Let u = the shell rate in still water (in miles per hour) and
let v = the rate of the current.


Then you have TWO "time" equations


 +  =     (1)    ("8 miles downstream and back in 1 1/2 hours")

 +  =    (2)    ("12 miles downstream and halfway back in 1 1/2 hours)


To solve the syste, introduce new variables  x = ,  y = .

Then the system takes the form

 8x + 8y = 

12x + 6y = .


Simplify it to

 x + y =     (3)

2x + y =     (4)


Subtract  eq(3)  from eq(4), eliminating "y".  You will get

 x =  -  =  -  =  = .


Then from (3)  y =  -  =  = .


It implies

u + v =  = 16,   (5)

u - v =  =  8.   (6)


Add equations (5) and (6), eliminating "v". You will get

2u = 16 + 8 = 24  ====>  u =  = 12.


Then from eq(5)  v = 16 - 12 = 4.


Answer.  The speed of the shell in still water is  12 mph.  The current rate is 4 mph.