am going to then go ahead and write the other equation just below that: 2x² - y = 4.
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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.
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Now, what I am going to do is subtract this second equation from the first.
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Rewriting that: this gives me 2x² - 2x²; these terms drop out.
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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...
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actually, there are slightly different coefficients, so I can't add those; this is 2y² - a negative
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(that becomes a positive) y, and then 10 - 4 is equal to 6.
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OK, so 2x² - 2x²--those drop out; I have a 2y² down here.
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I have a negative and a negative, gives me a positive y; and then this becomes a -4; so 10 - 4 is 6.
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So, I end up with 2y² + y = 6 by using elimination.
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Therefore, I have gotten rid of one of the variables.
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From there, I have a regular quadratic equation that I can solve: that gives me 2y² + y - 6 = 0.
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I am going to try to solve that by factoring; and this is going to give me (2y + something) (y - something).
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So, let's look at 6: if we take factors of 6, those are 1 and 6, and 2 and 3.
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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.
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Therefore, I am going to take + 3, - 2.
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But I have to take this 2y into account; so let's go ahead and see if this works.
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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.
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Let's actually make this a negative and see what happens.
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This gives me 2y², and then this is 4y, minus 3y; that gives me a y; and then, -3 times 2 is -6.
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So, this actually factored out, and it worked.
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Using the zero product property, let's go back up here and get (2y - 3) (y + 2) = 0.
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2y - 3 = 0, or y + 2 = 0; if either of these equals 0, the whole quantity equals 0.
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Solving for y gives me 2y = 3 or y = 3/2; solving for y gives me y = -2.
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Now, what I need to do is go back and find the corresponding x-value, so I have coordinate pairs as my answers.
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All right, let's first work with y = -2.
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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.
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x² + (-2)² = 5: that is x² + 4 = 5, or x² = 1.
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Therefore, x = 1 and -1; now, that is when y is -2.
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So, my solutions are (1,-2) and (-1,-2)--I have two solutions there.
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Now, I need to repeat that process for y = 3/2; let y equal 3/2, and then substitute in here.
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This is going to give me x² + (3/2)² = 5, so x² + 9/4 = 5.
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So, x² = 5 - 9/4; that gives me x² =...a common denominator of 4; it would be 20/4 - 94, so x² = 11/4.
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Therefore, x = ±√11/4; now, looking over here, when y is 3/2, x could be √11/4, or it could be -√11/4.
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So, I have another two members to my solutions: I have that x could be (√11/4,3/2), and (-√11/4,3/2)
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Here, we have a situation where we actually have four solutions.
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And I solved this by elimination; and then I had to go back and take each of the solutions I ended up with,
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-2 and 3/2 for y, and find corresponding solutions for x.
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And each of those yielded two values for x, so I ended up with 4 solutions altogether.
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And I can also illustrate that point using graphing--of why you can end up with 0, 1, 2, 3, or 4 solutions.
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Quadratic-quadratic systems: think about all of the different curves that you could come up with, and the possibilities.
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You could maybe have a circle and a parabola as a system that never intersect: this is 0 solutions.
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Perhaps you have an ellipse, and then another ellipse, like this: that gives you 1, 2, 3, 4 solutions.
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You may have an ellipse, and then a circle, right here; and this gives you two solutions.
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You could have, say, a parabola and an ellipse here that intersect at just one point.
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So, you can see how that, with various combinations, you could get 0, 1, 2, 3, or 4 solutions.
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And we just saw that demonstrated algebraically in the previous example--that we ended up with 4 solutions.
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Systems of quadratic inequalities: these are systems of inequalities that have two inequalities, and at least one of these is a quadratic inequality.
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So, for example, x²/9 + y²/4 < 1; and then that is considered along with (x - 1)² + (y - 3)² < 9.
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If you look at this, you have two quadratic equations; and you would recognize this
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as an ellipse in standard form; and this gives the equation for a circle in standard form.
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We have worked with some systems of inequalities before, and talked about how these can be solved by graphing.
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And we are going to do the same thing here.
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We are going to first graph the corresponding equation; and that will give us the boundary for the solution set.
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Then, we will use a test point to determine on which side of the boundary the solution set lies.
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We will do the same thing for the second equation that corresponds with the inequality:
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find the boundary; use a test point; find where the solution set is.
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And then, the overlap between those two solution sets gives you the solution for the system.
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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.
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So, the corresponding equation is going to be x²/9 + y²/4 = 1.
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Since this is an ellipse, looking at it in this form, I know that it has a center at (0,0).
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I know that A² = 9, so a = 3; and since the larger term is associated
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with the x² term, I also know that this has a horizontal major axis.
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The major axis is going to be oriented this way.
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B², right here, is equal to 4; therefore, B = 2.
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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).
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B is 2, so I am going to go up 2 and down 2; and this allows me to just sketch out the ellipse.
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All right, the next step is to use a test point, because now I have a boundary.
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And I actually need to be careful; I need to determine if the boundary should be dotted, or if it should be solid.
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And looking here, I actually have a strict inequality.
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What that tells me, recall, is that the boundary is not part of the solution set.
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And the way we make that known is by using a dotted or dashed line.
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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.
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OK, so I graph the boundary; I checked, and I had a strict inequality.
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Now, I am going to take a test point; this is a convenient test point,
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because the boundary has divided this into two regions: the region outside the ellipse, and the region inside it.
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And I need to determine where the solution set is.
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So, let's take the test point (0,0): I am going to put these values back into that original inequality.
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0 is less than 1; this is true; therefore, this test point is part of the solution set.
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So, the solution set must be inside this boundary.
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So, I graphed my first boundary, and I determined where the solution set for this inequality is.
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Looking at the second inequality: the corresponding equation will give me the boundary line for that.
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Looking at this, this is written in standard form for a circle.
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So, this is a circle with the center at (1,3), and r² = 9, so the radius equals 3.
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And this is a circle; and the center is at (1,3), so the center is right here.
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And this is also a strict inequality, so I am going to use the dotted line when I draw this boundary line.
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And the radius is 3: since the radius is 3, then that would be 4, 5, 6: that would go up to here.
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And then, this is at 1, so then I would have the end right here, here, here.
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OK, this is enough to get a rough sketch of the circle.
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So, first I will just draw it as solid, and then go back and make it dashed, since this is a strict inequality.
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All right, so this is a dashed line; it is a circle; it has a center at (1,3), and a radius of 3.
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So again, this boundary line is not part of the solution set.
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I am going to use another test point and insert it into that inequality to figure out where my solution set is.
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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).
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So, that is going to give me 1²/9 +...oops, I actually need to go back into that inequality...there is my equation
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for the circle, (x - 1)² + (y - 3)² < 9; my test point is (1,1).
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So, that is going to give me (1 - 1)² + (1 - 3)² < 9.
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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.
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Yes, this is true; therefore, this point is part of the solution set.
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So, the solution set lies inside the circle.
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The solution set for the system of inequalities is going to be this area here that is the overlap
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between the solution set for the circle and the solution set for the inequality involving the ellipse.
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And you can see, right here, the area where there is both red and black; it is that area of overlap.
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So again, this is similar to methods we have used before, involving solving systems of inequalities,
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where we graph the boundary line for one inequality; we graph the boundary line for the other inequality;
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we use test points to find the solution sets for each inequality; and then, we determine the area of overlap
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between those two solution sets, and that is the solution set for the system of inequalities.
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And here, we are doing that with quadratic inequalities involving a circle and an ellipse.
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Example 1: this is a linear-quadratic system: I have a linear equation here, and a quadratic equation here.
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Recall that the easiest way to solve for these is by substitution.
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Therefore, I am going to isolate x; I am going to rewrite this as x = y + 2.
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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.
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So, wherever there is an x, I am going to then put y + 2 instead; and I need to square that in this case.
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This equals 36; I am going to write this out as y² + 4x + 4 + y² = 36.
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OK, y² + y² gives me 2y² + 4y (we are working with y here) + 4 = 36.
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Now, it is a regular quadratic equation that I can just solve as I usually would.
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And I see here that I have a common factor of 2; let's first go ahead and subtract 4 from both sides.
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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.
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And now, I am going to solve it as I would any other quadratic equation: y² + 2y - 16 = 0.
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You could try this out, but it actually doesn't really factor out.
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This is a situation where we have to go back to the quadratic formula, y = -b ±√(b² - 4ac), divided by 2a.
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It is a little more time-consuming, but it will get us the answer when factoring doesn't work.
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So, let's rewrite this up here: y² + 2y - 16 = 0.
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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.
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Simplifying: y = -2 ±√...-2² is 4; -4 times 1 is -4; -4 times -16 is + 64
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(those negatives, times each other, become positive) all over 2.
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Therefore, y = -2±√...that is 64; -4 times 1 times -16 is 64, plus 4 is 68, divided by 2.
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Now, you actually could determine that 68 is equal to 4 times 17; therefore, √68 equals the perfect square of 4, times 17.
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So, I can then pull this 2 out by taking the square root of 4, which is 2; and it becomes 2√17.
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So, I am going to do that over here, as well.
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All right, you can factor out a 2, and those will cancel; I am just going to do that over here.
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y = 2(-1) ± 1√17/2; that will cancel.
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What you are left with -1 ± 1√17, divided by 2.
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OK, so this is not an easy problem, because you ended up having to use the quadratic formula.
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So, let's look at what we actually have: we have y = -1 + 1√17, and we also have y = -1 - √17.
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So, we don't really need this 1 here.
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The next step is to go back and substitute in: fortunately, we have an easy equation here
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that we can substitute: x - y = 2, which is the same as x = y + 2.
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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."
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Therefore, x = 2 - 1 (is 1)...1 + √17.
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That gives us an ordered pair: (1 + √17, -1 + √17).
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That is an ordered pair; then I am going to take this second possibility, where y equals -1 - √17.
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And I am going to substitute that into this, as well: x = -1 - √17 + 2.
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This is going to give me x = -1 + 2...that is going to give me positive 1...so 1 - √17.
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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.
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So, it gets a little confusing with all of those signs; you want to make sure that you are careful
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to check your work, and that you don't have any of the signs mixed up.
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But what it came down to is using the quadratic equation to determine that y = -1 + √17, and y = -1 - √17.
00:28:31.200 --> 00:28:39.000
Then, you take each possibility, starting with the first one; substitute for y in this equation,
00:28:39.000 --> 00:28:44.500
which is the same as this (just rearranged); and substitute this to determine the corresponding value of x
00:28:44.500 --> 00:28:50.100
(because remember, the solution is going to be an ordered pair, an x and a corresponding y value).
00:28:50.100 --> 00:28:54.400
So, I did that for this first one; then I went and took the second one, repeated that process,
00:28:54.400 --> 00:29:02.800
and got that, when y is -1 - √17, x is 1 - √17.
00:29:02.800 --> 00:29:15.400
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.
00:29:15.400 --> 00:29:21.200
Example 2: we are going to solve...this is a quadratic-quadratic system, because I have a second-degree equation here and here.
00:29:21.200 --> 00:29:23.100
I have two second-degree equations.
00:29:23.100 --> 00:29:28.300
Recall that the best way to solve these is by elimination.
00:29:28.300 --> 00:29:40.900
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,
00:29:40.900 --> 00:29:46.400
because right now, it is kind of messy; it is bigger numbers than I need to be working with.
00:29:46.400 --> 00:29:57.000
So, 2 goes into 8 four times; 2 goes into 2 once; and 2 goes into 40 twenty times.
00:29:57.000 --> 00:30:05.600
All right, now I am going to go ahead and take equation 2, 4x² + y² = 100, and rewrite it down here.
00:30:05.600 --> 00:30:19.100
And I see that, if I subtract this second equation, the y² terms will drop out, because they already have the same coefficient.
00:30:19.100 --> 00:30:28.500
I am going to rewrite this as 4x² + y² = 100; plus...I am going to make this a plus,
00:30:28.500 --> 00:30:43.300
and then apply that as -4x; minus y²; and then I have a -20 here; I am just adding the opposite of each term.
00:30:43.300 --> 00:30:56.800
So here, I end up with 4x² - 4x (these are different, so I can't just combine them);
00:30:56.800 --> 00:31:02.800
y² - y²...the y²'s drop out; 100 - 20 is 80.
00:31:02.800 --> 00:31:07.000
All right, now I am going to divide both sides by a factor of 4 to simplify this.
00:31:07.000 --> 00:31:15.100
This gives me x² - 4 = 20; now it is just a quadratic equation that I need to solve.
00:31:15.100 --> 00:31:22.300
x² - 4x - 20 = 0: let's rewrite that up here.
00:31:22.300 --> 00:31:28.600
x² - 4x - 20 = 0: and let's hope that we can solve this by factoring,
00:31:28.600 --> 00:31:32.900
instead of having to go back to the messy quadratic formula.
00:31:32.900 --> 00:31:45.400
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...
00:31:45.400 --> 00:31:52.600
this is just x; we divided by 4 to give us x², x, and 20, which will make this even easier to factor.
00:31:52.600 --> 00:32:00.000
OK, so since this is -1 for a coefficient, I want some factors that are close together, like 4 and 5.
00:32:00.000 --> 00:32:06.600
And if I take 4 - 5, I am going to get -1; so I am going to use this combination.
00:32:06.600 --> 00:32:19.300
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,
00:32:19.300 --> 00:32:28.200
minus the inner terms--that is 4x + -5x gives me -x; -5 times 4 is -20.
00:32:28.200 --> 00:32:36.200
The zero product property tells me that, if x - 5 equals 0, or x + 4 = 0, this whole thing will equal 0.
00:32:36.200 --> 00:32:42.600
So, solving for x will give me these two solutions: x = 5 and x = -4.
00:32:42.600 --> 00:32:48.200
OK, I have the x-values: the next thing is to find the y-values.
00:32:48.200 --> 00:32:53.300
So, I need to go back and substitute into one of these equations.
00:32:53.300 --> 00:33:06.600
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.
00:33:06.600 --> 00:33:17.100
So, starting out, I am letting x = -4, and then using this 8x + 2y² = 40.
00:33:17.100 --> 00:33:27.800
So, this is 8 times -4, plus 2y², equals 40; that is -32 + 2y² = 40.
00:33:27.800 --> 00:33:42.800
Adding 32 to both sides gives me 2y² = 72; dividing both sides by 2 gives me y² = 36 (72/2 is 36).
00:33:42.800 --> 00:33:55.200
Therefore, if I take the square root of 36, I get that y = ±6; so y = 6, and y = -6, when x = -4.
00:33:55.200 --> 00:33:59.200
OK, let's write some ordered pairs up here as solutions.
00:33:59.200 --> 00:34:07.500
When x = =4, you could equal 6; when x equals -4, another solution could be that y = -6.
00:34:07.500 --> 00:34:13.300
I am going to repeat this process with x = 5, substituting into this equation.
00:34:13.300 --> 00:34:18.600
This is going to give me 8(5) + 2y² = 40.
00:34:18.600 --> 00:34:32.300
That is 40 + 2y² = 40; 40 - 40 is 0; 2y² = 0; divide both sides by 2; I get y² = 0.
00:34:32.300 --> 00:34:36.600
The square root of 0 is 0, so I only get one solution for y here.
00:34:36.600 --> 00:34:45.100
So, when x is 5, y is 0; I ended up with 3 solutions for this system of quadratic equations,
00:34:45.100 --> 00:34:53.800
because it turned out that I had two values for x; one of these values of x, substituting in, yielded 2 values for y.
00:34:53.800 --> 00:35:03.700
The second value for x yielded only one result for y; so I have three solutions here.
00:35:03.700 --> 00:35:10.100
Another system: this is another quadratic-quadratic system, so I am going to use the approach of elimination.
00:35:10.100 --> 00:35:15.600
Before I do anything, though, I can simplify these, because there are common factors.
00:35:15.600 --> 00:35:23.800
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.
00:35:23.800 --> 00:35:34.000
So, I am going to subtract 4x² from both sides; so it is -4x² + 4y² = -28.
00:35:34.000 --> 00:35:40.900
OK, this is still equation 1; I am going to divide both sides by the common factor of 4.
00:35:40.900 --> 00:35:46.900
That is going to give me -x² + y² = -7.
00:35:46.900 --> 00:35:52.100
For the second equation, I have a common factor of 5; so I will divide both sides by 5
00:35:52.100 --> 00:35:59.200
to get x² + y²... 125 divided by 5 is 25.
00:35:59.200 --> 00:36:06.300
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.
00:36:06.300 --> 00:36:15.200
y² + y² is 2y²; 25 - 7 is 18.
00:36:15.200 --> 00:36:23.500
Just solve for y²: y² = 18/2, so I divided both sides by 2; y² is 9.
00:36:23.500 --> 00:36:28.400
Therefore, y = ±3, by taking the square root of 9.
00:36:28.400 --> 00:36:35.900
That means that y = 3, and y could also equal -3.
00:36:35.900 --> 00:36:44.400
Now, I need to go back and substitute into one of these equations when y = 3, and figure out what x is.
00:36:44.400 --> 00:36:51.100
Then, I need to see, when y is -3, what x is going to be.
00:36:51.100 --> 00:37:01.200
Let's see, the easiest one to work with would be this: and I could go back in and use the top one,
00:37:01.200 --> 00:37:07.700
but since I divided both sides by the same thing, I didn't really change this equation.
00:37:07.700 --> 00:37:11.900
And it is a lot easier to work with this without these larger coefficients.
00:37:11.900 --> 00:37:17.100
So, what I am going to do is say, "Let's let y equal 3."
00:37:17.100 --> 00:37:26.800
And then, I am going to look at this: x² + y² =...actually, this first one; the first one is a smaller number, over here.
00:37:26.800 --> 00:37:31.500
I am going to say -x² + y² = -7.
00:37:31.500 --> 00:37:34.500
And let's rearrange this a bit, because we are looking for x.
00:37:34.500 --> 00:37:41.200
So, let's move this y² to the other side; and now I am stuck with a bunch of negatives.
00:37:41.200 --> 00:37:49.900
And what I can do is just multiply both sides of the equation by -1, and that gives me x² = y² + 7.
00:37:49.900 --> 00:37:55.500
Now, this is the one I am going to substitute back into: again, you could have taken either of these forms;
00:37:55.500 --> 00:38:01.100
but I just took this and made it easier to work with, and solved for x².
00:38:01.100 --> 00:38:11.300
I am going to substitute 3 in wherever this is a y: so x² = 3² + 7; therefore, x² = 9 + 7.
00:38:11.300 --> 00:38:27.100
So, x² = 16, which means that x = ±4; so x could equal 4, and x could equal -4, when y is 3.
00:38:27.100 --> 00:38:30.900
So, let's start our solutions up here.
00:38:30.900 --> 00:38:42.500
When x is 4, y is 3; when x is -4, y is 3; that is two solutions so far.
00:38:42.500 --> 00:38:46.400
That is when y is 3; but recall, y can also equal -3.
00:38:46.400 --> 00:38:59.700
So, when y is -3, I am going to go back into that equation and substitute -3 and see what I get for x.
00:38:59.700 --> 00:39:12.300
(-3)² is 9; 9 + 7 is 16; so again, I get x = ±4; so x can equal 4, and x can equal -4.
00:39:12.300 --> 00:39:18.300
But this time, y is -3; so this is two different solutions from what I had up here.
00:39:18.300 --> 00:39:26.600
So, when x is 4, y is -3; that point is a solution for this system of equations.
00:39:26.600 --> 00:39:32.400
When x is -4, y is -3; so there are four solutions.
00:39:32.400 --> 00:39:36.900
If you graphed this out, you would find that these intersected at four points.
00:39:36.900 --> 00:39:44.800
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.
00:39:44.800 --> 00:39:52.700
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.
00:39:52.700 --> 00:39:59.000
Then, you added them together; the x² terms dropped out, which allowed you to just solve for y.
00:39:59.000 --> 00:40:11.100
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,
00:40:11.100 --> 00:40:22.000
right here, to find corresponding values for x; that yielded these two solutions; and then y = -3,
00:40:22.000 --> 00:40:29.400
when I found the x-values that corresponded to that...I got two more solutions, for a total of four solutions.
00:40:29.400 --> 00:40:40.600
This time, we have a system of quadratic inequalities: I have x² ≤ y, and then 4x² + 4y² < 36.
00:40:40.600 --> 00:40:46.300
So, remember, we are going to solve these by graphing the corresponding equation to find the boundary line.
00:40:46.300 --> 00:40:52.500
And then, we are going to use test points to find the solution sets for each, and then the solution set for the system.
00:40:52.500 --> 00:41:04.900
Starting out with x² ≤ y: this is x² = y--that is the corresponding equation.
00:41:04.900 --> 00:41:22.500
Finding a few points: x and y--recall that y = x², so when x is 0, 0² gives you 0, so y is 0.
00:41:22.500 --> 00:41:33.300
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.
00:41:33.300 --> 00:41:43.600
When x is -1, -1 times -1 is 1; when x is -2, -2 times -2 is 4; all right.
00:41:43.600 --> 00:41:53.000
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).
00:41:53.000 --> 00:42:04.100
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).
00:42:04.100 --> 00:42:11.500
I could have also just graphed half, and used reflection symmetry to graph the other part of the parabola.
00:42:11.500 --> 00:42:20.100
OK, now, this is less than or equal to; so I am actually using a solid line for my boundary,
00:42:20.100 --> 00:42:23.900
because the boundary line is included as part of the solution set.
00:42:23.900 --> 00:42:31.900
Now, I have the graph of the boundary; and I need to use a test point to determine where my solution set lies.
00:42:31.900 --> 00:42:35.000
Does it lie inside the parabola, or outside?
00:42:35.000 --> 00:42:40.200
And I am going to use the test point right here, (0,2); that would be convenient to work with.
00:42:40.200 --> 00:42:53.900
x² ≤ y; and I have a test point at (0,2); so when x is 0, that would give me 0² ≤ 2.
00:42:53.900 --> 00:43:00.900
Is 0 less than or equal to 2? Yes, this is true; therefore, the test point is part of the solution set.
00:43:00.900 --> 00:43:14.600
So, this solution set for this first inequality lies inside the parabola.
00:43:14.600 --> 00:43:20.300
That is shaded in; and it is going to include the boundary.
00:43:20.300 --> 00:43:28.600
This second equation that I have is 4x² + 4y² < 36.
00:43:28.600 --> 00:43:40.900
So, let's rewrite this down here, and find the corresponding equation, 4x² + 4y² = 36.
00:43:40.900 --> 00:43:50.300
And as you can see, there is a common factor of 4; so I am dividing both sides by that.
00:43:50.300 --> 00:44:03.400
Looking at this, you can see that this describes a circle; and the circle has a center at (0,0).
00:44:03.400 --> 00:44:18.300
r² = 4; therefore...actually, that should be 9, because I divided both sides by 4; so 36/4 is 9: r² = 9.
00:44:18.300 --> 00:44:26.000
Therefore, r = 3; so this equation--all I did is divide both sides by 4, and I can see now that I have
00:44:26.000 --> 00:44:34.600
an x² and y² term here on the same side of the equation, and they are equal to 9.
00:44:34.600 --> 00:44:42.900
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.
00:44:42.900 --> 00:44:45.900
So, the center of the circle is here; the radius is 3.
00:44:45.900 --> 00:44:52.500
There would be a point there...the edge of the circle there...there...and there.
00:44:52.500 --> 00:45:06.400
Then, you can fill in; now, this is a strict inequality, so I am just using a dashed, or dotted, line, as the border.
00:45:06.400 --> 00:45:11.000
I am not going to make this a solid line, because this border is not part of the solution set.
00:45:11.000 --> 00:45:20.700
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.
00:45:20.700 --> 00:45:30.500
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.
00:45:30.500 --> 00:45:36.200
So, I go back up here to the original, 4x² + 4y² < 36.
00:45:36.200 --> 00:45:42.600
So, 4 times 0² + 4 times 0² < 36.
00:45:42.600 --> 00:45:52.100
This just gives me 0: 4 times 0 is 0; and this is 0²...4 times 0 is also 0.
00:45:52.100 --> 00:46:06.000
Is 0 less than 36? Yes, so again, I have a solution set that is on the inside of this boundary.
00:46:06.000 --> 00:46:11.300
OK, so this is the solution set for the second inequality; this is the solution set for the first inequality.
00:46:11.300 --> 00:46:20.100
And the solution set for the system is going to be right in here, where the blue and the black overlap.
00:46:20.100 --> 00:46:26.100
This boundary is included in part of the solution set: the boundary of this circle is not.
00:46:26.100 --> 00:46:34.500
So, we solved this by graphing the boundary for the first inequality, then the boundary for the second inequality,
00:46:34.500 --> 00:46:39.100
and using test points to find the solution sets for the individual inequalities.
00:46:39.100 --> 00:46:46.500
And then, the overlap between the two is the solution set for the entire system.
00:46:46.500 --> 00:46:54.000
So today, we worked on systems involving quadratic equations--systems of equations where one was linear,
00:46:54.000 --> 00:47:01.800
one was quadratic, or both were quadratic, as well as some systems of quadratic inequalities.
00:47:01.800 --> 00:47:04.000
And that concludes today's lesson; thanks for visiting Educator.com.