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

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

7 ай 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

7 ай 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

7 ай 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

2 ай бұрын

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

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

7 ай 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

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).

I watched this video twice. Algebraic magic.

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

@anglaismoyen

7 ай бұрын

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

@ashutoshsethi6150

7 ай бұрын

Stem redux, just later in life.

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.

Really a beautiful explanation, congratulations from Italy.

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)}

U gives better education than our schools

@azzteke

7 ай бұрын

"U gives" is no English.

@itachu.

7 ай бұрын

​@@azztekehe's african

@akashchowdhury7918

7 ай бұрын

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

@luladrgn9155

7 ай бұрын

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

@RuthvenMurgatroyd

7 ай бұрын

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

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

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

7 ай бұрын

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

Amazing

You sir are a Maths Super Action Hero.

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.

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

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.

Awesome!!!! Simple awesome!

In the equation x^2+y^2-5x-5y+12=0, the x and y coefficients are equal, 1. Moreover, the coefficient of xy is 0. So, it's easy to say without even doing math that it's the equation of a circle.

quicker. fast is velocity and quick is time

Very nice. 150% speed is a good pace.

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

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?

Teacher thnks

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!

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

7 ай 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.

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

Ай бұрын

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

From where are you sir.

@itachu.

7 ай бұрын

Straight from the heavens

@dandeleanu3648

7 ай бұрын

He is a professional no matter where he is from

@munimahmed7877

Ай бұрын

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

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

@richardbraakman7469

7 ай 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

7 ай бұрын

@@richardbraakman7469 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 ...

This is my country :o

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

@vafasadrif12

5 ай бұрын

Perhaps i can help. At first we take 1/2 to the other side so sqrt(log(x)) = log(sqrt(x)) + 1/2 Then we raise both sides to the power of two so logx = (log(sqrt(x)) +1/2)² = log(sqrt(x))² + log(sqrt(x)) + 1/4 Now we can substitute log(sqrt(x)) as y: Let y = log(sqrt(x)) Using the rules of logarithm log(x) would be equal to 2y 2y = y² + y + 1/4 0=y² - y + 1/4 Using the quadratic formula we have y = 1/2 Note that the equation has only one root as the delta would be .equal to zero log(sqrt(x)) = 1/2 0.5log(x) = 0.5 log(x) = 1 x = 10 And that is our answer

@zypherdoesstuffonline5712

4 ай бұрын

@vafasadrif12 ty

@zypherdoesstuffonline5712

4 ай бұрын

@@vafasadrif12 ty

Lol. I watch every video at 2x speed.