A very nice olympiad question | Solve x+y^2=y^3; y+x^2=x^3 | Algebra |
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- Опубліковано 27 вер 2024
- See the way I breakdown the solution of this question. There is a lot you can learn from this video.
How to solve x+y^2=y^3; y+x^2=x^3
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Решение из сельской местности. Пропущено два действия.
It is symmetrical variables so x=y
X+x^2=x^3
So x=0 or
1+x=x^2
Nice one
another great challenge!
By comparison, how did
(y-x)(-1+y+x) = (y-x)(y²+xy+x²) results to y-x =0 ?
I tnhought we just have to cancel y-x on both sides and only have one case which is -1+y+x = y²+xy+x²?
Both sides of the equation have a common factor of y - x. If we divide it out, we will be left with only one case and miss out on other vital solutions. The only way we can not divide it out is if y - x = 0, but this would also set both sides of the equation to 0 and thus create a valid solution.
If you haven't seen maths solved this way, I am sure you've learnt a new thing in this video. Thanks.
(x - y) - (x² - y²) = - (x³ - y³)
x = y
x³ - x² - x = 0
x(x² - x - 1) = 0
*x = y = 0*
*x = y = (1 ± √5)/2*
1 - (x + y) = - (x² + xy + y²)
x² + xy + y² - (x + y) + 1 = 0
2x² + 2xy + 2y² - 2(x + y) + 2 = 0
(x + y)² + x² - 2x + 1 + y² - 2y + 1 = 0
(x + y)² + (x - 1)² + (y - 1)² = 0
Interesting