Me with this ! in still water, boat a travels 20km/hr and boat b travels 25km/hr. boat a travels 4 hours upstream against the current. boat b travels 2 hours downstream with the current. boat a travels 10 km further than boat b.

determine 2 equations that could be used to solve for the speed of current "c"
and "d" the distance traveled by boat a. then determine the distance boat a travels.

D=rt

distance of boat a is 10 more than b
da=10+db

boat atime=4
at=4
boat btime=2
bt=2

c is current

rate against current is r-c
rate with current is r+c
a goes agains
b goes with
a rate=20-c
ar=20-c
b rate=25+c
br=25+c

so

ad=bd+10

ad=ar*at
ad=(20-c)(4)

bd=br*bt
bd=(25+c)(2)
sub

(20-c)(4)=(25+c)(2)+10
divide both sides by 2
(20-c)(2)=(25+c)+5
distribute
40-2c=25+c+5
40-2c=30+c
minus 30 both sides
10-2c=c
add 2c both sides
10=3c
divide both sides by 3
3.3333=c

curent=10/3kmph

equations are
4(20-c)=d
2(25+c)+10=d

current=10/3km/hr

62926281916 inches of rain: )