1. **State the problem:** Solve the system of equations: $$y = -2x^2 + 4x - 1$$ $$y = 2x - 5$$ 2. **Set the equations equal to each other** since both equal $y$: $$-2x^2 + 4x - 1 = 2x - 5$$ 3. **Bring all terms to one side to form a quadratic equation:** $$-2x^2 + 4x - 1 - 2x + 5 = 0$$ $$-2x^2 + (4x - 2x) + (-1 + 5) = 0$$ $$-2x^2 + 2x + 4 = 0$$ 4. **Divide the entire equation by $-2$ to simplify:** $$\cancel{-2}x^2 + \cancel{2}x + \cancel{4} = 0 \Rightarrow x^2 - x - 2 = 0$$ 5. **Factor the quadratic:** $$x^2 - x - 2 = (x - 2)(x + 1) = 0$$ 6. **Set each factor equal to zero and solve for $x$:** $$x - 2 = 0 \Rightarrow x = 2$$ $$x + 1 = 0 \Rightarrow x = -1$$ 7. **Find corresponding $y$ values by substituting $x$ into $y = 2x - 5$:** - For $x = 2$: $$y = 2(2) - 5 = 4 - 5 = -1$$ - For $x = -1$: $$y = 2(-1) - 5 = -2 - 5 = -7$$ **Final solutions:** $$(2, -1) \text{ and } (-1, -7)$$