1. The problem is to analyze the system of inequalities: $$y \leq x + 10$$ $$y > 6x + 2$$ 2. These inequalities represent two regions on the coordinate plane. The first inequality includes all points on or below the line $y = x + 10$. The second inequality includes all points strictly above the line $y = 6x + 2$. 3. To find the solution region, we need to find where these two conditions overlap. 4. First, find the intersection point of the two lines by setting: $$x + 10 = 6x + 2$$ 5. Solve for $x$: $$x + 10 = 6x + 2$$ $$x - 6x = 2 - 10$$ $$\cancel{5}x - \cancel{5}x = -8$$ $$-5x = -8$$ $$x = \frac{-8}{-5} = \frac{8}{5} = 1.6$$ 6. Substitute $x = 1.6$ into one of the lines to find $y$: $$y = 1.6 + 10 = 11.6$$ 7. The intersection point is $(1.6, 11.6)$. 8. The solution region is all points $y$ such that: $$6x + 2 < y \leq x + 10$$ and $x$ values where this is possible, which is for $x < 1.6$ because for $x > 1.6$, $6x + 2$ is greater than $x + 10$. Final answer: The solution set is the region between the lines $y = 6x + 2$ (not including the line) and $y = x + 10$ (including the line) for $x < 1.6$.