hi guys so we are in the final section of chapter 9 and today we're gonna be talking about solving nonlinear systems of equations so you'll see a bunch of different ways that we can solve them so here we go so the first technique that we could use to solve a linear system is by graphing so when you see something like this the easiest way to go about doing it so just go ahead and put that into desmos on your computer and what we're looking for is we're looking for the the point of intersection here so if we have y equals 2x squared plus 5x minus 1 and y equals x minus 3 let's take a look and put that into decimals so we have 2x squared plus 5x minus 1 and then we have X minus 3 so what we're looking for here is we're looking for the point of intersection and you can see that if we look here these two graphs intersect right here at that one particular point which is at negative 1 comma negative 4 so our only solution to this particular set of equations would be negative 1 negative 4 so what we're looking for is we're looking for the point of intersection so if we look here our only point of intersection which we just saw was right there so our solution here is negative 1 comma negative 4 what we're looking for we're looking for a system as we're looking for what the x value is equal to which in this case is negative 1 and what the Y value is equal to and which in this case is negative 4 so you can either write it that way or you can write it as a point doesn't matter well take a look at one or two more of these so let's look at the second one which is y equals x squared plus 4x minus 4 and 2x minus 5 so I'm gonna get rid of these x squared plus 4x minus 4 and then our second equation was 2x minus 5 so from looking at looking for the point of intersection there's only one point again here this one and that happens to be at negative 1 comma negative 7 so my only solution to this particular system would be at negative 1 comma negative 7 now both of these only have one solution but it's possible that you can have more than one solution ok so let's take a look at say our third one which is 3x minus 15 and x squared minus 2x minus 7 so 3x minus 15 and x squared minus 2x minus 7 3x minus 15 and x squared minus 2x minus 7 so we're looking here guys at this particular graph you'll notice that they don't intersect anywhere okay so in this particular system of equations there's no solution because these graphs are not intersecting anywhere on Earth in our coordinate plane so this would have no solution so what you're looking for is you're looking for that point of intersection sometimes there's two solutions so say I change this a little bit maybe I make this 3x plus maybe something like this notice here we have two solutions we have a solution here or we have a point of intersection in the solution here so you can have no solution you can of one solution you could have two solutions it all depending on where those point of intersections are occur let's keep going and here the graphs yeah we've another way that we could solve these and this is by substitution remember best it's a substitution we had systems what we're doing is we're substituting what this Y is in over here so we substitute that in so we're gonna end up getting we're end up getting negative 2 X plus 3 is equal to x squared plus X minus 1 now let's solve this so I have to set it equal to 0 add the 2x to both sides and I'm going to subtract the 3 to both sides so now I'm going to end up getting 0 its equal to 0 is equal to x squared plus 3x minus 4 so now you can think to yourself okay I have a quadratic let me think of the different ways I can solve a quadratic easiest way is to factor I'm gonna see if I can factor this if not I'm going to use quadratic formula or go on a different technique so let's use a quad let's use factoring because we can find something that's gonna multiply to negative 4 and add 2030 that would be 4 and negative 1 so this will factor 2x plus 4 times X minus 1 is equal to 0 using the zero product property set both of these equal to 0 add 1 and you're gonna get X equal to negative 4 and X equal to negative this is gonna be subtract that X minus 1 equals 0 this isn't me X is equal to a positive 1 now you're not done though because your solution to all of these is a point when you're asked to solve the system the solution is a point so we need to find the Y values to both of these X values so how are we gonna do that we're going to plug in our X values back in so we know that Y is equal to negative 2 times negative 4 plus 3 so Y is gonna be equal to 11 so I'm gonna point it negative 4 comma 11 I'm gonna plug in a positive 1 is equal to negative two times positive 1 plus 3 so y is equal to positive 1 so I'll have another solution at 1 comma 1 and those would be my two solutions now I couldn't solve this by graphing if I wanted to so if I want to go back to desmos and i'm x squared plus X minus 1 so I have x squared I can graph so let's graph that x squared y squared X minus 1 and then our other function was negative 2x plus 3 they get up to X plus 3 look we have a solution that at negative 4 11 which we got and our other solution at 1 1 which is the two points of intersection so obviously desmos and graphing is the easiest technique here but you're not always gonna have the luxury of using desmos you could use a graphing calculator as well but obviously you might not have a graphing calculator so it's important to use some of these other techniques as well so that's how substitution would work but again you can always just check your answers using decimals now we also have another technique which is elimination so remember with elimination you want to eliminate one of the variables so I know if I take a look at both of these I can just subtract these two equations and the Y's are gonna cancel last and then what I'm going to be left with is x squared minus nothing is going to be x squared negative 3 minus a negative 3 is gonna cancel out as well and then negative 2 minus 8 is going to be a positive 6 so we end up getting x squared plus 6 is equal to 0 and then I can go ahead and I can solve that ok unit square roots so I'm gonna get x squared is equal to negative 6 and let's take a look at what happens here so let's graph these y equals x squared minus 3x minus 2 so x squared minus 3x minus 2 and let's take a look at negative 3x minus 8 and if we take a look at both of these you'll notice that this is another one of those cases where they're never crossing which should make sense to you because when we try to solve by elimination we end up getting x squared equals a negative which we know that that can't happen because x squared can never be equal to a negative so we end up getting no real solutions here and we could see that by by both looking at the graph because there's no intersection and by trying to solve this using elimination okay so we're using any one of these techniques remember substitution you're taking this value and you're plugging it in there elimination we're trying to eliminate one of the variables by easiest way subtracting both of these and we can get rid of the Y's every time but again we could use graphing to help us in any of these cases here we're asked to approximate the solutions of the system to the nearest thousandth so again we're gonna solve this by graphing so we're gonna take a look at desmos again so 1/2 x squared plus 3 and 3 to the X 1/2 x squared three and three to the X you three to the X so obviously these graphs are a little bit different but we're looking for that point of intersection and here's that point of intersection you'll notice that it's a decimal so we want to round that and it said round to the nearest hundredth so that's two decimal places so our solution is gonna be one point one nine comma three point seven one so our solution which we see right here one point one nine comma three point seven one or test thousands rather so we round to three decimal places so one point one nine four comma three point seven one three is your solution let's look at example five here which is very similar so when you have two things equal to each other you're just gonna graph them separately so I'm gonna treat them as if it's a system or you're graphing them as two separate equations and we're gonna see what they end up equaling so we're gonna graph negative two times 4 to the X plus 3 so negative 2 times 4 to the X plus 3 and then we're gonna graph we're gonna graph our right side which is 0.5 x squared minus 2x so 0.5 x squared minus 2x this we can see has two solutions they're crossing at two separate points here so one of my solutions is that negative one comma two point five and the other solution at point four six eight comma negative 0.8 to seven so even though they're set equal to each other we can just graph them separately and we'll get our two points of intersection there okay so our two solutions in this case again we're negative one comma two point five and then point four six eight comma negative 0.8 to silver this is showing you on the graphing calculator happen how you can go about doing that all right so that's all for this lesson make sure the key thing here know the different techniques so you know how to solve by substitution you know how to solve by elimination you know how to solve using desmos and graphing which will be the easiest technique for you use one key thing when you are using des modes make sure that you're specifying that you're using that and you were looking for the intersection points remember you can have one point of intersection you could have no intersection points in which case there's no real solution or you could have two points of intersection like we do here good luck guys and if you have any questions come to the Google mean see ya