1. **State the problem:** We have two rectangles, one inside the other. The larger rectangle has side lengths $2x + 1$ and $4x + 3$. The smaller rectangle inside has side lengths $2x + 6$ and $x + 2$. We need to find the difference in their perimeters. 2. **Recall the formula for the perimeter of a rectangle:** $$\text{Perimeter} = 2(\text{length} + \text{width})$$ 3. **Calculate the perimeter of the larger rectangle:** $$P_{large} = 2((2x + 1) + (4x + 3))$$ Simplify inside the parentheses: $$(2x + 1) + (4x + 3) = 2x + 4x + 1 + 3 = 6x + 4$$ So, $$P_{large} = 2(6x + 4) = 12x + 8$$ 4. **Calculate the perimeter of the smaller rectangle:** $$P_{small} = 2((2x + 6) + (x + 2))$$ Simplify inside the parentheses: $$(2x + 6) + (x + 2) = 2x + x + 6 + 2 = 3x + 8$$ So, $$P_{small} = 2(3x + 8) = 6x + 16$$ 5. **Find the difference in perimeters:** $$\text{Difference} = P_{large} - P_{small} = (12x + 8) - (6x + 16)$$ Simplify: $$12x + 8 - 6x - 16 = (12x - 6x) + (8 - 16) = 6x - 8$$ **Final answer:** The difference in the perimeters of the rectangles is $$6x - 8$$.