Their ages form an arithmetic sequence with n=15 terms and common
difference d=1.5.  So we use the formula:

an = a1 + (n-1)d

a15 = a1 + (15-1)(1.5)

a15 = a1 + (14)(1.5)

a15 = a1 + 21

The eldest's age is a15
The youngest's age is a1

So we have the system of two equations in two unknowns: 

  

Solve by substitution or elimination and get

a1 = 3 and a15 = 24

The youngest is 3 and the eldest is 24

-------------------------
Checking:

Their ages are 

3, 4.5, 6, 7.5, 9, 10.5, 12, 13.5, 15, 16.5, 18, 19.5, 21, 22.5, 24

That's 15 ages, all 1.5 years apart and 24 is 8 times 3.
So it checks.

Edwin