1. **State the problem:** We are given the line $p$ with equation $6x + 3y = 15$. We need to find the equation of a line parallel to $p$ that passes through the point $(-1, 1)$. 2. **Find the slope of line $p$:** Rewrite $6x + 3y = 15$ in slope-intercept form $y = mx + b$. $$3y = -6x + 15$$ $$y = -2x + 5$$ The slope $m$ of line $p$ is $-2$. 3. **Slope of the parallel line:** Parallel lines have the same slope. So, the slope of the new line is also $-2$. 4. **Use point-slope form to find the new line's equation:** The point-slope form is: $$y - y_1 = m(x - x_1)$$ Using point $(-1, 1)$ and slope $-2$: $$y - 1 = -2(x - (-1))$$ $$y - 1 = -2(x + 1)$$ $$y - 1 = -2x - 2$$ $$y = -2x - 1$$ 5. **Final answer:** - The slope of the new line is $-2$. - The equation of the new line is $y = -2x - 1$.