Question 201288
Let {{{f}}} = fish's speed in still water in mi/hr
Let {{{c}}} = speed of current in mi/hr
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Given:
{{{2}}} hrs = time to go {{{8}}} mi downstream
{{{14}}} hrs = time to go {{{8}}} mi upstream
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{{{8 = (f + c)*2}}} mi
{{{8 = (f - c)*14}}} mi
and
(1) {{{8 = 2f + 2c}}} 
(2) {{{8 = 14f - 14c}}}
Multiply both sides of (1) by {{{7}}} and
add the equations
(1) {{{56 = 14f + 14c}}} 
(2) {{{8 = 14f - 14c}}}
{{{28f = 64}}}
{{{f = 2.286}}} mi/hr answer
check answer:
{{{8 = (f + c)*2}}}
{{{8 = (2.286 + c)*2}}}
{{{8 = 4.571 + 2c}}}
{{{2c = 3.429}}}
{{{c = 1.714}}} mi/hr
and
{{{8 = (f - c)*14}}} 
{{{8 = (2.286 - 1.714)*14}}} 
{{{8 = .572*14}}}
{{{8 = 8.008}}} close enough