Question 638031
The distance between city A and B is 28 km. 
Two hikers left from each city (One from city A to city B and the other from
 city B to city A) heading towards each other at the same speed.
The hiker from city A rested for an hour after having walked 9 km and then
 continued at a speed greater than 1 km/hour than his previous speed, then
 rested again 4 km away from city B. 
There he met the hiker from city B, who walked at a constant speed without
 resting until the two met.
 What were the speeds of the two hikers?
:
Let s = speed of the two hikers
then
(s+1) = speed of the A after his 1 hr rest
:
Find the distance hiker A traveled after his rest, to meet B, 4 km from City B
28-9-4 = 15 km
:
Write a time equation; time = dist/speed
{{{9/s}}} + {{{15/(s+1)}}} + 1 = {{{4/s}}}
:
{{{9/s}}} + {{{15/(s+1)}}} = {{{4/s}}} - 1
:
{{{(9(s+1)+15s)/(s(s+1))}}} = {{{(4-s)/s}}}
:
{{{(9s+9+15s)/(s(s+1))}}} = {{{(4-s)/s}}}
:
{{{(24s+9)/(s(s+1))}}} = {{{(4-s)/s}}}
Multiply both sides by s
{{{(24s+9)/((s+1))}}} = (4 - s)
24s + 9 = (s+1)(4-s)
FOIL the right side
24s + 9 = 4s - s^2  + 4 - s
24s + 9 = 3s - s^2 + 4 
Combine on the left
s^2 + 24s - 3s + 9 - 4 = 0
s^2 + 21s + 5 = 0
Actually this equation has two negative solutions so the given scenario is not possible.
It seems unreasonable, to find a speed, where A travels 24 km (+ a rest),
while B travels 4 km