so we're going to consider solving systems of non-linear equations and we can solve these systems by either using substitution or elimination sometimes you have to do something a little bit clever or creative to solve the system and I'll work through various examples for you to pick up some techniques so first example we have the system expert minus y squared equals 21 and X plus y equals 7 now the elimination method will not help us here because notice in the first equation both the variables are squared and in the second equation they're raised to the first power so I can't do anything to cancel out the variables if I were to try to add the equations together so that means I'm stuck using substitution let's solve for X in the second equation I could solve for y instead it doesn't matter and then I'm gonna substitute this expression in for X into the first equation so I'm gonna have 7 minus y squared minus y squared equals 21 then I'm gonna multiply that expression out so I'll have 49 minus 14 wife minus 14y plus y squared minus y squared equals 21 and then notice Y squared cancels out here so I just have 49 minus 14y equals 21 so negative 14 y is negative 28 which means y equals 2 cool now I just need X but that's pretty easy because I know X is 7 minus y so X is gonna be 7 minus 2 which is 5 so my solution set contains the ordered pair 5 2 and we don't classify these systems as independent dependent consistent all that stuff that was for the previous section only studied linear systems but in this case just the solution set is sufficient and we just study Tonique so the first equation you should recognize as the graph of a hyperbola and the second one is a line so you could try graphing them and verify that that's the solution ok let's look at this next example example B we have 2y squared plus 3x y minus 3 XY plus 6y plus 2x plus 4 is 0 okay what a mess and then we have 2x minus 3y plus 4 equals 0 well it's gonna be super-difficult if i were to try to solve for X or Y in the first equation so we're not doing that we're gonna solve for either X or Y in the second equation and then use substitution so let's just solve for X since that term is positive so I'll have 2x equals 3y minus 4 which means x equals 3 halves y minus 2 now I'm going to substitute this in for X into the first equation so right there and right there so now I'm gonna have 2 y squared minus 3 times 3 halves y minus 2 times y plus 6 y plus 2 times 3 halves Y minus 2 plus 4 is 0 okay not the cutest equation you've ever run into but it is doable okay so let's start distributing we have 2y squared minus 9 halves well this Y let's distribute that while we're at it too right we're in precalc now we can do this all at once - 9 halves y squared then this is gonna be plus 6y then I have another plus 6y plus 3 y minus 4 plus 4 is 0 so let's see what happens the fours are gone Bam Bam and then these are like terms so that's gonna give me a negative five-halves Y squared and then I have 6y 6 y 3y so that's 15 y is 0 now how do I solve from here this is a quadratic equation there's only two terms of so I'm just gonna factor out a Y and then I have negative five-halves Y plus 15 is 0 so either Y is 0 or negative five-halves Y is negative 15 which means Y would equal let's see negative 15 times 2 is 30 divided by 5 is 6 so I get Y is either 0 or 6 now what do I do with that this hasn't happened before that's because these are nonlinear well I have to substitute each of these values of Y in to figure out what the corresponding value of X says from right here okay so 4y equals zero I'd get X is three-halves times 0 minus 2 which would give me negative 2 and if y is 6 then I get X is three-halves times 6 minus 2 that's going to be 18 divided by 2 which is 9 minus 2 which is 7 so your solution contains the ordered pairs negative 2 0 as well as sub in comma 6 okay what a thing of beauty alright moving on we have y squared minus x squared plus 4 equals 0 and then 2 x squared plus 3 y squared equals 6 so again we can use substitution or you could do elimination in this case if you wanted so let me show you elimination just to spice things up the first equation I'm gonna rewrite it it is negative x squared plus y squared equals negative 4 and notice I have a 2 x squared plus 3 y squared equals 6 so I'm gonna multiply this by 2 and now it's gonna become negative 2 x squared plus 2 y squared equals negative 8 and then I'm gonna add to it the equation below 2 x squared plus 3 y squared equals 6 so if I add this together the X Squared's cancel out and then I got 5 y squared equals negative 2 now wait a minute can I have Y squared equaling a negative number no that's not possible not at least I won't get a real solution and don't worry we're not finding complex or imaginary solutions in this case so this this problem is done there's no real solution so you could write the empty set you could write a literal empty set I've seen that before okay we disappointed I'm so sorry let's do another one 2 over x squared minus 3 over Y squared plus 1 is 0 and then 6 over x squared plus 7 over y minus 7 over y squared plus 2 equals 0 ok so a few different ways you can go about doing this one we can try using elimination again how about we multiply this top equation by negative 3 so then it's gonna become negative 6 over x squared plus 9 over y squared minus 3 equals 0 and then I have 6 over x squared minus 7 over y squared plus 2 equals 0 so now I'm gonna add them together and then notice the 6 over x squared and negative 6 over x squared cancel out so I'm left with 2 over y squared minus 1 equals 0 which means 2 over Y squared equals 1 so Y squared equals 2 so y equals plus or minus rad - ok that's good now I just need to figure out what X is gonna be so you can use either equation let's just stick with the first one so I know 2 over x squared minus 3 over now Y squared regardless if it's plus or minus Rab - it's gonna be 2 plus 1 is zero so negative 3 halves plus 1 that's negative 1/2 and then if I add it over to the other side I'm going to get 2 over x squared equals the positives on 1/2 and then cross multiplying I get x squared equals 4 so x equals plus or minus 2 and I get this solution if Y was either positive rad 2 or negative rad 2 so that means I have four solutions in total right did you count them up so X is positive 2 if Y is positive rad 2 or X could be negative rad - I mean negative 2 if Y was negative rad 2 and similarly X could be positive 2 if Y was negative rad 2 or negative 2 before I was positive rad 2 did I get them off negative negative positive negative negative positive positive positive yep that's all of them so let's put them in our solution set there's four of them I'm gonna write them out separately so there's no confusion as to the number of solutions that you have and where's the last one negative 2 rad - thanks okay good technically you would be able to write plus or minus 2 and plus or minus rad - since there were all 4 combinations but if you only had the positives together and the negatives together or some other weird combo if you didn't have all 4 possible pairings then you couldn't do this notation but in this case because we had all four options as part of our solution set that's fine okay good last example here we go so we have x squared minus XY minus 2 y squared is 0 and then XY plus X plus 6 equals 0 so I don't know maybe the first thing that jumps out at you is oh I want to add them together cancel out that X Y so let's see what would happen if you did them together so you would get x squared then these would cancel then you'd have plus X minus 2 y squared plus 0 and then from there what in the world do you do you have two variables that's not a good situation so then maybe you go back to that second equation and you go oh well maybe I can solve for one of the variables so I know that XY plus X would be negative 6 which means XY equals negative 6 minus X which means that y equals negative 6 minus x over X maybe I'll just substitute this guy in for my Y okay let's see what happens so we have x squared plus X minus 2 times negative 6 minus x over x squared plus 6 is 0 and then you're a brave math student sure you keep going so you have x squared plus X minus 2 times this is gonna be 36 plus 12x plus x squared over x squared plus 6 is 0 and then you know you want to clear the fractions out so you multiply everything by an x squared and you realize things are getting out of control so you have Baxter four plus X cubed Oh now minus 72 minus 24 X minus two x squared plus six x squared is zero and you're SuperDuper stuck so what's the other thing to do maybe something you wouldn't have noticed or you factor and solve the first equation yes it occurred so the first equation was x squared minus XY minus 2 y squared is 0 so you can factor this can't you you're gonna have X minus 2 y times X plus y is 0 oh my goodness which means that you get X in terms of Y X is either 2 Y or X is negative Y but this isn't bad because remember your second equation was X y plus X plus 6 is 0 so you can just substitute these values in to the other equation so say I'm substituting in x equals 2y then I'd have to Y times y plus 2 y plus 6 is 0 so then I get 2y squared plus 2y plus 6 is 0 let's divide this by 2 so I'd get Y squared plus y plus 3 is 0 that doesn't factor so we're gonna try the quadratic formula so opposite of B plus or minus square root b squared minus 4ac and right away maybe you notice oh the discriminant it's gonna be negative which means there's no real solution not no real solution overall just no real solution if x equals 2y so that's not possible okay so now I'm gonna try it if x equals negative Y so then I'd have negative Y times y plus negative y plus 6 is 0 then you get negative y squared minus y plus 6 is 0 let's multiply through by a negative so it's easier to factor so we'll have Y squared plus y minus 6 is 0 which is gonna give us Y plus 3 times y minus two is zero so then I'm left with y equals negative three so nice or y equals two and remember X is equal to negative Y right so that means if Y is negative three X is three and if Y is 2 X is negative 2 so those are only two solutions 3 negative 3 and negative 2 2 okay so maybe you wouldn't think to factor the first equation at first glance right but what happens with these problems is you'll go down some paths and when you get really stuck and it looks like cruel and unusual punishment then you think to yourself maybe there was some trick I missed from the very beginning that would save me a lot of heartache ok so one of those tricks could be factoring and making a clever substitution so that concludes the video hope you enjoyed it don't forget to subscribe comment and leave me feedback as to what sort of videos you'd like to see next