1. The problem is to find the solution region for the system of inequalities: $$y \leq 2x + 1$$ $$y > 3 - x$$ 2. These inequalities represent two half-planes on the coordinate plane. The first inequality includes points on or below the line $y = 2x + 1$. 3. The second inequality includes points strictly above the line $y = 3 - x$. 4. To find the solution region, we need to find where these two conditions overlap. 5. First, find the intersection of the boundary lines by solving: $$2x + 1 = 3 - x$$ 6. Add $x$ to both sides: $$3x + 1 = 3$$ 7. Subtract 1 from both sides: $$3x = 2$$ 8. Divide both sides by 3: $$x = \frac{2}{3}$$ 9. Substitute $x = \frac{2}{3}$ into one of the lines to find $y$: $$y = 2\left(\frac{2}{3}\right) + 1 = \frac{4}{3} + 1 = \frac{7}{3}$$ 10. The lines intersect at the point $\left(\frac{2}{3}, \frac{7}{3}\right)$. 11. The solution region is the set of points $\left(x,y\right)$ such that: $$y \leq 2x + 1$$ $$y > 3 - x$$ and lies between these two lines, including the boundary line $y = 2x + 1$ but excluding the boundary line $y = 3 - x$. Final answer: The solution is the region between the lines $y = 2x + 1$ (inclusive) and $y = 3 - x$ (exclusive), intersecting at $\left(\frac{2}{3}, \frac{7}{3}\right)$.