1. **Problem statement:** We are given the ordered numbers: $x$, $(x + 1)$, $5$, $(2x + 1)$, $(y + 2)$, $7$, $8$, and $(3x + 2)$. The mean of these numbers is 6.5, and the median is 7. We need to find the values of $x$ and $y$. 2. **Formula for mean:** $$\text{Mean} = \frac{\text{Sum of all numbers}}{\text{Number of numbers}}$$ 3. **Calculate the sum using the mean:** There are 8 numbers, so: $$\frac{x + (x+1) + 5 + (2x+1) + (y+2) + 7 + 8 + (3x+2)}{8} = 6.5$$ Multiply both sides by 8: $$x + (x+1) + 5 + (2x+1) + (y+2) + 7 + 8 + (3x+2) = 52$$ 4. **Simplify the sum:** Combine like terms: $$x + x + 1 + 5 + 2x + 1 + y + 2 + 7 + 8 + 3x + 2 = 52$$ Group $x$ terms: $$x + x + 2x + 3x = 7x$$ Constants: $$1 + 5 + 1 + 2 + 7 + 8 + 2 = 26$$ So the equation becomes: $$7x + y + 26 = 52$$ Subtract 26 from both sides: $$7x + y = 26$$ 5. **Median condition:** The median of 8 numbers is the average of the 4th and 5th numbers when the numbers are ordered. Median = 7, so: $$\frac{\text{4th number} + \text{5th number}}{2} = 7$$ $$\text{4th number} + \text{5th number} = 14$$ 6. **Order the numbers to find the 4th and 5th:** The numbers are: $$x, x+1, 5, 2x+1, y+2, 7, 8, 3x+2$$ We need to find $x$ and $y$ such that the median is 7. 7. **Assuming $x$ is such that the numbers in order place 7 as median:** Try to find $x$ so that the 4th and 5th numbers sum to 14. 8. **Try $x=2$:** Calculate numbers: $$2, 3, 5, 5, y+2, 7, 8, 8$$ Order: $$2, 3, 5, 5, 7, 8, 8, y+2$$ If $y+2$ is greater than 8, the 4th and 5th numbers are 5 and 7, sum 12, not 14. 9. **Try $x=3$:** Numbers: $$3, 4, 5, 7, y+2, 7, 8, 11$$ Order: $$3, 4, 5, 7, 7, 8, 11, y+2$$ If $y+2$ is greater than 11, 4th and 5th numbers are 7 and 7, sum 14, median 7. 10. **Use equation from step 4:** $$7x + y = 26$$ Substitute $x=3$: $$7(3) + y = 26$$ $$21 + y = 26$$ $$y = 5$$ 11. **Check $y+2$:** $$5 + 2 = 7$$ So the numbers are: $$3, 4, 5, 7, 7, 7, 8, 11$$ Ordered: $$3, 4, 5, 7, 7, 7, 8, 11$$ 4th and 5th numbers are both 7, median is 7. 12. **Final answer:** $$x = 3, \quad y = 5$$