A “Simple” Equation That Isn’t

Added: 2026-05-10

[00:00:00]Hello everyone.

[00:00:02]In this video we're going to be solving a very interesting, a very unique system of equations.

[00:00:09]We have x + y + xy = 195 and we have another equation, a quintic equation.

[00:00:17]So we're going to solve this system which means we're going to find xy values that satisfy both of these equations. So how do you do something like this, right? That's pretty complicated, isn't it?

[00:00:29]So let's go ahead and approach this from two different angles. First of all, I want to go ahead and use the first equation to isolate one of the variables. And I'd like to do that with y so I can use substitution.

[00:00:45]Let's go ahead and factor out y here and write it as y * 1 + x + x = 195.

[00:00:54]So these two terms have y in them so I factor the y out. And next step is going to be subtracting x from both sides so that we can isolate y. To isolate y, we need to divide by 1 + x on both sides, right?

[00:01:11]And that'll give us y in terms of x which is 195 - x / 1 + x. Awesome. Now this is something we can definitely use and substitution as you know is a really good method, right?

[00:01:27]Hopefully.

[00:01:29]For systems. One of the methods at least, right? We sometimes use elimination, sometimes we just take the equations, add them up, sometimes we use inequalities, there is quite a few different ways to approach it and you can learn by uh solving different types of systems.

[00:01:45]And I have quite a few systems on my channel, go ahead and check out the playlists. I also have another channel that focuses on complex numbers which is a + bi. Also make sure to check it out and let me know what you think. Okay?

[00:02:01]Great. So let's go ahead and see how we can use this equation along with the second one. That's my cat by the way. I don't think he's hungry and he has food but he likes attention like any other cat I guess, right? Anyways. So let's go ahead and proceed with the problem. Now if you go ahead and plug this in here what happens? Let's go ahead and do that. 195 - x over 1 + x to the fifth power + 3 * the same expression to the third power and then this is going to equal x to the fifth + 3 x to the third.

[00:02:43]Well, it kind of looks like this is a quintic equation, right? The problem is it isn't. Why? Because when you expand this you're going to get 195 - x to the fifth power or at least when you start simplifying and then you're going to have other terms. And of course to get rid of the fraction you need to multiply both sides by 1 + x to the fifth power which is going to give you actually a decic equation which means you're going to have x to the 10th power. I don't think you want to solve that. There are no formulas for a decic equation, there are no formulas for a nonic octic heptic, hectic, whatever.

[00:03:23]Unfortunately, after the quartic we have no general formulas. And there's no quintic formula either.

[00:03:29]But I know some people are going to say, "Oh, we use certain radical." Oh, no, no, no.

[00:03:33]Just accept the fact that there are no quintic formulas.

[00:03:37]That's it. Period. Now anyways, that's a different story. But let's go back here.

[00:03:42]Can you solve this equation? Are you serious? I don't think so. I mean you're serious but I don't think you want to solve an equation that looks like this.

[00:03:51]So you're probably thinking at this point, hopefully I could make you think that there must be an easier way to do it. And guess what?

[00:04:01]That happens with my videos most of the time because the first method is kind of like painful, no pain, no gain style and the second method is usually more elegant. And I'm I'm hoping that you're going to agree with me on this.

[00:04:15]Sometimes you get to choose but this time I think we'll have an agreement, right? Consensus. Great. Let's go ahead and rewrite this system and see if we can approach this problem a little differently, okay? Now what was the original problem? x + y + xy was equal to something. I can't remember the right hand side because I modified the problem after I wrote it. 195. By the way, you can consider this a homemade problem because I thought about the idea but anyone can come up with a problem like this especially after I show you the strategy that I'm going to be using which is really, really cool, okay? Now there's also uh two ways to go about it so I could probably call this uh first one 2a and the second one, you know what is coming up, right? Hopefully. Okay. So 2a goes like this.

[00:05:08]We have an equation and here's what we can do. We can actually subtract everything on the right hand side. But I'm only working with the quintic because that's more complicated and I want to simplify it as much as possible, okay? So here's how it goes. We subtract x to the fifth from y to the fifth and from 3y cubed we subtract 3x. Why did we subtract like that? Because we want to factor this. Make sense?

[00:05:34]Now we're going to factor by grouping and of course this is the difference of uh fifth power so it factors like this and hopefully you know the formula. Uh you put the minus sign which is the same as this and then the second term or second factor is uh starts with the uh fourth power of y because you do need to get y to the fifth from here so you need y to the fourth and then every term you're going to have a plus sign everywhere because we started with a minus sign. The other factor has a minus sign. And then uh the powers of y are going to go down just like the binomial theorem except we don't have the binomial uh coefficients, okay?

[00:06:15]So y cubed is going to be accompanied by y and we're going to have y squared x squared and then y x cubed and finally x to the fourth power. That takes care of the sum, I mean the difference of two fifth powers. What about the second part? We're going to have put a plus sign here and if you just factor out the three here, you're going to realize, uh-oh, this is difference of two cubes so I can go right kind of write it as y - x + y squared + y x + x squared and the whole thing is equal to zero. Now what I want you to notice is the presence of a common factor which is good because we can factor that out.

[00:06:55]Let's do that and proceed with the rest.

[00:06:57]y to the fourth + y cubed x + y squared x squared + y x cubed + x to the fourth and then I'll continue with this. Of course there's a three, let's not forget that. 3y squared, 3yx + 3x squared. Uh-oh, such a gigantic factor, right?

[00:07:16]Great. Well, not so great because we have two factors, we have to consider both. But the second piece will actually show you what's going on. It's nice and I think much cleaner. And you got to decide which one you like better one obviously. But by setting this equal to zero I'm really getting something super duper nice. What is that? It is y = x.

[00:07:37]Wow. The second equation, even though it looked very complicated it gave us a really nice result, right?

[00:07:46]So if y is equal to x, just go back to the first equation which is x + y + xy is equal to 195 and let's go ahead and replace y with x, okay? If you do that we're going to get 2x from here and x squared from here.

[00:08:03]That gives us 195. Now this problem can be solved in so many ways so we have kind of like 2a1, 2a2. You we don't really get into all these branches but let me tell you something. One way to approach this problem is factor out the x and think about if there is any nice natural solutions. Then uh we're looking for two factors that differ by two and their product is now 195. So you think about how 195 can be factored, right? 195 is 39 * 5 and 39 is 13 * 3 and 3 * 5 is 15. So we can actually write this as 13 * 15 which means x can be 13 and x + 2 can be 15.

[00:08:49]So that works. x = 13 is definitely a solution. But the million dollar question is, is that the only solution?

[00:08:57]No. Because the factors are 13 and 15, they can also be -15 * -13. We negate and switch them because this is the smaller factor so it can be -15 as well.

[00:09:10]So there are two solutions for x. This is kind of like 2b one, right? Or 2a1. Now 2a2 is going to look like this. You can take this and notice that if I add one to both sides, I'll be completing the square, right? Add one, add one, this becomes x + 1 quantity squared and 196 happens to be 14 squared. And from here from difference of two squares or using the absolute value, it doesn't matter which one you like, x + 1 can be 14 or x + 1 can be -14 because there are two numbers whose square equals 14 squared, 14 and -14. Does that make sense? And x = 13 and x = -15 as before, those are going to be the two solutions to this equation or system, right? Now, we can easily find the Y values by substitution and guess what? Since Y and X are equal, if X is 13, Y is 13 and vice versa, okay? You get the idea. Now, this is only the first branch.

[00:10:14]Actually, not the first branch. The first branch of the second solution. So, it's a kind of like 2A. Now is the time for 2B and allow me to make the joke. 2B or not 2B, that's the problem. So, here's how it goes. We have the following system. So, let me rewrite that. We start with X + Y + XY is equal to 195 and then the second equation is going to be uh what was that?

[00:10:46]Y to the fifth plus 3Y cubed. Okay, Y to the fifth plus 3Y cubed is equal to X to the fifth plus 3X cubed. Now, here's what we're going to do.

[00:10:57]We're going to look at the system and take a good look and we're going to focus on the second equation again. I'm going to call this one, I'm going to call this second equation. Now, what is so special about this equation, right? Well, if I define a function f of t and define it as t to the fifth power plus 3t cubed. Of course, that's not a coincidence. We're trying to mirror these, right? And what happens with this function? Let's differentiate. If you differentiate this function, you get 5t to the fourth plus 9t squared. Now, unless t is zero, this function cannot be zero or negative. In other words, with the exception of t equals zero, of course, I'm going to have to exclude that and you know that is not a solution.

[00:11:48]Uh we can safely say that this function is greater than zero uh when t does not equal zero, right?

[00:11:56]What does that supposed to mean?

[00:11:57]It means f prime is positive, which implies that f is increasing, okay?

[00:12:06]And what does that mean when a function is increasing? That means it's always increasing and if it's always increasing, it can only have a single solution when you intersect it with the horizontal line. In other words, this is what I'm trying to say. Uh my function is a bijection. So, this means f of y, this means f of x. So, we have this, f of y equals f of x and this implies y equals x. Why? Because f is bijective.

[00:12:37]Make sense? Okay, this has to uh this has to result in this uh how do you say that? Okay, anyways, this has to be the conclusion, okay? That's what I was trying to say. So, again, plug it in and you're going to get the same exact same solutions.

[00:12:53]All right?

[00:12:55]And this brings us to the end of this video. Thanks for watching. I hope you enjoyed it. Please let me know. Don't forget to comment, like, and subscribe.

[00:13:02]I'll see you next time with another video.

[00:13:05]Until then, be safe, take care. Don't forget to check out CyberMath, the short channel, and A + BI and bye-bye.