Austrian Olympiad System of Equations

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Күн бұрын

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@rcnayak_58 Жыл бұрын

Dear Sir, this time I am trying to solve this problem perhaps in a simpler way. This type of equation is known as symmetrical equation, where exchanging x with y values does not alter the properties of the equations, if f(x) = f(y) then x = y. In this case x = y is a solution. Since x = y, let us replace all y as x in either of the two equations. Say the 1st equation: Now we have (x -1)(x²+6) = x (x² + 1). That is x³ - x² + 6x - 6 = x³ + x. This becomes x² - 5x + 6 = 0. The roots of x are 2 and 3. As they are symmetric, they can be arranged in 4 different ways. That is, (x , y) = (2 , 2) and (3, 3), (2 , 3) and (3, 2).

@SanePerson1 Жыл бұрын

I took this approach as well and, like you, I solved the equation for x and got x = 2 and 3. The problem is that you (and I) derived the equation to be solved, x² - 5x + 6 = 0, by setting x = y in either of the given equations. So the solutions (x , y) = (2 , 2) or (3, 3) are fine. (x , y) = (2 , 3) and (3, 2) are indeed solutions, but while our approach shows that if (x , y) = (2, 3) is a solution, (3 , 2) must also be a solution (and vice-versa), we haven't shown that either one is a solution.

@rcnayak_58 Жыл бұрын

Yeah. If we replace all x as y and solve it, we will get a similar solution y = (2,3) or (3,2). Therefore, we have x = 2 or x = 3 (we have already proved) and y = 2 or 3. Since the problem is symmetrical in x and y i.e., f( x,y) = f(y,x), we have 4 solutions as (x,y) or (y,x) = {(2,2), (3,3), (2,3), (3,2)}.

@theupson Жыл бұрын

@@rcnayak_58 you have literally said "since x=y... we get (2,3) as a solution". that does not follow. edit: to illustrate my objection to your reasoning, let me be concrete: (x-2)^2+(y-2)^2 = 4 and (x+y-3)*(x+y-2) = 0 has the same type of symmetry you have cited. neither are the solutions arranged in a rectangle, nor in fact are there any solutions of the form (x,x).

@hammadsirhindi1320 8 ай бұрын

Good approach❤. I have another question. x^2+y^2=a x^3+y^3=b

@eshuatbits 4 ай бұрын

@@theupsonnayak sahab was wrong

@sophisticatedplayer Жыл бұрын

When adding the equations, you can split 12 into 6 + 6, move y to the other side and factor the -1, so you get: x² - 5x + 6 = (y² - 5y + 6)(-1) (x - 2)(x - 3) = (y - 2)(y - 3)(-1)

@nanamacapagal8342 Жыл бұрын

I started with the assumption that since the two equations are symmetric, surely a solution existed where x and y are the same From that I got (2, 2) and (3, 3) Then I added the equations and landed in a similar spot as you said I managed to get here: If x^2 - 5x + 6 = 0 then so must y^2 - 5y + 6 thus, (2, 3) and (3, 2) The other possibility is x^2 - 5x + 6 = - (y^2 - 5y + 6)but that's difficult to evaluate Then maybe the subtraction method works

@shmuelzehavi4940 Ай бұрын

@@nanamacapagal8342 Your assumption is not correct as a general one, for every symmetric system of equations.

@nanamacapagal8342 Ай бұрын

@@shmuelzehavi4940 yeah fair maybe that was obvious

@RandomKid3489 Жыл бұрын

*”Those who stop learning, stop living.”*

@荻野憲一-p7o Ай бұрын

Notice the alternativity, and solve the equations for x-y, xy. Then, for x,y.

@andreabaldacci1142 Жыл бұрын

I went about it slightly differently. Expanding the original equations and subtracting the second from the first, gives 5x-y^2-12+5y-x^2=0. Switching signs, rearranging the terms and splitting 12 in 4+4+2+2 gives (y-2)^2+(x-2)^2-(x-2)-(y-2)=0, which can be rewritten as (y-2)(y-3)+(x-2)(x-3)=0. Setting y=2 gives x=2 and x=3, while setting y=3 gives x=2 and x=3.

@dougaugustine4075 5 ай бұрын

I watched it yet a third time. Still impressed and liked seeing the graph of the solution at the end: intersecting hyperbolas (or things that look like hyperbolas).

@amtep Жыл бұрын

I started by just watching math videos and enjoying the cleverness but now I've graduated to trying the problems myself first :)

@anglaismoyen Жыл бұрын

This is the way. Just wait until you unironically buy (or download) a textbook and work through it from cover to cover.

@ashutoshsethi6150 Жыл бұрын

Stem redux, just later in life.

@brunoporcu3207 Жыл бұрын

Really a beautiful explanation, congratulations from Italy.

@dan-florinchereches4892 3 ай бұрын

I think i have watched the video two times and solved in 2 ways which for me seem simpler. After addition and subtraction we are left with X^2-5x+6+Y^2-5y+6=0 And 2xy-x-y-7=0 First method takes ento account the equations are symmetric in x and y so: Let S=x+y and P=xy the equations become: S^2-2P-5S+12=0 and 2P-S-7=0 , so 2P = S+7 which we substitute in firs equation and solve a quadratic S^2-6S+5=0 so (S-1)(S-5)=0 which gives soluions (S,P) of (1,4) and (5,6) solving a quadriatic resulting from vietas formulas yields the real solutions for x,y (2,3) and (3,2) My second solution was to let the first equation become -(x-3)(x-3)-(y-2)(y-3)=0 And the second be X=(7+y)/(2y-1) Substituting in the original equation and factoring we get (Y-2)(Y-3)(-1-15/(2y-1)^2)=0 Last term is always the sum of two negative terms which is never zero I think this problem has something which baits you into solving it and you sir have some really clever manipulations sometimes.

@dorkmania 5 ай бұрын

Since both equations are symmetrical in x and y, it follows that there must exist a solution that satisfies x = y, thus (x - 1)(x² + 6) = x(x² + 1) => x³ - x² + 6x - 6 = x³ + x => x² - 5x + 6 = 0 Or x = {2, 3} = y We already know (2, 2) and (3, 3) are solutions, and if we plug in and check the remaining combinations of (2, 3) and (3, 2) we find that they satisfy the system.

@dougaugustine4075 7 ай бұрын

I watched this video twice. Algebraic magic.

@m.h.6470 10 ай бұрын

I would go a completely different route: Since the equations are exactly the same except the switched variables, you can say, that one valid solution is x = y. This means, you can have one equation with just one variable: (x - 1)(x² + 6) = x(x² + 1) x³ + 6x - x² - 6 = x³ + x |-x³ -x -x² + 5x - 6 = 0 |*-1 x² - 5x + 6 = 0 (x - 2)(x - 3) = 0 x ∈ {2, 3} and therefore also y ∈ {2, 3} You have 2 solutions for 2 variables, that is 4 solutions in total: (x, y) ∈ {(2, 2), (2, 3), (3, 2), (3, 3)}

@adw1z Жыл бұрын

For those wondering, a+b = -4 ==> a^2 + b^2 + 2ab = 16 Combined with a^2 + b^2 = 1/2 , ==> ab = 7.75 Combined with a+b = -4, this clearly isn't possible (a,b need to be same sign) which is why it doesn't give any extra solutions for a,b real; (to be pedantic, this is because a and b are roots of the quadratic: m^2 + 4m + 7.75 = (m+2)^2 + 3.75 > 0 for all m, and hence has complex roots, meaning a and b are complex)

@SalmonForYourLuck Жыл бұрын

Thank you for the explanation.... I understood everything

@weo9473 Жыл бұрын

U gives better education than our schools

@azzteke Жыл бұрын

"U gives" is no English.

@itachu. Жыл бұрын

@@azztekehe's african

@akashchowdhury7918 Жыл бұрын

you give he/she/it gives this is the correct grammar

@luladrgn9155 Жыл бұрын

kinda weird to criticise school when you don't know how grammar works

@RuthvenMurgatroyd Жыл бұрын

@@luladrgn9155 Ehh, if he is ESL he jind of gets a pass.

@elephantdinosaur2284 Жыл бұрын

Great video. Keep up the good work. A nice side problem is to show that x - y always divides p( x , y ) - p( y , x ) where p is a polynomial in two variables.

@senpaikunbi2552 Жыл бұрын

Hey sir, I have a faster solution for -2xy+x+y=7. When you have x^2+y^2-5x-5y+12=0, then x^2+y^2+2xy-x-y-7-5x-5y+12=0. This becomes (x+y)^2-6(x+y)+5=0 and now we have (x+y-1)(x+y-5)=0, so x+y=1 or x+y=5. When you have x+y, you can get xy and you con solve for x,y with Viète’s theorem. Thank you and have a nice day sir!

@BP-gn2cl 3 ай бұрын

If in first original equation x is replaced by y, we get second original equation. By using this fact after getting x=2, 3 and x=y, we can write (x, y) = (2, 3), (2, 2), (3, 2), (3, 3)

@miloradtomic Жыл бұрын

Respected Sir, I am pleased, because I create simular tasks for mi students in Serbia. It even bigger when they do it themselves. Wonderful.

@claudiopeixoto4463 Жыл бұрын

It suffices to get two new equations by adding and then subtracting the original equations. Solving the system, one gets x + y = 1 and x + y = 5.

@duroarezina 2 ай бұрын

I believe we have two more valid (imaginary) solutions to our equations here. Solving the second system using method: (x + y)^2 - 2xy - 5(x + y) +12 = 0 and 2xy = x + y + 7 one gets two results: x + y = 1 and x + y = 5. So, from x + y = 1 there are two more valid (though imaginary) solutions: x5, y5 = (1 + i * Sqrt(15))/2, (1 - i * Sqrt(15))/2 x6, y6 = (1 - i * Sqrt(15))/2, (1 + i * Sqrt(15))/2 I tried to verify these results by substitution into our two original equations and yes they work indeed! Since both equations are kind-of third degree ones (the maximum degrees are x^2*y and x*y^2) - to have six solutions (in our case 4 real ones and two imaginary ones) makes perfect sense. I hope this helps. P.S. I admit... I became addicted to your channel. :)

@shmuelzehavi4940 Ай бұрын

You're absolutely right. I've found these two imaginary solutions either (using the same method you did). I try to understand, where and why these solutions were missed within this presentation (even though the problem asked to solve for real solutions only).

@donwald3436 Жыл бұрын

15:12 lol it took me a minute to figure out that your 7\frac{1}{2} meant {\tt 7-1/2} not 7*1/2 lol, who uses that notation any more?

@rizviwasi 4 ай бұрын

I started watching your videos just for your beautiful smile and really got impressed with your style @PrimeNewtons

@vikasseth9544 Жыл бұрын

You sir are a Maths Super Action Hero.

@jweissen9024 2 ай бұрын

When I graph this system of equations, I am getting a slightly different graph. I get the intersection points, but there's an extra line at x = 1. That is also giving a solution, which was not mentioned. What's the deal with that, am I missing something?

@mplaw77 Жыл бұрын

Well done and interesting, I proceeded a little differently after several failed attempts.

@knupug Жыл бұрын

I'm wondering if I'm missing something. After multiplying out the original equations and adding them, you got x^2-5x+y^2-5y+12=0. My immediate thought was to turn that into (x^2-5x+6)+(y^2-5y+6)=0, which becomes (x-2)(x-3)+(y-2)(y-3)=0. That leads you to (2,2), (2,3), (3,2), and (3,3). The only other possible solutions would be when (x-2)(x-3)=-(y-2)(y-3). I'm not quite sure how to prove that's impossible when x is between 2 and 3 or y is between 2 and 3 ... the only ways to generate negative products.

@gghelis Жыл бұрын

"The only other possible solutions would be when (x-2)(x-3)=-(y-2)(y-3)" It's not "other", it's equivalent to (x-2)(x-3)+(y-2)(y-3)=0. It also has an infinite number of other possible solutions, other than (2,2), (2,3), (3,2), and (3,3). You seem to be trying to prove that (x-2)(x-3)+(y-2)(y-3)=0 on its own is equivalent to the original system, which it's obviously not.

@pascall4674 2 ай бұрын

When you look to factorize, just split 12 in two 6 and you get (x-3)(x-2)+(y-3)(y-2)=0 and get four pairs in a shot... but need to check for not 0+0 solutions

@LUFFY-o8f Жыл бұрын

Sir would you make a video on Darboux's theorem?

@isar7349 Жыл бұрын

From where are you sir.

@itachu. Жыл бұрын

Straight from the heavens

@dandeleanu3648 Жыл бұрын

He is a professional no matter where he is from

@munimahmed7877 6 ай бұрын

from the school of our dreams (which only exists in our dreams)

@JatinChawla-v5f 4 ай бұрын

What he showed at last(graph), is my first attempt to answer the question 🫣

@zypherdoesstuffonline5712 Жыл бұрын

Hello can you solve this question Sqrt(log x) - 1/2=log sqrt(x) It was on my math exam and i didnt know how to solve it

@zypherdoesstuffonline5712 10 ай бұрын

@vafasadrif12 ty

@zypherdoesstuffonline5712 10 ай бұрын

@vafasadrif12 ty

@НикитаКоданев-ф7м Жыл бұрын

Why are we not considering a+b+4=0?

@amtep Жыл бұрын

It means b = 4 - a. Substituting into a^2 + b^2 = 1/2 and expanding gives 2a^2 - 8a + 16 = 1/2 which has no real solutions

@tontonbeber4555 Жыл бұрын

@@amtep Yep you exclude in R, but it's interesting that the equation is 6th degree globally, so admits 6 solutions in C. 4 are real, 2 are not ... the 2 complex solutions are not so difficult to find too ...

@pietergeerkens6324 Жыл бұрын

Very nice. 150% speed is a good pace.

@vitotozzi1972 10 ай бұрын

Awesome!!!! Simple awesome!

@ratratrat59 Жыл бұрын

quicker. fast is velocity and quick is time

@BP-gn2cl 3 ай бұрын

And (2, 3) and (3, 2) satisfy x+y-2xy+7=0

@abdullahbarish8204 Жыл бұрын

Amazing

@tontonbeber4555 Жыл бұрын

xy2 + 6x - y2 - 6 = x2y + y x2y + 6y - x2 - 6 = xy2 + x (i)+(ii) 5(x+y) - (x2+y2) - 12 = 0 (x+y)2 - 2xy - 5(x+y) + 12 = 0 (iii) will be useful later ... (i)-(ii) xy(y-x) + 6(x-y) + (x2-y2) = xy(x-y) + (y-x) -2xy(x-y) + 7(x-y) +(x+y)(x-y) = 0 (x-y) (x+y-2xy+7) = 0 (a) x=y => in (1) or (ii) : x3 + 6x - x2 - 6 = x3 + x x2 - 5x + 6 = (x-2)(x+3) = 0 => sol (x,y) = (2,2) or (3,3) (b) x+y-2xy+7 = 0 2xy = x+y+7 in (iii) : (x+y)2 - (x+y) -7 - 5(5+y) + 12 = 0 (x+y)2 - 6(x+y) + 5 = 0 (x+y-1)(x+y-5) = 0 (b1) x+y=5 => xy = 6 => (x,y) = (2,3) or (3,2) (b2) x+y=1 => xy = 4 => x(1-x) = 4 => x2 - x + 4 = 0 no real solution So 4 real solutions : (2,2) (3,3) (2,3) (3,2) and probably 2 other complex solutions as global equation is 6th degree : x2 - x + 4 = 0 => x = (1+/-iV15)/2 so ((1+iV15)/2,(1-iV15)/2) and ((1-iV15)/2,(1+iV15)/2)) By the way, nice curves if you plot them

@munimahmed7877 6 ай бұрын

man! this must have taken you forever to write down...

@xyz9250 4 ай бұрын

Exactly how I solved it.

@BP-gn2cl 3 ай бұрын

This will save much time. Correct me if i am wrong