Math problem? Algebra 1

Connie and Ron walked to the dock at 3 miles per hour, jumped into the boat, and motored to Dillion at 11 miles per hour. If the total distance as 33 miles and the trip took 5 hours in all, how far did they go by boat?

Please dont guess, i need an answer that you know is right! Thanks :)

dock distance:a
boat distance:b
a/3 +b/11=5
a+b=33
a=33-b
(33-b)/3+b/11=5
then b=33*6/8=198/8=24.75 miles
quite easy u should work harder ;)

Distance = speed x time

call time walking w
call time boating b

We know that w+b=5

And to get the distance we know that

3w+11b=33

This is 2 equations and 2 unknowns

since we want b (how long they boated) we solve the first equation for w in terms of b

w=5-b

now sub into second for

3(5-b)+11b=33

15-3b+11b=33
8b=18
b=9/4 hours
to get distance we multiply by speed 11miles/hour

Distance by boat = 99/4 = 24.75 miles

set the total time they walked is "x" hour
the time of the boat travel is "y" hour

D=VT (distance=velocity*time)

3x+11y=33 ---a

x+y=5 ------> 3x+3y=15---b

then do the subtraction: a-b

you will get: 8y=18

y=9/4 (hr.)

so the answer is:

(9/4)*11=99/4 (miles)

Let x be the time they traveled by boat
then the time they walk is (5 - x)

Distance by walking: 3(5 - x)
Distance by boating: 11x
Total distance is 33
therefore

11x + 3(5 - x) = 33
11x + 15 - 3x = 33
8x = 18
x = 18/8 = 2.25 hours
11 * 2.25 = 24.75 miles <======