Question 412453
{{{3}}} mi/hr is the speed of current
Let {{{s}}} = speed of kayak in still water
going downstream:
speed = {{{s + 3}}}
going upstream:
speed = {{{s -3}}}
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given:
(1) {{{4 = (s - 3)*t
and
(2) {{{10 = (s + 3)*t}}}
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This is 2 equation and 2 unknowns, so it's solvable
(1) {{{4 = (s - 3)*t

(1) {{{4 = s*t - 3t}}}
(2) {{{10 = s*t + 3t}}}
Subtract (1) from (2)
(2) {{{10 = s*t + 3t}}}
(1) {{{-4 = -s*t + 3t}}}
{{{6 = 6t}}}
{{{t = 1}}} hr
Now I plug this into (1) and solve for {{{s}}}
(1) {{{4 = s*1 - 3*1}}}
(1) {{{4 = s - 3}}}
(1) {{{s = 7}}}
The speed of the kayak in still water is 7 mi/hr
check answer:
(2) {{{10 = (s + 3)*t}}}
(2) {{{10 = (7 + 3)*1}}}
{{{10 = 10}}}
OK