Solve the system of equations: 3) $$\begin{cases} 3x - y = 10 \ x^2 + xy - y^2 = 20 \end{cases}$$

Let's solve the system of equations step by step: 1. From the first equation, we can express $$y$$ in terms of $$x$$: $$y = 3x - 10$$. 2. Substitute this expression for $$y$$ into the second equation: $$x^2 + x(3x - 10) - (3x - 10)^2 = 20$$ $$x^2 + 3x^2 - 10x - (9x^2 - 60x + 100) = 20$$ $$x^2 + 3x^2 - 10x - 9x^2 + 60x - 100 = 20$$ $$-5x^2 + 50x - 100 = 20$$ $$-5x^2 + 50x - 120 = 0$$ 3. Divide the equation by -5: $$x^2 - 10x + 24 = 0$$ 4. Now, let's solve the quadratic equation $$x^2 - 10x + 24 = 0$$. We can factor this quadratic equation as: $$(x - 6)(x - 4) = 0$$ 5. This gives us two possible values for $$x$$: $$x - 6 = 0 \Rightarrow x_1 = 6$$ $$x - 4 = 0 \Rightarrow x_2 = 4$$ 6. Now, let's find the corresponding values for $$y$$ using the expression $$y = 3x - 10$$: For $$x_1 = 6$$: $$y_1 = 3(6) - 10 = 18 - 10 = 8$$ For $$x_2 = 4$$: $$y_2 = 3(4) - 10 = 12 - 10 = 2$$ Thus, the solutions for the system of equations are $$(x_1, y_1) = (6, 8)$$ and $$(x_2, y_2) = (4, 2)$$. Answer: $$(6, 8)$$ and $$(4, 2)$$