Question 269393
Let {{{w}}} = wind speed
Let {{{b}}} = bike's speed with no wind
The trip that takes less time is with the wind
The 2 equations to use are:
{{{d = (b - w)*t[1]}}} (against the wind)
{{{d = (b + w)*t[2]}}} (with the wind)
given:
{{{d = 30}}}
{{{t[1] = 5}}} hrs
{{{t[2] = 3}}} hrs
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The only unknowns are {{{b}}} and {{{w}}},
so I've got 2 equations and 2 unknowns
(1) {{{30 = (b - w)*5}}}
(1) {{{30 = 5b - 5w}}}
(1) {{{6 = b - w}}}
and
(2) {{{30 = (b + w)*3}}}
(2) {{{30 = 3b + 3w}}}
(2) {{{10 = b + w}}}
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Add the equations
(2) {{{10 = b + w}}}
(1) {{{6 = b - w}}}
{{{16 = 2b}}}
{{{b = 8}}}
and, since
{{{10 = b + w}}}
{{{10 = 8 + w}}}
{{{w = 2}}}
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(a) 
The bicyclist's average speed for the 1st leg:
{{{b - w = 8 - 2}}}
{{{b - w = 6}}} mi/hr
The bicyclist's average speed for the 2nd leg:
{{{b + w = 8 + 2}}}
{{{b + w = 10}}} mi/hr
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(b)
{{{b - w = 6}}}
and
{{{b + w = 10}}}
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(c)
In still air the cyclist's speed is {{{b = 8}}} mi/hr
The wind seed is {{{w = 2}}} mi/hr