Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Systems of nonlinear equations in two variables. We need to solve for one variable using elimination or substitution method. We also must make sure to check our answers. We may get what's called extraneous roots, roots that appear to work but don't really. So let's look at some examples. The first example is going to be a line and a parabola intersecting. If we have a line and we have a parabola, we know that a line and a parabola could intersect in two spots. A line and a parabola might intersect in one spot, and a line and a parabola might never intersect. So in this problem, we're going to stick the Y equal X ^2 + 1 into the first equation. This is substitution method. So X minus the quantity X ^2 + 1 equal -1. If I distribute the negative out, we get negative X ^2 - 1, and I'm going to take this -1 to the other side by adding one the -1 and +1 cancel. If we factor out in X, we get X times the quantity 1 -, X equaling 0, so X = 0 or X equal 1. We need to put that back into one of the two originals to find our Y. So Y equal 1 when X is 0 and y = 2 when X is 1. You must check these Does 0 - 1 equal -1 and does 1 equal 0 ^2 + 1 and the answer is yes for both those here. Does 1 - 2 equal -1 and does 2 equal 1 ^2 + 1 and the answer is yes for both of those. So this is an example similar to this first picture, and we could graph it more accurately if we wanted, but there will be two intersection points in that example. This next one is a line and a circle and a line and a circle could intersect twice, could intersect once, and it could not intersect at all. That's why we need to do the checks. So we're going to solve one or the other of these equations. I'm going to take this top one and solve for Y. So Y is going to be negative two X + 4. So when we put it into the second equation, we get X + 1 ^2 plus instead of Y. I'm going to put a negative two X + 4 - 2 ^2 equaling 4. Now we're going to foil this out so we can combine our like terms. This one would have reduced to -2 X +2 quantity squared. So first out, our inner last, I'm going to get four X ^2 minus eight X + 4 if we subtract the four over so we have 0 on one side. Now we're going to combine our like terms so we get 5X squareds -6 X and a positive one that factors into five X -, 1 and X - 1. So X is going to be 1/5 or X is going to be 1. To find our Y, we need to substitute it back in -2 * 1/5 + 4 and -2 * 1 + 4. So this is going to be negative 2/5 + 20 fifths or 18 fifths. So this is a .1 fifth 18 fifths. This other one Y is going to equal -2 + 4 or Y equal 2. So the .12 when we check those they both work because 2 * 1/5 + 18 fifths. That's a times in there equal 4. This ends up being 20 fifths, which is 4, and then 1/5 + 1 ^2 + 18 fifths -2 ^2 equal 4. This is 6 fifths squared. This is 8 fifths squared, so 36 + 64 / 25 100 / 25 is 4. So that first point definitely works. We can do the same kind of check for the second point. So does 2 * 1 + 2 equal 4? And that one's pretty easy to see. Does 1 + 1 ^2 + 2 - 2 ^2 equal 4? That one also is pretty easy to see. This next one we're going to do via elimination, and so we're going to take this top equation. Actually, let's take the bottom equation, it doesn't matter, and multiply it by 4. If I multiply this bottom equation by 4, I get twelve X ^2 + 4 Y squared equaling 124. If we leave the top equation the way it is, then when we add them we get 13 X squared. the Y squareds are going to cancel, giving us 100 and 17117 / 13. That 913 * 9 seven carry two. Yeah, so X ^2 is going to equal 9. So X is going to equal positive or negative sqrt 9. So X is going to be positive, negative or three, three. Once we get the X, we need to figure out the Y. We could come back here and we could multiply the top equation by -3 So negative three X ^2 + 12 Y squared equaling 21. Three X ^2 + y ^2 equal 31. 13 Y squared would equal 52 Y squared is 4 Y equals positive. Negative sqrt 4 Y equal positive -2. We need to check all these, and when we do, we're actually going to find we have 4 points. We're going to have 3-2. We're going to have three -2 we're going to have -3 two and -3 negative 2. This was the intersection of a hyperbola and an ellipse. So an ellipse and a hyperbola, and it's going to have four points of intersection 1234. This could have had 4321 or zero depending on how the ellipse and the hyperbola what the equations were. Our next example, if we multiplied the top one through by a -1, we'd get negative X ^2 -, y ^2 equaling 5. I'm going to leave the bottom one alone, although I am going to foil out that second, the Y -, 8 ^2 the second term, so that when I add these, the X squareds cancel. The Y squareds are also going to cancel and we're going to get -16 Y plus 64 equaling 46. Oops, I forgot to multiply by a negative up there. So that's going to equal 36 negative all the way through. Left this alone, distributed it out -16 Y plus 6441, so the X ^2's canceled, the Y ^2's canceled -16 Y plus 64 equaling 36 so -16 Y. If we take the 64 crossed, we take 64 -, 36, which is going to give us -28 So then if we divide by 1628 sixteenths, those are both divisible by 4, so 7 fourths. If we then substitute back in to get our X, we're going to have X ^2 + 49 sixteenths. I'm going to change 5 into sixteenths, so 80 sixteenths. If we subtract 49 sixteenths, we're going to get 31 sixteenths. So X is going to be positive negative sqrt 31 / 4. So we're going to have sqrt 31 / 4, 7 fourths and negative sqrt 31 / 4 seven fourths. We need to make sure to check them. Thank you and have a wonderful day.