Step 1: Split the compound inequality into two separate inequalities. The given inequality is $\frac{1}{2}(5x-4) \le x + 5\frac{1}{2} < 3x + 1\frac{1}{2}$. This can be written as: *a) $\frac{1}{2}(5x-4) \le x + 5\frac{1}{2}$ *b) $x + 5\frac{1}{2} < 3x + 1\frac{1}{2}$ Step 2: Solve the first inequality. $$ \frac{1}{2}(5x-4) \le x + 5\frac{1}{2} $$ Convert the mixed number to an improper fraction: $5\frac{1}{2} = \frac{11}{2}$. $$ \frac{1}{2}(5x-4) \le x + \frac{11}{2} $$ Multiply the entire inequality by 2 to eliminate fractions: $$ 2 \times \frac{1}{2}(5x-4) \le 2 \times x + 2 \times \frac{11}{2} $$ $$ 5x-4 \le 2x + 11 $$ Subtract $2x$ from both sides: $$ 5x - 2x - 4 \le 11 $$ $$ 3x - 4 \le 11 $$ Add 4 to both sides: $$ 3x \le 11 + 4 $$ $$ 3x \le 15 $$ Divide by 3: $$ x \le 5 $$ Step 3: Solve the second inequality. $$ x + 5\frac{1}{2} < 3x + 1\frac{1}{2} $$ Convert mixed numbers to improper fractions: $5\frac{1}{2} = \frac{11}{2}$ and $1\frac{1}{2} = \frac{3}{2}$. $$ x + \frac{11}{2} < 3x + \frac{3}{2} $$ Multiply the entire inequality by 2 to eliminate fractions: $$ 2 \times x + 2 \times \frac{11}{2} < 2 \times 3x + 2 \times \frac{3}{2} $$ $$ 2x + 11 < 6x + 3 $$ Subtract $2x$ from both sides: $$ 11 < 6x - 2x + 3 $$ $$ 11 < 4x + 3 $$ Subtract 3 from both sides: $$ 11 - 3 < 4x $$ $$ 8 < 4x $$ Divide by 4: $$ \frac{8}{4} < x $$ $$ 2 < x $$ Step 4: Combine the solutions. From Step 2, we have $x \le 5$. From Step 3, we have $x > 2$. Combining these two conditions gives $2 < x \le 5$. Step 5: Find the integral values of $x$. The integral values of $x$ that are greater than 2 and less than or equal to 5 are 3, 4, and 5. The integral values are $\boxed{3, 4, 5}$. 3 done, 2 left today. You're making progress.