Question 162664
Let time for 1st part  = {{{t}}} hrs
Then the time for the 2nd part will be {{{3 - t}}}hrs
The speed for the 1st part is {{{s}}}
The speed for the 2nd part is {{{s - 5}}} mi/hr
The distance for the 1st part was {{{69}}} mi
The distance for the 2nd part was {{{2}}} mi
(1) {{{69 = s*t}}}
(2) {{{2 = (s - 5)*(3 - t)}}}
{{{2 = 3s - 15 - st + 5t}}}
{{{2 = 3s - 15 - 69 + 5t}}}
{{{3s = 2 + 15 + 69 - 5t}}}
{{{3s = 86 - 5t}}}
Substituting (1) into this,
{{{3s = -5*(69/s) + 86}}}
Multiply both sides by {{{s}}}
{{{3s^2 = -345 + 86s}}}
{{{3s^2 - 86s + 345 = 0}}}
Use quadratic equation
{{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 3}}}
{{{b = -86}}}
{{{c = 345}}}
{{{s = (86 +- sqrt( 86^2-4*3*345 ))/(2*3) }}}  
{{{s = (86 +- sqrt(7396 - 4140))/6}}}
{{{s = (86 +- sqrt(3256))/6}}}
{{{s = (86 + 57.06)/6}}}
{{{s = (86 - 57.06)/6}}}
I'll pick the answer that makes the most sense
{{{s = 23.84}}}
and
{{{s = 4.82}}}(I can't go 5 mph slower than this, so I'll pick
the 1st answer)
{{{s - 5 = 18.84}}}
The canoeist went 23.84 mi/hr on the 1st part of the trip and
18.84 mi/hr on the 2nd part
check answer:
(1) {{{69 = s*t}}}
(2) {{{2 = (s - 5)*(3 - t)}}}
{{{69 = 23.84t}}}
{{{t = 2.89}}}hrs
(2) {{{2 = 18.84*(3 - 2.89)}}}
{{{2 = 18.84*.11}}}
{{{2 = 2.07}}} (error is due to rounding off)
Hope I got it right!