Can You Solve This Problem?

suppose jay has travlled x miles,ben has travelled 8-x miles and they meet after time t

for jay: d=x mile, t=t hour and r=5mph
therefore x=5 t
or t=x/5 ...(1)

for ben: d=(8-x) mile, t=t hour and r=7mph
therefore 8-x=7 t
or t=(8-x)/5 ...(2)
from eqns.1 and 2
we have
x/5 = (8-x)/7
or 7x = 40-5x
or x = 40/12 = 3.333 miles
so jay travels 3.333 miles
and ben travels =8-3.333=4.67miles,so ben travels longer distance
combined distance travelled is 8 miles
time at which they meet = 3.333/5=.6666 hrs or 66.66 mins

Let x be Jay's distance and y be Ben's distance.
x + y = 8 mile -- (1)
Since they meet, it took same time:
t = d/r = x/5 = y/7 => 7x = 5y --- (2)
Solving two simultaneous equations (1) and (2) gives:
7x = 5(8-x) => 7x = 40 - 5x => 12x = 40 => x = 3.33333... = 3 1/3
∴ y = 4.66666... = 4 2/3
t = x/5 = (3 1/3)/5 = (10/3)/5 = 2/3 (hr) = 40 (min)

It's been awhile; but, let me try: After 40"(40/60)x, Jay traveled 3 1/3 mi. and Ben traveled 4 2/3 miles. The total distance is 8 and the time is 40 minutes.
Check: d=rt 3 1/3mi=5mph(40"=2/3hr)
4 2/3mi=7mph(40"=2/3hr)

That can't be right though considering that in one hour, they each would individually have traveled 7 and 5 miles (Their MPH rate) and would have traveled a total distance of 12 miles, not 8. It has to be lower than an hour for them to meet within those 8 miles.

Jay and Ben will meet 40 minutes after they start running. Ben will have ran the longer distance, 4.67 miles, rounded to the nearest hundredth. Combined, they will have traveled 8 miles.