1. **Problem:** Write down the parametric equations of the line $2x - y - 3 = 0$. 2. **Formula and rules:** A line in the plane can be written parametrically as: $$x = x_0 + at, \quad y = y_0 + bt$$ where $(x_0, y_0)$ is a point on the line and $(a,b)$ is a direction vector parallel to the line. 3. **Find a point on the line:** Set $x=0$ in the equation: $$2(0) - y - 3 = 0 \implies -y - 3 = 0 \implies y = -3$$ So, point $P_0 = (0, -3)$ lies on the line. 4. **Find the direction vector:** Rewrite the line as: $$y = 2x - 3$$ The slope is $2$, so a direction vector is: $$(1, 2)$$ 5. **Write parametric equations:** $$x = 0 + 1 \cdot t = t$$ $$y = -3 + 2 \cdot t = -3 + 2t$$ **Final parametric equations:** $$\boxed{x = t, \quad y = -3 + 2t}$$ This completes the first problem.