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A nice math problem to solve| Solve for x and y 

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29 сен 2024

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Комментарии : 23   
@Mohammad1Ahmad
@Mohammad1Ahmad 17 дней назад
The equation has no real solutions... but it accepts complex solutions.
@habeebalbarghothy6320
@habeebalbarghothy6320 16 дней назад
It's very clear that this eq. cannot be solved
@kennethkan3252
@kennethkan3252 16 дней назад
x=y=無限大 1/無限大=無限小。 無限小+無限小 =無限小
@SidneiMV
@SidneiMV Месяц назад
1/(x + y) = (x + y)/xy xy = (x + y)² y = kx kx² = x²(k + 1)² k² + k + 1 = 0 k = (-1 ± i√3)/2 *y = x(-1 ± i√3)/2* [ complex solution only ]
@imetroangola4943
@imetroangola4943 Месяц назад
It's the perfect solution, since the domain of x and y were not specified in the video. Indeed, this is the solution.
@daveh1924
@daveh1924 28 дней назад
hi I have a little lost, why y=kx?
@SidneiMV
@SidneiMV 27 дней назад
@@daveh1924 parametrization. it's a common way to find a solution.
@daveh1924
@daveh1924 27 дней назад
@@SidneiMV thanks for the reply. Yes your method indeed can perfectly find a particular solution if y and x are proportional.
@alanyoung1711
@alanyoung1711 20 дней назад
@@daveh1924 Any 2 non-zero numbers, real of imaginary are always proportional. Just divide on by the other to get the proportion. This is a fully general solution.
@alanyoung1711
@alanyoung1711 20 дней назад
This is a disappointing video - taking 5 minutes to reach an incorrect conclusion. There are infinitely many solutions to this, indeed for any non-zero choice of x one can find a y that makes it work and vice versa. @SidneiMV is perfectly correct, writing out in more detail: Assume x and y are non-zero, write (1) y=kx for some abitrary non-zero k, so the problem becomes (2) 1/(x+kx) = 1/x + 1/kx Cross multiply (2) by k(1+k) and bring everything to the same side, note that the denominator is non-zero by choice of x and k, and get (3) x^2(k^2+k+1)=0 Since x is non-zero, we must have: k^2+k+1=0. Put this into the standard quadratic formula [-b+/-sqrt(b^2-4ac)/2] to get (4) k = (-1+i*sqrt(3))/2 and (5) k = (-1-i*sqrt(3))/2 Pick ANY non-zero x (can be complex), use (1) to calculate y. Put these values into the original formula and verify the solution (try x=1 as a simple example). Side bar: solutions (4) and (5) are the two complex roots of 1, each of which is the complex conjugate of the other. One can observe this by noting that (k^3-1)=(k-1)(k^2+k+1). Nothing magic, just a happy co-incidence for this problem; if it was solve 1/(x+2y)=1/x+1/y then same methodology would give a different value of k.
@dvthiep8878
@dvthiep8878 16 дней назад
Có gì đó sai, nếu x=y= 0 thì 1/x và 1/y ko đúng
@arseniylanin
@arseniylanin 12 дней назад
What about comlex numbers?
@imetroangola4943
@imetroangola4943 Месяц назад
The emblematic problem in your videos is that unfortunately you do not say the domain of the variables x and y. What you used is true, if x and y are real numbers.
@snarkybuttcrack
@snarkybuttcrack 24 дня назад
what's the complex solution then?
@timetraveller6643
@timetraveller6643 24 дня назад
Say your name out loud to understand why no-one is going to take you seriously.
@roryd4852
@roryd4852 17 дней назад
Except that the author did not specify x's and y's domain, a very clever solution.
@imetroangola4943
@imetroangola4943 14 дней назад
​@@timetraveller6643 Do you have a problem? It looks like you do.
@imetroangola4943
@imetroangola4943 14 дней назад
​@@roryd4852The solution is simple, mathematically speaking. I don't take away the merit of the person who makes the video, my tip is to help the channel to improve more and more!
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