Question 193861
The equation is 
distance = rate x time but with a modification:
Let {{{r[s]}}}= rate of the boat in still water in mi/hr
Let {{{r[c]}}}= rate of the current in mi/hr
Let {{{t[u]}}}= time going upstream in hrs
Let {{{t[d]}}}= time going downstream in hrs
{{{r[s] - r[c]}}}= rate of boat against the current in mi/hr
{{{r[s] + r[c]}}}= rate of the boat going with the current in mi/hr
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So, going upstream, I can write:
(1) {{{d = (r[s] - r[c])*t[u]}}}
And going downstream, I can write:
(2) {{{d = (r[s] + r[c])*t[d]}}}
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Given:
{{{d = 5}}} mi
{{{t[u] = t[d] + 20}}} hrs
{{{r[c] = 4}}} mi/hr
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Now I can rewrite (1) and (2)
(1) {{{5 = (r[s] - 4)*(t[d] + 20)}}}
(2) {{{5 = (r[s] + 4)*t[d]}}}
This is 2 equation with 2 unknowns, so it's solvable
(1) {{{5 = (r[s] - 4)*(t[d] + 20)}}}
(1) {{{5 = r[s]*t[d] - 4t[d] + 20r[s] - 80}}}
(1) {{{r[s]*t[d] + 20r[s] = 4t[d] + 85}}}
(1) {{{r[s]*(t[d] + 20) = 4t[d] + 85}}}
And, from (2)
(2) {{{5 = (r[s] + 4)*t[d]}}}
(2) {{{5 = r[s]*t[d] + 4t[d]}}}
(2) {{{r[s]*t[d] = - 4t[d] + 5}}}
(2) {{{r[s] = -4 + 5/t[d]}}}
substitute this in (1)
(1) {{{(-4 + 5/t[d])*(t[d] + 20) = 4t[d] + 85}}}
(1) {{{-4t[d] + 5 - 80 + 100/t[d] = 4t[d] + 85}}}
(1) {{{8*(t[d])^2 + 160t[d] - 100 = 0}}}
(1) {{{2*(t[d])^2 + 40t[d] - 25 = 0}}}
Using quadratic equation:
{{{t[d] = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 2}}}
{{{b = 40}}}
{{{c = -25}}}
{{{t[d] = (-40 +- sqrt( 40^2-4*2*(-25) ))/(2*2) }}} 
{{{t[d] = (-40 +- sqrt( 1600 + 200 ))/4 }}} 
{{{t[d] = (-40 +- 42.426)/4 }}}
{{{t[d] = 2.426/4}}}
{{{t[d] = .6065}}} hrs
I want to find {{{r[s]}}}
(2) {{{r[s] = -4 + 5/t[d]}}}
(2) {{{r[s] = -4 + 5/.6065}}}
(2) {{{r[s] = -4 + 8.244}}}
(2) {{{r[s] = 4.244}}} mi/hr answer
check answer:
(1) {{{5 = (r[s] - 4)*(t[d] + 20)}}}
(1) {{{5 = (4.244 - 4)*(.6065 + 20)}}}
(1) {{{5 = .244*20.6065}}}
(1) {{{5 = 5.027}}} (rounding error?)
and
(2) {{{5 = (r[s] + 4)*t[d]}}}
(2) {{{5 = (4.244 + 4)*.6065}}}
(2) {{{5 = 8.244*.6065}}}
(2) {{{5 = 5}}}