Question 82909
The time spent driving plus the 2 hours at her mom's adds up to the
elapsed time of 3 hours (noon to 3 PM).
If s is the average speed to her mom's, then s+5 is the average speed
from her mom's to Sue's
{{{20/s + 2 + 15/(s+5) = 3}}}
This is just {{{t[1] + 2 + t[2] = 3}}}
{{{20/s + 15/(s+5) = 1}}}
Multiply both sides by {{{s(s+5)}}}
{{{20(s+5) + 15s = s(s+5)}}}
{{{20s + 100 + 15s = S^2 + 5s}}}
{{{s^2 - 30s - 100 = 0}}}
Use the formula
{{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 1}}}
{{{b = -30}}}
{{{c = -100}}}
{{{s = (-(-30) +- sqrt( (-30)^2-4*1*(-100) ))/(2*1) }}}
{{{s = (30 +- sqrt(900 + 400 ))/ 2 }}}
{{{s = (30 +- sqrt(1300 ))/ 2 }}}
{{{s = (30 +- 10*sqrt(13 ))/ 2 }}}
{{{s = 15 +- 5*sqrt(13)}}}
{{{s = 15 + 5*3.6055}}}
{{{s = 15 + 18.0278}}} (note that the other answer is negative and
therefore makes no sense)
{{{s = 33.028}}}
But the average speed going to Sue's is {{{s + 5}}}
{{{s + 5 = 33.028 + 5}}}
{{{s + 5 = 38.028}}} mph answer
check answer
{{{20/s + 2 + 15/(s+5) = 3}}}
Is this true when {{{s = 33.028}}}?
{{{20/33.028 + 2 + 15/38.028 = 3}}}
{{{.6056 + 2 + .3944 = 3}}}
{{{1 + 2 = 3}}}
OK