Can you please help with correct format for the following?: re: d = rt equation

Question

A boat traveled 40 miles down a river in 2 1/2 hours. The return trip going upstream took twice as long. What is the rate of the boat in still water?

Appreciate any help on the set up.

Thank you,

Answer 1

Easiest way is to use 2 variables

let x be the speed of the boat in still water
let y be the speed of the current

so going downstream the boat is going x+y mph
going against the current it is x-y mph

now use your formula, we need t = d/r

time upstream = 40/(x-y)
time downstream = 40/(x+y)

then 40/(x-y) + 40/(x+y) = 5/2

I would multiply each term by 2(x+y)(x-y)
or 2(x^2 - y^2)

80(x+y) + 80(x-y) = 5(x^2 - y^2)
80x + 80y + 80x - 80y = 5x^2 - 5y^2
simplify and divide by 5
32x = x^2 - y^2

but we are also told that
40/(x-y) = 2(40/(x+y))
1/(x-y) = 2/(x+y)
2x-2y = x+y
x = 3y

sub back in:
32(3y) = 9y^2 - y^2
96y = 8y^2
12y = y^2
y = 12
then x = 36

the boat can go 36 mph in still water, the current is 12 mph

check:
time to go upstream = 40/24 = 5/3
time to go downstream = 40/48 = 5/6

5/3 + 5/6 = 5/2 , as required