Find all real pairs (x,y)

This is a competition problem from Russia. Kindly share alternative strategies to the one I used if there are. I am also not sure if there are other solutions.

[Jeremy-id]

Thank you for another wonderful and blessed video.

An alternative way of presenting the algebra to solve these equations, as suggested in at least one other comment, is to use polar coordinates.

I guess because it's equivalent, this method still leads to a quartic equation which must be solved, and so is no more efficient than your solution. However, it is presented below for interest and comparison.

With either approach, it seems clear to me that the solutions you found are the only ones. This is reflected in my approach by the lack of other solutions for tan(theta) from the quadratic factor of the quartic equation, and for r from the values of tan(theta) which are solutions to it.

Let
x = r cos(theta)
y = r sin(theta)
where r > 0 (as you note in your solution, x=y=0 is not a solution, as both equations are undefined there, so r = 0 is excluded).

Substituting these into both equations and simplifying gives
(r^2+3) ~ tan(theta) = 3r*sec(theta) (1)
and
(r^2-3)*tan(theta) = 1 (2).

From (2), r^2 = cot(theta)+3 - call this (*)
and substituting this for r^2 into (1) gives
cot(theta)+6-tan(theta) = 3r*sec(theta) - call this (**).

Squaring this equation, again substituting for r^2 from (*) on RHS, replacing sec^2 by {1 + tan^2) on RHS, and simplifying gives
= 0.

Multiplying this equation by tan^2(theta), rearranging and letting t = tan(theta) gives
the quartic equation
26t^4 + 21t^3 ~ 7t^2 ~ 3t - 1 = 0.
which factorises as
(2t-1)(t+1)(13t^2+4t+1) = 0.

The quadratic factor has discriminant 16-4(13)(1) = -36 < 0 and so gives no real solutions for t. Hence the only real solutions for t = tan(theta) are 1/2 and -1.

Substituting these values into (*) gives corresponding values of r^2 of 2+3=5 and -1+3=2, and hence values of r or root(5) and root(2), respectively.

Because of its squaring above, it must be checked that both of these are in fact solutions of (**) and not spurious. But this can be verified for both:

For t=1/2
LHS(**) = 2+6-1/2=15/2 and RHS(**) = 3*root(5)*(root(5)/2) = 15/2

For t=-1
LHS(**) = -1+6+1 = 6 and
RHS(**) = 3*root(2)*(root(2)/1) = 6.

Hence the only solutions for (x, y) =
(r cos(theta), r sin(theta)) are
(root(5)*cos(arctan(1/2)), root(5)*sin(arctan(1/2))) = (2, 1)
and
(root(2)z*cos(arctan(-1)), root(2)*sin(arctan(-1))) = (1, -1).

Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ♥️

[randomjin]

Multiply the 2nd by i (imaginary unit) and add to 1st to get 3 = x + (3x-y)/(x²+y²) + iy - i(x+3y)/(x²+y²) = (3-i)(x-iy)/(x²+y²) + x + iy = (3-i)(x-iy)/(x-iy)(x+iy) + (x+iy) = (3-i)/(x+iy) + (x+iy); We could cancel x-iy since x = y = 0 is not a solution. Now setting x + iy = t we arrive at t² - 3t + (3-i) = 0; D = -3+4i = (1+2i)², so t₁ = 1-i, t₂ = 2+i meaning x₁ = 1, y₁ = -1, and x₂ = 2, y₂ = 1. All operations we did were linear and we didn't lose nor introduced extraneous roots. Since a quadradic can only have two solutions, we can only have two pairs in (x, y) which we've just found.

[ThePhotonMan]

I love algebraic manipulation problems like this! Great video

[holyshit]

Because of x^2+y^2 then parametrization of circle could work
Parametrization of circle will work but it may not be the quickest way to solve this system

[laitinlok]

Rearranging the variable in equation and replace (x^2+y^2) to (x+3y)/y in equation 1 should help solving it faster.

[jay_sensz]

When you look at the plot of the two implicit functions in Wolfram Alpha, there is some funny behavior around x=0, y=0.
The two lines do appear to intersect at a right angle at x=0, y=0 along the tangents (3x-y)=0 and (x+3y)=0, but there's obviously a gap in the plot at exactly x=0, y=0.
A better asymptotic approximation for the first function is `(3x-y-3*(x^2+y^2)) = 0` (to account for the 3 on the RHS), which is equivalent to the circle described by `(x-1/2)^2 + (y+1/6)^2 = 5/18`.
Note that the center of that circle (x=1/2, y=-1/6) lies exactly on the tangent of the second function (x+3y)=0.

[camilotiznado]

at the beginning you asumed that (0, 0) cannot be a solution because you will get a "0/0" kind fraction, but for example, sin(x)/x, which is a "0/0" fraction for x=0, is not undefined, actually is equal to 1. So from this I think is not completly correct to asume (in general), that a value is not a solution because of the 0/0 argument.

[Maths__phyics]

Hello, I am from Azerbaijan, I have discovered the laws of physics, I want to introduce them to America, please help me. I follow you on you tube

[Modo]

Plotting the graphs on Desmos, these seem like the only two answers. However, it appears like there is also a supposed third intersection at (0, 0) but this is undefined due to starting conditions. Overall, though, this is definitely a really interesting graph to look at.

[YuvrajKharat-iv]

Sir please make more videos on such problem this are so great 🥰👌

[larzcaetano]

Can you do a video on how to derive the Laplace Transform?
I have been asking everywhere for something like that and no one has been able to do so.

[dske]

Thank you for the kind explanation 🎉
More difficult problem could be nice

[Nothingx]

This is a real strawberry 🍓 😍 flavor 😋

[Lightseeker-jp]

Actually found another method. Solution: => y^3+x^2y-x-3y=0 => y(x^2+y^2)=x+3y. At this point we can divide both sides by y only if y ≠ 0. It's easy to see that y ≠ 0 by plugging in this value into the second equation. So x^2+y^2=(x+3y)/y. Besides, x+3y ≠ 0, since x^2+y^2 ≠ 0 (it will be relevant a bit later). Substitute this expression to the first equation to get => x^2-y^2+6xy=3(x+3y). Add this new equation with x^2+y^2=(x+3y)/y to get 2x^2+6xy=3(x+3y)+(x+3y)/y => 2x(x+3y)-(3+1/y)(x+3y)=0 => (x+3y)(2x-3-1/y)=0. So either x+3y=0 or 2x-3-1/y=0. Since x+3y≠0 (we stated that earlier), then 2x-3-1/y=0 and x=(1+3y)/2y. From now on I used the same strategy as shown in the video.

[Ron_DeForest]

I’m at a bit of a loss. The part where 1 = 12y ^ 2…

Why did you make the denominator equal to the numerator? You had to divide the denominator into the numerator.

[kevinonealsokengachigui]

Hi prof, please is it possible to have some videos from the begin of function please

[LovePullups]

You forgot the "let's get into the video" part

[Ranoake]

4 roots since highest power is 4. But only 2 real roots since the other 2 powers of y (4y^2 + 1 = 0) require i for the solution.

[Lightseeker-jp]

At 9:27 You should mention that y ≠ 0. It's because we have x=(3y+1)/2y where 2y cannot be 0, thus y cannot be 0.

[duckyoutube]

Question for everyone.

If b/a
And c/b
Then c/a?