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).
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.
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)}.
@@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).
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)
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 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.
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)}
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.
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)
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.
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.
"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.
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
@@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 ...
