Question 990904
 Kiran drove from Tortula to Cactus, a distance of 212 mi.
 She increased her speed by 13 mi/h for the 385 mi trip from Cactus to Dry Junction.
 If the total trip took 10 h, what was her speed from Tortula to Cactus?
:
let s = speed from T to C
then
(s+13) = speed from C to D
:
Write a time equation, time = dist/speed
{{{212/s}}} + {{{385/((s+13))}}} = 10 hrs
multiply equation by s(s+13)
s(s+13)*{{{212/s}}} + s(s+13)*{{{385/((s+13))}}} = 10s(s+13)
Cancel the denominators
212(s+13) + 385s = 10s^2 + 130s 
212s + 2756 + 385s = 10s^2 + 130s
597s + 2756 = 10s^2 + 130s
Arrange as a quadratic equation
10s^2 + 130s - 597s - 2756 = 0
10s^2 - 467s - 2756 = 0
Use the quadratic formula to find s: a=10; b=-467; c=-2756
I got a positive solution of 52 mph from T to C
;
:
you can check this for yourself
{{{212/52}}} + {{{385/65}}} =