<PRE><B>Question 162019</B>
ROSE DROVE HER CAR FROM HER HOME TO THE PROVINCE AND BACK, A TOTAL DISTANCE OF
 120KM. HER AVERAGE SPEED RETURNING, WAS 3KPH SLOWER THAN HER AVERAGE SPEED
 GOING, TO THE PROVINCE. IF HER TOTAL DRIVING TIME WAS 9 HOURS,
WHAT WAS HER AVERAGE SPEED IN GOING TO THE PROVINCE?
:
From the given information we can assume that one way; 120/2 = 60 km
:
Let s = her speed going 
and
(s-3) = her speed returning
:
Write a time equation: Time = {{{dist/speed}}}
Go time + return time = 9 hrs
{{{60/s}}} + {{{60/((s-3))}}} = 9
:
Eliminate the denominators, multiply equation by s(s-3), results:
60(s-3) + 60s = 9(s(s-3))
;
60s - 180 + 60s = 9(s^2 - 3s)
:
120s - 180 = 9s^2 - 27s
:
0 = 9s^2 - 27s - 120s + 180
A quadratic equation:
9s^2 - 147s + 180 = 0
This can be factored to
(3s - 45)(3s - 4) = 0
Two solutions
3s = 4
s = {{{4/3}}}
and
3s = 45
s = 15 km/h this seems to be the most reasonable
:
Speed going: 15 km/h; and speed returning 12 km/h
:
Check in original equation
60/15 + 60/12 = 
4 + 5 = 9 hrs</PRE>