1/x+y = 1/x + 1/y

In school you were taught that 1/x+y is not the same as 1/x + 1/y, but for which x and y is it actually true? Watch this video and find out!

[ethannguyen]

Society if this was true for all values of x and y

[RCSmiths]

*Looks at thumbnail*

"Wait that's heresy" xD

[factsheet]

Such a nice excrise!
This could definitely appear in like a 12th grade exam where they learn about complex numbers, maybe even as a question on a complex analysis 1 exam who knows!

[o]

That sad, I'm disappointed I thought that there was a real solution.

[drhubblebubble]

Wow the equilateral part was really fucking dope!

[GammaFZ]

did not expect that result, that’s pretty amazing

[chimetimepaprika]

I love this type of stuff. Matrix representation in analytic geometry and of quadratic forms and spectral theory fill me with glee.

[BabyXGlitz]

also a trivial solution when x= infinity which results in 0 = 0

[MrCigarro]

Really, Really interesting. I've been following your videos in complex numbers and everytime I've to spend some hours until I understand your solutions. Thank you, that keeps me away from covid.

[aboodahmad]

Perfect video:)
Why if y also =x(-1+-√3 •i)/2
then we say that x=x((-1+-√3 •i)/2)^2
Then imagenry have 4 real solutions why??

[mathematics]

Sir I want to ask something that you please solve some of the questions of jee advanced they are really good and there are many students who are preparing for that exam. It will be very helpful and the questions will be too good. I HOPE A REPLY FROM YOU SIR ....:)

[mangaram]

Those are cube roots of I mean, solutions of x^3=1.
Edit: Except '1' of course.

[sheshakrishna]

e power +_2pi /3 is omega and omega square

[manavbakshi]

From the equation x^2 + xy + y^2 = 0, we can multiply x - y on both sides, to get x^3 - y^3 = 0, therefore x = y. But this doesn’t work as on putting these values into the question, we don’t get the right answer!

[Jen-bihv]

what if the equation was 1/x + 1/y = 2/x+y ? how would we solve that?

[chrisbrice]

fun problem! brings to light the relevance of euler's formula:D (Not to mention the solution looks much nicer this way!)

[Kdd]

It _might_ seem that x = 0 and y = 0 will work as the real solutions but in fact they don't coz if you use those values in 1/x+y = 1/x +1/y, division by zero is not allowed! So it has no real sols

[brunolaugier]

But I should want the values of x and y ?

[MoodyG]

hey Dr Peyam, can you help me integrate e^(-x-a/x)/(b+c/x) from 0 to +infinity ? a, b, and c are constants.

[ukaszskiba]

Your solution is very chaotic. You write some implications but not in all cases. You should use equivalence in all transformations of equation. Only then your solution would be correct. If you afraid of <=> use implication. In Real number equation implies y=0 and x=0 and then you should check if (x, y) =(0, 0) implies the equation and get a contradiction.