 This video will talk about systems of nonlinear equations. So let's think about how many solutions we would even have if we're talking about nonlinear. We've talked about linear and realized that most of the time we're going to have one. But if I have a line in a circle, how many would I have? Well if I have a circle and I have a line that looks like this, then that would mean that I have no solutions. And if I have a line that's going to be tangent to my circle, that means that we have one solution. And then my only other possibility is if my line goes through my circle, and that will give me two solutions. So I have zero, one, two, or possible solutions. What about a circle and a parabola? We could have any combination. We could have a line and a parabola, a line and a circle, a circle and a parabola, a circle and a circle, a parabola and a parabola. We could even have a log and a parabola, all kinds of things. But we're just taking a couple into consideration here. So if I have a parabola that looks like this, and then I have a circle below it, say, well that would give me zero solutions. Could I have one? Well if I have a circle with a parabola that the vertex is tangent to that circle, I could have one. Could I possibly have two? Well let's think, a circle. And what if this vertex of my parabola was inside the circle so that only my legs went through the circle? Well that would give me two. Could I possibly have three? Sure. Shift that up. Remember these parabolas could be flipped upside down too and do these same kinds of things. I'm just doing one direction here. In fact I'll switch it just to be different. What if I had my vertex right there and then I went through my circle like this and I'd have three solutions. I could do four if my vertex was outside of my circle but it went all the way through it. So that would give me four. And five I think would be a little difficult since I only have two legs of my parabola. So when we're solving these things, we have multiple solutions but we're going to use the same methods that we did for linear equations. So let's try one. This says substitution. So I see that this is a y equal. So I want to put this one into this equation. So here's my y. Minus one is equal to ln x plus 12. And some of you are beginning to see that these definitely look different than the linear ones. So what is y equal to? It's equal to ln x squared and then plus one. Well if I drop my parentheses because there's nothing on the outside to distribute I can see that I'm going to have positive one and negative one that cancel each other out. So I'm left with ln x squared on this side and I'm equal to ln of x plus 12 on this side. Well I have that uniqueness property. Same log, same base of my log. So I can just say that x squared is equal to x plus 12 and we have the quadratic. So we can say that we have x squared minus x minus 12 bringing both those terms to the left hand side. And I'm going to use my quadratic formula on it. Find out that when I have a is one, b is negative one and c is negative 12 and it tells me that my x's are four and negative three. So I have two x's but remember a system is ordered pairs. So I know I'm going to have x equal four but I've got to find out what y goes with it. And then I also know that x is going to be equal to negative three and I need to know the y that goes with it. So let's go back here. Remember we can plug it back in to any one of our original equations. I'm going to go back to that top one. It looks a little bit nicer for me. And I'm going to say that y is equal to ln and then I've got my four that I have to square plus one. I go to my calculator and ln of four squared close the parenthesis plus one tells me that I have 3.77 approximately. So now I have negative three that I want to work with and I'm going to do the same thing. I'm going to say y is equal to ln of negative three squared but it's inside plus one. So this will be nine when I squared ln of and then I need another parenthesis because my base on that square is actually negative three and then squared and then close my argument plus one. And I find out that y in this case is going to be 3.197 so we'll say two zero. And those are my two answers. So then it does say that we should check by graphing and I already put these into my calculator. What I did was I said this was y is equal to ln x plus 12 but then I add the one to the other side. So here's my two equations. ln x squared plus one. That's the top equation. Then ln x plus 12 and then plus the one when I brought it over and I get this graph. So I can see that I have two intersecting points right there. So second trace five. And notice I don't have anything in my y right here so I need to move my cursor until I do and then I can just press enter, enter, enter for the first one. And there's my four and my 3.77 and then I do second trace five again and if you get this trick it's helpful but if you don't it's okay. If I start moving trying to get closer to there you see I'm moving along this bunny looking graph. And you see that that's the ln x squared plus one. If I toggle down and say my down arrow now I have y2 and I'm looking at the other graph so I'll be on this log graph and I can move to the left but I get there a lot quicker because I am just on the simpler curve and I press enter three times and I get my negative three and my 3.197 or 3.2. Elimination remember we said we had to line up like things and have opposite coefficients on them. Well this is a linear and this is a circle so what are we going to do there? Well let's look at it. I can solve this thing for y so y is going to be equal to 5 plus 2x and I can solve this one for y squared is equal to 85 minus x squared. Now if I tried to get to y here I'd have a square root and that makes things really difficult but I see this y and I could make it y squared. If I square both sides I could make this left hand side become y squared. So squaring this side I get y squared. Squaring this side remember it's 5 plus 2x times 5 plus 2x so 5 times 5 will be 25 to take the first term and square it and then I'm going to have 5 times 2x twice so that would be 10 times 2 or plus 20x and then I'm going to have my 2x times my 2x which would give me plus 4x squared so now I can say underneath this bottom equation that my new equation that's going to combine with it is going to be y squared is equal to 25 and I'm going to rearrange this so that things line up so 4x squared and then plus 20x I'm just going to kind of switch these two terms so things line up nicely and if I can get them to be opposites then nice things happen so I'm going to multiply through by a negative in here and when I multiply through in a negative that becomes a negative y squared and a negative 25 and a negative 4x squared and a negative 20 and now I can add so y squared plus negative y squared is 0 85 plus a negative 25 is going to be 60 negative 4x squared and a negative 4x squared will be minus 5x squared and then minus 20x because there's nothing to combine with that one and you can see that we have a quadratic if it helps you we could say negative 5x squared minus 20x plus 60 it's a positive 60 there and I'm just going to use my quadratic formula so that I don't care that that's a negative 5x squared so negative 5 is my a negative 20 is my b and 60 is my c and I find out that x is equal to negative 6 and x is equal to 2 but once again I have to find out what my y is that goes with those so I'm going to go back up to this linear equation and say that y is equal to 5 plus 2 times my negative 6 well that's negative 12 plus 5 so y is going to be equal to negative 7 so up here I'm going to write my ordered pair of negative 6 for x and negative 7 for y and then I also have an ordered pair that starts with 2 for the x and then I'll work my problem so y is equal to 5 plus 2 times my 2 which is x so y is equal to 9 so I also have the ordered pair 2, 9