If x⁴ + y⁴ + x² + y² = $$17\frac{1}{{16}}$$ and x² - xy + y² = $$5\frac{1}{4},$$ then one of the values of (x - y) is:

Solution (By Examveda Team)

$$\eqalign{ & {x^2} + {y^2} - xy = 5\frac{1}{4} \cr & {x^2} + {y^2} - xy = \frac{{21}}{4}........\left( {\text{i}} \right) \cr & {x^4} + {y^4} + {x^2}{y^2} = 17\frac{1}{{16}} \cr & {x^4} + {y^4} + {x^2}{y^2} = \frac{{273}}{{16}} \cr & {\left( {{x^2} + {y^2}} \right)^2} - {\left( {xy} \right)^2} = \frac{{273}}{{16}} \cr & \left( {{x^2} - xy + {y^2}} \right)\left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \cr & \frac{{21}}{4}\left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \cr & \left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \times \frac{4}{{21}} \cr & \left( {{x^2} + xy + {y^2}} \right) = \frac{{13}}{4}........\left( {{\text{ii}}} \right) \cr & {\text{solve }}\left( {\text{i}} \right) + \left( {{\text{ii}}} \right) \cr & 2\left( {{x^2} + {y^2}} \right) = \frac{{21}}{4} + \frac{{13}}{4} \cr & 2\left( {{x^2} + {y^2}} \right) = \frac{{17}}{2} \cr & \left( {{x^2} + {y^2}} \right) = \frac{{17}}{4} \cr & {\text{solve }}\left( {\text{i}} \right) - \left( {{\text{ii}}} \right) \cr & - 2xy = \frac{{21}}{4} - \frac{{13}}{4} \cr & - 2xy = \frac{8}{4} \cr & - 2xy = 2 \cr & xy = - 1 \cr & {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr & {\left( {x - y} \right)^2} = \frac{{17}}{4} - 2 \times \left( { - 1} \right) \cr & {\left( {x - y} \right)^2} = \frac{{17}}{4} + 2 \cr & {\left( {x - y} \right)^2} = \frac{{25}}{4} \cr & x - y = \frac{5}{2} \cr} $$