```Question 524416
You need 2 equations: 1 for uphill and
1 for downhill
{{{d = 15}}} mi is the same for both trips
Let {{{t}}} = the time to go uphill in hrs
{{{ 2 - t }}} is the speed going downhill in hrs
Let {{{s}}} = the speed going uphill
{{{ s + 20 }}} is the speed going downhill
----------------------------
Going uphill:
(1) {{{ 15 = s*t }}}
Going downhill:
(2) {{{ 15 = ( s+20 )*( 2-t ) }}}
-----------------------
(2) {{{ 15 = 2s + 40 - s*t - 20t }}}
(2) {{{ 20t - 2s = 25 - s*t }}}
Substitute (1) into (2)
(2) {{{ 20t - 2s = 25 - 15 }}}
(2) {{{ 20t - 2s = 10 }}}
-----------------
(1) {{{ 15 = s*t }}}
(1) {{{ t = 15/s }}}
Substitute this into (2)
(2) {{{ 20*(15/s) - 2s = 10 }}}
(2) {{{ 300 - 2s^2 = 10s }}}
(2) {{{ 2s^2 + 10s - 300 = 0 }}}
(2) {{{ s^2 + 5s - 150 = 0 }}}
{{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 1 }}}
{{{ b = 5 }}}
{{{ c = -150 }}}
{{{s = (-5 +- sqrt( 5^2-4*1*(-150) ))/(2*1) }}}
{{{s = (-5 +- sqrt( 25 + 600 )) / 2 }}}
{{{s = (-5 +- sqrt( 625 )) / 2 }}}
{{{s = (-5 +- 25) / 2 }}}
{{{ s = 20/2 }}} ( can't use the negative root )
{{{ s = 10 }}}
The ride uphill is 10 mi/hr
check:
(1) {{{ 15 = s*t }}}
(1) {{{ 15 = 10t }}}
(1) {{{ t = 1.5 }}} hrs
and
(2) {{{ 15 = ( s+20 )*( 2-t ) }}}
(2) {{{ 15 = ( 10+20 )*( 2-t ) }}}
(2) {{{ 15 = 30*( 2-t ) }}}
(2) {{{ 15 = 60 - 30t }}}
(2) {{{ 30t = 60 - 15 }}}
(2) {{{ 30t = 45 }}}
(2) {{{ t = 1.5 }}} hrs
OK```