Question 657891
Let s be the speed of the slower plane and s+75 be the speed of the faster plane.  

The time it takes the slower plane to make the flight is {{{1000/s}}}
The time it takes the faster plane to make the flight is {{{1000/(s+75)}}}

The difference in these two times is 3 hours, so
{{{1000/s - 1000/(s+75) = 3}}}
{{{((s+75)/(s+75))*(1000/s) - (s/s)*(1000/(s+75)) = 3}}}
{{{(1000*s+75000)/(s^2+75*s) - (1000*s)/(s^2+75*s) = 3}}}
{{{(1000*s + 75000 - 1000*s)/(s^2 + 75*s) = 3}}}
{{{75000/(s^2 + 75*s) = 3}}}
{{{(s^2 + 75*s)*(75000/(s^2 + 75*s)) = (s^2 + 75*s)*3}}}
{{{75000 = 3*s^2 + 225*s}}}
{{{25000 = s^2 + 75*s}}}
{{{s^2 + 75s - 25000 = 0}}}
We can solve for s using the quadratic equation {{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} with a=1, b=75, and c=-25000
{{{s = (-75 +- sqrt( 75^2-4*1*(-25000) ))/(2*1) }}}
{{{s = (-75 +- sqrt( 5625+100000 ))/2 }}}
{{{s = (-75 +- sqrt( 105625 ))/2 }}}
{{{s = (-75 +- 325)/2 }}}
So s = -200 or s = 125.  We'll ignore the negative answer because the planes are not flying backward.
So the slower plane is flying at 125 mph and completes the trip in 8 hours.
The faster plane is flying at 200 mph and completes the trip in 5 hours.