Question

Travel: It took a motorboat 1 hour longer to travel 36 miles against he current than to go 36 miles with the current. The rate of the current was 3 mph. Find the rate of the boat in calm water.

Answer 1

answer is 15 and the check is easy

Answer 2

the work is less so, but i can write it for you if you like

Answer 3

pls? I had y+3*1 and y-3*1 but I somehow think that is wrong :)

Answer 4

ok we can work it out like this

Answer 5

distance is 36
rate with current is x + 3
rate against current is x - 3 and using
\[T=\frac{D}{R}\] we know the time with current is
\[\frac{36}{x+3}\] and the time against the current is
\[\frac{36}{x-3}\]

Answer 6

and we know that the time against the current is one hour more than the time with the current, so we set
\[\frac{36}{x+3}+1=\frac{36}{x-3}\] and solve for x

Answer 7

so far so good?

Answer 8

ok so i was almost there. I'm following your logic

Answer 9

ok good. again the time against the current is one more than the time with, so if we add 1 to the time with the current we get the time against the current. that is what the equation says. now we have to solve for x which takes a bit of algebra. i will write it

Answer 10

\[\frac{36}{x+3}+1=\frac{36+x+3}{x+3}=\frac{39+x}{x+3}\]
so we get
\[\frac{39+x}{x+3}=\frac{36}{x-3}\]

Answer 11

cross multiply to get
\[(x-3)(x+39)=36(x+3)\]

Answer 12

what happened to the +1? Does get ignored? I understand the 36+3 and the 39-3.

Answer 13

multiply out to get
\[x^2-36x-117=36x+108\] giving
\[x^2-225=0\] or \[x = \sqrt{225}=15\]

Answer 14

we added the one to
\[\frac{36}{x+3}\][

Answer 15

again
\[\frac{36}{x+3}+1=\frac{36}{x+3}+\frac{x+3}{x+3}=\frac{36+x+3}{x+3}=
\frac{x+39}{x+3}\]

Answer 16

ok...it just clicked. :)

Answer 17

got it?

Answer 18

got it! TY!!!!!!

Answer 19

yw. btw answer is easy to check 15+3=18 and 36/18=2
15-3=12 and 36/12=3, one hour longer

Answer 20

TY! You did a great job of explaining this to me!