1. **State the problem:** Solve the system of inequalities by graphing: $$y \leq \frac{1}{10}x + 6$$ $$y < -4x + 2$$ 2. **Understand the inequalities:** - The first inequality is $y \leq \frac{1}{10}x + 6$, which means the solution includes the line $y = \frac{1}{10}x + 6$ (solid line) and all points below it. - The second inequality is $y < -4x + 2$, which means the solution includes points strictly below the line $y = -4x + 2$ (dotted line). 3. **Graph each line:** - For $y = \frac{1}{10}x + 6$, the slope is $\frac{1}{10}$ and the y-intercept is 6. - For $y = -4x + 2$, the slope is $-4$ and the y-intercept is 2. 4. **Shade the regions:** - Shade below the solid line $y = \frac{1}{10}x + 6$. - Shade below the dotted line $y = -4x + 2$. 5. **Find the solution set:** - The solution to the system is the intersection of the shaded regions, i.e., points that satisfy both inequalities simultaneously. **Final answer:** The solution set is all points below or on the line $y = \frac{1}{10}x + 6$ and strictly below the line $y = -4x + 2$. This can be expressed as: $$\{(x,y) \mid y \leq \frac{1}{10}x + 6 \text{ and } y < -4x + 2\}$$