A System of Equations from Moscow | Incorrect Solution

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[visheshlonial]

Hey Syber, actually you have made an error because in the numerator you had (1-sqrt(x^2-y^2)) whereas in the denominator you had sqrt(1-(x^2-y^2)) and this is not the square root of the earlier expression because if you square this you get (1-(x^2-y^2)).

[gropius]

Let c = sqrt(x²-y²) and let d = sqrt(1-x²+y²).
Then (x-cy)/d=a and (y-cx)/d=b .
Solve these two straightforwardly for x and y in terms of a, b, c, and d, noting that 1-c² = d².
This yields x = (a + bc)/d and y = (b + ac)/d.
Square the set of equations from the 2nd line and subtract to (again using 1-c² = d²) arrive at x² - y² = a² - b².
Use this last fact to rewrite c and d in terms of a² and b².

[abdoshaat]

There's problem in the square roots

[awodey]

Demonstration ok til 2'02, then false when introducing "c" at 2'23. I solve the system of equations by a different way and found : x=(a+b.sqrt(a^2-b^2)) / sqrt(1-a^2+b^2) and y=(b+a.sqrt(a^2-b^2)) / sqrt(1-a^2+b^2) with conditions for values a an b (1>a^2-b^2>=0). This works for me, for example a=sqrt(3) and b=sqrt(5/2). Hope this helps.

[SG]

Besides the mistake everybody mentioned already: If you divide both equation you need to analyse the case that the divisor equals 0. Otherwise you can lose solutions on your way. In your case it would require to analyse x-y=0 and sqrt(1-x^2+y^2)=0. Although I am not sure if you reach this point without the mistake everybody mentioned already.

[leickrobinson]

I hope you decide to do this one again with a more correct solution, because it is an interesting problem. :-D

[yoav]

I got: x=(a+bsqrt(a^2-b^2))/sqrt(1-(a^2-b^2)), I checked for a=0.5 and b=0 in wolfram and got the right solution. The way i solved it was squaring both equations and subtract to get x^2-y^2=a^2-b^2 so y=sqrt(x^2-(a^2-b^2)) then plug it in the first eq and you get It was very long calculations..

[mbmillermo]

@SyberMath -- At 2:10 - "One is the square of the other". Nope. You should delete this one and redo it!

[zawatsky]

Подождите! На 2:31 единица в числителе _за_ корнем, а в знаменателе - _под_ ним. Как можно их приравнивать друг к другу?! Не припомню такого правила, где 1-√c=√(1-c). Попробовал пару подстановок - не сходится!

[maxxie]

Either you skipped a step in the explanation, or this is very wrong...

[rajeshbuya]

I thought the solution would be a piece of k 😊😊

[yt-]

enteresan sorularda hep bir hata çıkıyor