Canada | A Nice Algebra Problem | Math Olympiad

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[@comdo777]

asnwer=1/ /3

[@comdo777]

asnwer=(1+/3)(1-/3) isit

[@SidneiMV]

x² - 3x = y
y² - 3y = x
x² - 3x - y = 0
y² - 3y - x = 0
x² - y² - 2x + 2y = 0
(x + y)(x - y) - 2(x - y) = 0
(x - y)(x + y - 2) = 0
[ I ] x - y = 0 => y = x
x² - 3x = x
x² - 4x = 0
x(x - 4) = 0
*x = 0 => y = 0*
*x = 4 => y = 4*
[ II ] x + y - 2 = 0 => y = 2 - x
x² - 3x = 2 - x
x² - 2x - 2 = 0
(x - 1)² - 3 = 0
(x - 1)² = 3
x = 1 ± √3
*x = 1 + √3 => y = 1 - √3*
*x = 1 - √3 => y = 1 + √3*

[@giannaleoci2328]

x(x-3)=y
y(y-3)=x
x^2-3x=y
y^2-3y=x
x^2-3x-y^2+3y=y-x
(x^2-y^2)-3(x-y)+(x-y)=0
(x-y)(x+y)-2(x-y)=0
(x-y)(x+y-2)=0
x-y=0 x+y-2=0
x=y x=2-y
1) x=y
x^2-3x=x
x^2-4x=0
x(x-4)=0
x=y=0 x=y=4
2) x=2-y
y^2-3y=2-y
y^2-2y-2=0
y=[2+-rq(4+8)]/2
y=1+-rq3
x=2-y=2-(1+-rq3)
x=1-+rq3