1. **State the problem:** Find the equation in the form $y = mx + 6$ of the line containing the point $(2, 5)$ and perpendicular to the line $2x + y = -3$. 2. **Identify the slope of the given line:** Rewrite $2x + y = -3$ in slope-intercept form $y = mx + b$. $$y = -2x - 3$$ The slope $m$ of this line is $-2$. 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another with slope $m$ is the negative reciprocal, so $$m_{perp} = \frac{1}{2}$$ 4. **Use the point-slope form:** The line passes through $(2, 5)$ with slope $\frac{1}{2}$. $$y - 5 = \frac{1}{2}(x - 2)$$ 5. **Simplify to slope-intercept form:** $$y - 5 = \frac{1}{2}x - 1$$ $$y = \frac{1}{2}x + 4$$ 6. **Adjust to the form $y = mx + 6$:** We want the constant term to be 6, so rewrite as $$y = \frac{1}{2}x + 6 - 2$$ or equivalently, $$y = \frac{1}{2}x + 6 - 2$$ This shows the line with $y = mx + 6$ form is not exactly the line through $(2,5)$ perpendicular to $2x + y = -3$ unless the constant is 4. Therefore, the equation of the line perpendicular to $2x + y = -3$ passing through $(2,5)$ is $$y = \frac{1}{2}x + 4$$ which cannot be written exactly as $y = mx + 6$ without changing the line. **Final answer:** $$y = \frac{1}{2}x + 4$$