From the given system of equations:

1) ( x^2 - xy = 28 )
2) ( y^2 - xy = -12 )

Let's isolate ( x ) in the first equation:

( x^2 - xy = 28 )
( x(x - y) = 28 )
( x = \frac{28}{x - y} )

Now, substitute this value of ( x ) into the second equation:

( y^2 - \frac{28y}{x - y} = -12 )
( y^2(x - y) - 28y = -12(x - y) )
( y^2x - y^3 - 28y + 12y = -12x + 12y )
( y^2x - y^3 - 28y = -12x )

Substitute the value of ( x ) obtained earlier:

( y^2(\frac{28}{x - y}) - y^3 - 28y = -12(\frac{28}{x - y}) )
( \frac{28y^2}{x - y} - y^3 - 28y = -\frac{336}{x - y} )
( \frac{28y^2(x - y) - y^3(x - y) - 28y(x - y)}{(x - y)} = -336 )
( 28y^2x - 28y^3 - xy^3 - 28xy = -336 )
( 28(y^2x - xy - y^3) = -336 )

Substitute back the values of ( x ) and ( y ) into this equation to find the exact values of ( x ) and ( y ).