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