1. **State the problem:** Given the system of equations $$2x + y = \frac{1}{\sqrt{5 - 2}}, \quad 2x - y = \frac{1}{\sqrt{5 + 2}}$$ find the value of $$4x^2 - 4x - y^2 + 2y.$$ 2. **Simplify the denominators:** $$\sqrt{5 - 2} = \sqrt{3}, \quad \sqrt{5 + 2} = \sqrt{7}.$$ So the system becomes $$2x + y = \frac{1}{\sqrt{3}}, \quad 2x - y = \frac{1}{\sqrt{7}}.$$ 3. **Add the two equations to eliminate $y$: ** $$ (2x + y) + (2x - y) = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} $$ $$ 4x = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} $$ $$ x = \frac{1}{4} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right).$$ 4. **Subtract the second equation from the first to eliminate $x$: ** $$ (2x + y) - (2x - y) = \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} $$ $$ 2y = \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} $$ $$ y = \frac{1}{2} \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right).$$ 5. **Calculate $4x^2$: ** $$4x^2 = 4 \left( \frac{1}{4} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right) \right)^2 = \cancel{4} \times \frac{1}{\cancel{16}} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right)^2 = \frac{1}{4} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right)^2.$$ 6. **Calculate $-4x$: ** $$-4x = -4 \times \frac{1}{4} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right) = - \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right).$$ 7. **Calculate $-y^2$: ** $$-y^2 = - \left( \frac{1}{2} \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right) \right)^2 = - \frac{1}{4} \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right)^2.$$ 8. **Calculate $2y$: ** $$2y = 2 \times \frac{1}{2} \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right) = \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right).$$ 9. **Sum all parts: ** $$4x^2 - 4x - y^2 + 2y = \frac{1}{4} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right)^2 - \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right) - \frac{1}{4} \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right)^2 + \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right).$$ 10. **Group terms: ** $$= \frac{1}{4} \left[ \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right)^2 - \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right)^2 \right] - \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right) + \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right).$$ 11. **Use difference of squares: ** $$ (a+b)^2 - (a-b)^2 = 4ab $$ where $a=\frac{1}{\sqrt{3}}$, $b=\frac{1}{\sqrt{7}}$. So, $$ \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right)^2 - \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right)^2 = 4 \times \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{7}} = \frac{4}{\sqrt{21}}.$$ 12. **Substitute back: ** $$= \frac{1}{4} \times \frac{4}{\sqrt{21}} - \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{7}} \right) + \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right) = \frac{1}{\sqrt{21}} - \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{7}}.$$ 13. **Simplify the last terms: ** $$- \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = 0,$$ $$- \frac{1}{\sqrt{7}} - \frac{1}{\sqrt{7}} = - \frac{2}{\sqrt{7}}.$$ 14. **Final answer: ** $$4x^2 - 4x - y^2 + 2y = \frac{1}{\sqrt{21}} - \frac{2}{\sqrt{7}}.$$