1. **Stating the problem:** Solve the system of equations for $x$ and $y$: $$\begin{cases} xy = 3 \\ x^2 + y^2 = 10 \end{cases}$$ 2. **Use the identity:** $$ (x + y)^2 = x^2 + 2xy + y^2 $$ This helps relate $x + y$ to $xy$ and $x^2 + y^2$. 3. **Calculate $(x + y)^2$:** Multiply the first equation by 2 and add to the second: $$ 2xy + (x^2 + y^2) = 2 \times 3 + 10 = 6 + 10 = 16 $$ So, $$ (x + y)^2 = 16 $$ 4. **Find $x + y$:** $$ x + y = \pm 4 $$ 5. **Form quadratic equations:** Since $x$ and $y$ are roots of the quadratic equation $$ t^2 - (x + y)t + xy = 0 $$ we have two cases: - For $x + y = 4$: $$ t^2 - 4t + 3 = 0 $$ - For $x + y = -4$: $$ t^2 + 4t + 3 = 0 $$ 6. **Solve the quadratics:** - For $t^2 - 4t + 3 = 0$: $$ (t - 1)(t - 3) = 0 \implies t = 1, 3 $$ - For $t^2 + 4t + 3 = 0$: $$ (t + 1)(t + 3) = 0 \implies t = -1, -3 $$ 7. **Write solution pairs:** From the roots, the four solution pairs $(x,y)$ are: $$ (1, 3), (3, 1), (-1, -3), (-3, -1) $$ **Final answer:** $$ \boxed{\{(1,3), (3,1), (-1,-3), (-3,-1)\}} $$