 In this video, we provide the solution to question number 24 from the practice final exam for math 1050. We're given a system of equations. They're non-linear equations though. Well, at least the second one is right. The first equation is X plus Y equals five. That is the graph of a line, but the second one Y equals three minus X squared. That's the graph of a parabola. We have a quadratic function going on there. So we have to be a little more careful when we solve systems involving non-linear equations. Substitution or elimination can still be appropriate here as long as we use them as such. Notice that with the second equation, it's already set up as Y equals. So it seems like it's screaming at me to do substitution. I just substitute this Y into the other Y, right? So that gives us that X plus, well, instead of Y, we get three minus X squared. This is equal to five. We now have a quadratic equation. Maybe we can solve this quadratic equation. Let's move everyone to the right-hand side here. So we end up with zero is equal to, negative X squared when we move to the right becomes a positive X squared. That's actually why I've moved everything to the right. I like my leading coefficient to be positive. Some of them are optimists there. The X then moves over and becomes a negative X, and then you get a five minus three, which is a positive two, like so. So then we try to factor this one. Factors of two that have to be negative one, your really only option would be negative one and negative two, which gives us a negative three. So it turns out that we don't have any rational solutions there. So we'll just go to the quadratic formula. X equals negative B. So you get a negative negative one, plus one there. Plus or minus the square root of B squared, which is one, minus four AC. So four times one times two is an eight, all over two A. Simplifying this, you'll notice you get one plus or minus the square root of negative seven over two, or better yet, one plus or minus I root seven over two. This right here, these are not real numbers, right? Because you can see the imaginary part right here. So as we're trying to graph these things, we have this line that's been graphed. We have this parabola that's trying to be graphed. We actually get something like this, if we drew this a little bit better. But the point is, because X turned out to be a non-real value, it's actually suggesting to us that these two graphs, the line and the parabola, actually have no intersection, which is what the system of equations is trying to find, that's the solution there. So it turns out that we actually have no solution, because we're only looking for real solutions here. There's no solution to this non-linear system, for which we can just stop there, say there's no solution, the system's inconsistent because the parabola does not intersect the line.