Your starting equations are

    x   + y   =  10,      (1)

    x^3 + y^3 = 300.      (2)


In equation (2), use the decomposition of the sum of cubes in the left side

    x^3 + y^3 = (x+y)*(x^2 - xy + y^2).


In this decomposition, replace the factor  (x+y)  by 10, based on equation (1).
You will get then

    x^2 - xy + y^3 = 30.   (3)    (after dividing both sides by 10)


So, now you have equivalent system of equations

    x + y = 10,            (4)

    x^2 - xy + y^2 = 30,   (5)


but the degree is lowered from 3 to 2, which is good.


Now, from equation (4) express  y = 10-x  and substitute it into equation (5).  You will get

    x^2 - x(10-x) + (10-x)^2 = 30,

    x^2 - 10x + x^2 + 100 - 20x + x^2 = 30,

    3x^2 - 30x + 70 = 0.


Use the quadratic formula

     =  =  = .


Two values for x are   = 6.290994449 (approximately)  and   = 3.709005551  (approximately).


The ordered pairs are  (x,y) = ( ,  ) = ( , )

                  and  (x,y) = ( ,  ) = ( , ).


You may check that  6.290994449^3 + 3.709005551^3 = 300.0000000205... , so the approximate solution is very good.


You also may check that exact solutions (x,y) satisfy equations (1) and (2) precisely.