Let d=f(D) where d is Doug's age in dog years and D is the dog's actual age. and f is a function relating D to d.
7(D+R)=F-1 where R=Russel's age and F is Fred's age (all in normal years). F=7(D+R)+1.
d=R+13. In 6 years' time F+6=4d/3, assuming d is the dog's present age (not in 6 years' time).
Replacing d by f(D): f(D)=R+13, F+6=4f(D)/3, F=4f(D)/3-6.
Therefore,
7(D+R)+1=4f(D)/3-6, 7(D+R)=4f(D)/3-7. D+R=4f(D)/21-1, D+R=4(R+13)/21-1.
We have two unknowns but only one equation and f(D) has been eliminated. So we cannot find D and R uniquely without further information. We can relate D and R:
D+R=4R/21+52/21-1=4R/21+31/21, (21D+17R)/21=31/21, 21D+17R=31. We know that no ages can be negative, so that limits D to less than 2 years old (1.48) and R to less than 2 years old (1.82). Both ages can't be in whole numbers of years. If D=⅔ years old (8 months) then 21D=14, so 17R=31-14=17, R=1 year old. D+R=1⅔=5/3. That seems to be the best solution. d=f(D)=f(⅔)=13+1=14 years old (Doug's age in dog years). (Fred's age, F=7(D+R)+1=7(5/3)+1=38/3=12⅔ years old. So in this problem 8 months is equivalent to 14 years in dog years.)