Question 129878
Distance(d) equals Rate(r) times Time(t) or d=rt;r=d/t and t=d/r

Let r=speed of plane A
Then r+50=speed of plane B

Time for plane A=d/r=2800/r
Time for plane B=d/r=2000/(r+50)

Now we are told that the travelling time for plane B is three hours less than the travelling time for plane A, so our equation to solve is:

2000/(r+50)=(2800/r)-3  multiply each term by r(r+50)

2000r=2800(r+50)-3r(r+50)  get rid of parens

2000r=2800r+140000-3r^2-150r   subtract 2000r from each side

2000r-2000r=2800r-2000r+140000-3r^2-150r collect like terms

0=650r+140000-3r^2  multiply each term by -1

3r^2-650r-140000=0  quadratic in standard form  solve using the quadratic formula

{{{r = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{r = (650 +- sqrt( 422500+4*3*140000 ))/(2*3) }}} 
{{{r = (650 +- sqrt( 2102500 ))/(6) }}} 
{{{r = (650 +-1450)/(6) }}}

Disregard the negative value for r; rates are positive in this case
{{{r = (650 +1450)/(6) }}}
{{{r = (2100)/(6) }}}
{{{r = 350 }}} km/hr-------------------speed of plane A

{{{r+50=350+50=400}}}km/hr--------------------speed of plane B

CK

Time for plane A=2800/350=8 hr
Time for plane B=2000/400=5 hr
8-5=3
3=3

Hope this helps---ptaylor