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)
@nanamacapagal834210 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
@rcnayak_5810 ай 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).
@SanePerson110 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_589 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)}.
@theupson9 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).
@hammadsirhindi13205 ай бұрын
Good approach❤. I have another question. x^2+y^2=a x^3+y^3=b
@eshuatbitsАй бұрын
@@theupsonnayak sahab was wrong
@andreabaldacci114210 ай 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.
@dougaugustine40752 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).
@SamarBist-b8l10 ай бұрын
*”Those who stop learning, stop living.”*
@richardbraakman746910 ай бұрын
I started by just watching math videos and enjoying the cleverness but now I've graduated to trying the problems myself first :)
@anglaismoyen10 ай бұрын
This is the way. Just wait until you unironically buy (or download) a textbook and work through it from cover to cover.
@ashutoshsethi61509 ай бұрын
Stem redux, just later in life.
@brunoporcu320710 ай бұрын
Really a beautiful explanation, congratulations from Italy.
@dorkmania2 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.
@weo947310 ай бұрын
U gives better education than our schools
@azzteke10 ай бұрын
"U gives" is no English.
@itachu.10 ай бұрын
@@azztekehe's african
@akashchowdhury791810 ай бұрын
you give he/she/it gives this is the correct grammar
@luladrgn915510 ай бұрын
kinda weird to criticise school when you don't know how grammar works
@RuthvenMurgatroyd9 ай бұрын
@@luladrgn9155 Ehh, if he is ESL he jind of gets a pass.
@dougaugustine40754 ай бұрын
I watched this video twice. Algebraic magic.
@dan-florinchereches4892Ай 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.
@m.h.64707 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)}
@elephantdinosaur22849 ай 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.
@adw1z10 ай 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)
@SalmonForYourLuck10 ай бұрын
Thank you for the explanation.... I understood everything
@claudiopeixoto44639 ай 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.
@BP-gn2cl24 күн 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)
@miloradtomic10 ай 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.
@senpaikunbi25529 ай 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!
@knupug10 ай 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.
@gghelis10 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.
@rizviwasi2 ай бұрын
I started watching your videos just for your beautiful smile and really got impressed with your style @PrimeNewtons
@isar734910 ай бұрын
From where are you sir.
@itachu.10 ай бұрын
Straight from the heavens
@dandeleanu364810 ай бұрын
He is a professional no matter where he is from
@munimahmed78774 ай бұрын
from the school of our dreams (which only exists in our dreams)
@vikasseth95449 ай бұрын
You sir are a Maths Super Action Hero.
@mplaw7710 ай бұрын
Well done and interesting, I proceeded a little differently after several failed attempts.
@PureHanbali9 ай бұрын
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.
@LUFFY-o8f10 ай бұрын
Sir would you make a video on Darboux's theorem?
@donwald34369 ай 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?
@zypherdoesstuffonline571210 ай 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
@vafasadrif127 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
@zypherdoesstuffonline57127 ай бұрын
@vafasadrif12 ty
@zypherdoesstuffonline57127 ай бұрын
@@vafasadrif12 ty
@НикитаКоданев-ф7м10 ай бұрын
Why are we not considering a+b+4=0?
@richardbraakman746910 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
@tontonbeber455510 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 ...
@pietergeerkens632410 ай бұрын
Very nice. 150% speed is a good pace.
@BP-gn2cl24 күн бұрын
This will save much time. Correct me if i am wrong
@JatinChawla-v5fАй бұрын
What he showed at last(graph), is my first attempt to answer the question 🫣
@ratratrat599 ай бұрын
quicker. fast is velocity and quick is time
@vitotozzi19728 ай бұрын
Awesome!!!! Simple awesome!
@BP-gn2cl24 күн бұрын
And (2, 3) and (3, 2) satisfy x+y-2xy+7=0
@abdullahbarish820410 ай бұрын
Amazing
@juma412710 ай бұрын
Teacher thnks
@tontonbeber455510 ай 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
@munimahmed78774 ай бұрын
man! this must have taken you forever to write down...
@xyz92502 ай бұрын
Exactly how I solved it.
@TrinityRed2 ай бұрын
Resolution too complicated. Adding the equations and you will solve it in 5 minutes.
@eduardionovich4425Ай бұрын
Вторая половина решения (после рассмотрения случая х=у) страшно длинна и не рациональна.