I don't see any place where anyone has shown two 5's instead of
one 5; there are supposed to be
four 5's!
In the book's picture, they show every value as "5 ... 5", for example, putting the number of them below; I showed what they mean, by actually writing out that many of each value, but it's easier to keep track of if you write what they wrote, while visualizing it as what I wrote.
Jomo took a non-algebraic approach equivalent to what they did; do you follow his explanation?
Here's a slower explanation of their work:
There are 4+8+x, that is, 12+x, numbers less than 3; and 6 numbers equal to 3; and 5+4, that is, 9, numbers greater than 3. In order to get a median of 3, the middle number, which is the median, must have equal numbers on either side.
The first "3" in the list has 12+x numbers on its left, and 14 numbers on its right (namely, five 3's and 9 others). So if this is the middle number, we need to have 12+x = 14. That's their left equation. It yields x=2 as the largest possible value of x, because this way we are moving the 3's as far as possible to the right.
Similarly, the last "3" in the list has 17+x numbers on its left (namely, five 3's and 12 others), and 9 numbers on its right. So if this is the middle number, we need to have 17+x = 9. That's their right equation. It yields x=-8; but we can't have a negative number, so the smallest value of x is 0.
In fact, if x=0, there are no 2's, and that list is
0 0 0 0 1 1 1 1 1 1 1 1 3 3 3 3 3 3 4 4 4 4 4 5 5 5 5
There are 4+8+6+5+4 = 27 numbers here, so the middle one is at position (27+1)/2 = 14, with 13 on its left and 13 on its right. I marked that one in bold above; that's a 3, so we have the desired median.
