Question 181929
Two Air Force planes, one a jet and the other propeller-driven plane, left an
 air base at the same time and flew to another base 600 miles away.
 The jet's average speed was 300 miles per hour greater than that of the
 propeller-driven plane and its flying time was 2 and 2/3 hours less.
 Find the average speed of each plane.
:
Let s = speed of the prop plane
then
(s+300) = speed of the jet plane
:
Write a time equation; time = {{{dist/speed}}}
Prop time - jet time = 2{{{2/3}}} hrs
:
{{{600/s}}} -{{{600/((s+300))}}} = {{{8/3}}}
:
Multiply equation by 3s(s+300) to get rid of those annoying denominators
600*3(s+300) - 600*3s = 8s(s+300)
:
1800s + 540000 - 1800s = 8s^2 + 2400s
:
do the math and arrange as a quadratic equation
8s^2 + 2400s - 540000 = 0
;
simplify, divide by 8
s^2 + 300s - 67500 = 0
:
With a little sweat we can get this to factor
(s + 450) (s - 150) = 0
the positive solution:
s = +150 mph is the prop plane
:
I'll let you find the jet plane speed
:
:
Check solution in original equation