Question 322395
Distance(d) equals Rate (r) times Time(t) or d=rt; r=d/t and t=d/r
Let r1=rate of stream
And r2=rate of the boat
r2-r1=rate upstream
r2+r1=rate downstream
On the first run:
time upstream=30/(r2-r1)
time downstream=44/(r2+r1)
and we are told that the above times equals 10 hrs, so:
30/(r2-r1) + 44/(r2+r1)=10-----------------------eq1
On the second run:
time upstream=40/(r2-r1)
time downstream=55/(r2+r1)
And these times equals 13 hours, so:
40/(r2-r1)+55/(r2+r1)=13----------------------eq2
In eq1, multiply each term by (r2-r1)(r2+r1) and we get:
30(r2+r1)+44(r2-r1)=10(r2^2-r1^2) simplify
30r2+30r1+44r2-44r1=10r2^2-10r1^2
74r2-14r1=10r2^2-10r1^2-----------------eq1a
Let try to simplify eq2 in the same manner
40(r2+r1)+55(r2-r1)=13(r2^2-r1^2)
40r2+40r1+55r2-55r1=13r2^2-13r1^2
95r2-15r1=13r2^2-13r1^2------------eq2a

From eq1a, (r2^2-r1^2)=(74r2-14r1)/10.  Substitute this into eq2a:
95r2-15r1=13(74r2-14r1)/10) multiply each side by 10
950r2-150r1=962r2-182r1 simplify
-150r1+182r1=962r2-950r2
32r1=12r2 or
8r1=3r2
r1=(3/8)r2 and r1^2=(9/64)r2^2; substitute these into eq1a
74r2-14(3/8)r2=10r2^2-10(9/64)r2^2 multiply each term by 64
4736r2-336r2=640r2^2-90r2^2 simplify
4400r2=550r2^2 divide each side by 10r2
440=55r2
r2=8 mph-----------------------------------------speed of boat
r1=(3/8)r2=(3/8)*8=3 mph ------------------------speed of stream

CK
30/(8-3)=6hr
44/(8+3)=4hr
10=10
40/5=8 hr
55/11=5 hrs
13=13

You can look it over.  There may be an easier way.
Hope this helps---ptaylor