Find all real pairs (x,y)

[@Jeremy-i1d]

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 7+cot^2(theta)-26tan^2(theta)+3cot(theta)-21tan(theta) = 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 ♥️

[@ThePhotonMan110]

I love algebraic manipulation problems like this! Great video

[@michaelgolub2019]

Though it has already been mentioned, I'd like to point that to multiply by y we have to ensure y is not 0. If we plug y=0 into the second equation we have also x=0, but (0,0) is not a solution as you have shown before. So, y is not 0.

[@victorchoripapa2232]

Hello teacher. I hope you are fine. And thank you very much for this exercise. One multiplies an equation by this or that, and ends up in such a way that when adding or subtracting member by member with another equation, one already believes that everything has become extremely easy. I never saw coming the strategy you used when I previously tried to do it on my own before watching the video.

[@AndrejPanjkov]

Eliminate x^2+y^2 - gives you a biquadratic in x and y, which is a conic section. in this case, a hyperbola with centre (3/2,0) and skew asymptotes (but they are perpendicular) Your simplification is also a hyperbola with horizontal and vertical asymptotes. (They are the x-axis and the vertical line x=3/2) The solutions are the intersections of these two hyperbolas. Graphically, you can see that there will be exactly two solutions.

[@motogee3796]

What I did is to take x=Ky, eliminate x and y to get a biQuadratic in k which luckily can easily be factored into 2 quadratics one of which has no real roots and the other yields 2 values for k. They give 2 final answers (1,-1) & (2,1). There is no loss of generality in this approach as k is a variable which changes for each solution (x,y) that satisfy the original equations.

[@rohangt1]

Hey man. Love your videos. I am wondering if you could do a simple video on proving quadratic formula using completing the square method.

[@randomjin9392]

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.

[@AubreyForever]

I wish I could understand what you mean by cancelling x-iy.. I didn't see it at both sides of the equations above.

[@coendert2010]

Very nice!

[@Jeremy-i1d]

Beautiful solution, thank you

[@Jeremy-i1d]

@@AubreyForeverthe top of the term containing x^2+y^2 is (3-i)*(x-iy) and the bottom factorises as (x+iy))*(x-iy). Hence x-iy (which is non-zero) cancels

[@AubreyForever]

@@Jeremy-i1d I am just learning imaginary numbers. Is the imaginary unit just i?

[@dske9146]

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

[@XTRA14LIVE]

The system of equations given is not symmetric, so the typical strategies of adding or subtracting the equations may not work. To generate similarities between the two equations, the first equation is multiplied by y and the second equation is multiplied by x. Adding the modified equations together simplifies the problem and leads to an easier-to-solve equation in terms of x and y. Isolating x in terms of y and substituting it back into one of the original equations allows for finding the values of y, which can then be used to determine the corresponding values of x. The final solution involves two pairs of x and y values, and the author suggests there may be additional solutions that could be found using other strategies.

[@YuvrajKharat-i1v]

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

[@holyshit922]

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

[@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.

[@jay_sensz]

Wolfram alpha query if you want to look at it yourself: plot x+(3*x-y)/(x^2+y^2) = 3, y-(x+3*y)/(x^2+y^2) = 0, (3*x-y-3*(x^2+y^2)) = 0, 3*x-y=0, (x+3*y)=0, -2

[@laitinlok1]

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

[@Comedy_boy_30]

Plz solve the this eqn by another method 😢

[@Modo942000]

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.

[@FIST_3807]

Awesome Video ❤

[@SALogics]

Very nice problem! ❤❤

[@Nothingx303]

This is a real strawberry 🍓 😍 flavor 😋

[@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.

[@Lightseeker1-j5p]

Actually found another method. Solution: y-(x+3y)/(x^2+y^2)=(y^3+x^2y-x-3y)/(x^2+y^2)=0 => 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+(3x-y)/((x+3y)/y)=x+y(3x-y)/(x+3y)=(x^2+3xy+3xy-y^2)/(x+3y)=3 => 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.

[@slavinojunepri7648]

Did your strategy produce more than two real solutions? PrimeNewton's gives only two, and he's wondering if his approach is complete enough to find all real solutions.

[@Lightseeker1-j5p]

@@slavinojunepri7648 My strategy produced exact same solutions as PrimeNewton's. Only (2;1) and (1; -1).

[@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.

[@Lightseeker1-j5p]

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.

[@Lightseeker1-j5p]

Besides, it's not correct to divide 3y+1 by 2y without checking a possible case when y=0

[@jam9339]

Great video!

[@nicolasb11]

Well done

[@Serg-U]

Thanks!

[@kevinonealsokengachigui8735]

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

[@FactRain2239]

Can you please provide the proof of the Basel problem?

[@nebulousquartz2712]

Thank you ❤

[@surendrakverma555]

Thanks Sir

[@duckyoutube6318]

Question for everyone. If b/a And c/b Then c/a?

[@BartBuzz]

Nice!

[@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.

[@ebiebrahimian4248]

Y/X =t. Y=tX And see what happend!!

[@ShrishtiSinghrajputjairajp-r1p]

Plz . Can u elaborate and solve without using prime Newton method 😢😢

[@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

[@DhirajKumar-g]

Really can you show me your work on Physics laws

[@Maths__phyics]

@@DhirajKumar-g no

[@DhirajKumar-g]

But why if you invented the new theory , thats great i also do , but i confusing in my inventions, let compare both invention ​@@Maths__phyics

[@Maths__phyics]

@@DhirajKumar-g why my trust, what do you do, where are you from?

[@DhirajKumar-g]

@@Maths__phyics I am from India INDIA

[@Metaverse-d9f]

12:41 It's "since", not "sine"

[@Logia_]

...rly?...wow...

[@slavinojunepri7648]

Looks like we've got some English language policing going on here, typical of people who rush on simple syntax errors to bring down others.

[@jonahs92]

He's a true mathematician, he unconsciously writes about trigonometric functions 😂

[@LovePullups]

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

[@maburwanemokoena7117]

me rushing to the comments for alternative solutions😂

[@camilotiznado5325]

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.

[@absolutezero9874]

No reply

[@obeyy0urmaster]

Where was the intro my man ? :( !!!!

[@slavinojunepri7648]

That's what I noticed. A viewer who hates the African rhythm in the usual introduction made a negative comment about it, describing it hypocritically as "dramatic". That may have influenced PrimeNewton's decision not to use it.

[@PrimeNewtons]

Oh no! I was tired when I edited the video. I simply forgot.