x # y = x ( x − y ) x # ( x # y ) x # y = x ( x − y ) = x 2 − x y x # ( x # y ) = x # ( x 2 − x y ) = x [ x − ( x 2 − x y )] = x ( x − x 2 + x y ) = x 2 − x 3 + x 2 y
x 2 − x 3 + x 2 y = 8 x 2 ( 1 − x + y ) = 8 x 2 ( 1 − x + y ) = 2 2 ⋅ 4 ⟺ x 2 = 2 2 an d 1 − x + y = 4 x = 2 an d y = 4 − 1 + x x = 2 an d y = 3 + 2 x = 2 an d y = 5
if x # y = x(x-y) => x # (x # y) = x # [x(x-y)] =x[ x - x(x-y)] = x( x - x^2 +x*y ) = x^2 - x^3 + ( x^2 ) * y;
The expression x # ( x # y ) evaluates to x 2 − x 3 + x 2 y using the defined operation. Additionally, it can equal 8 when the equation x 2 ( 1 − x + y ) = 8 holds true for certain values of x and y . Both interpretations are consistent as they follow from the defined operation and basic algebra.
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