Algebra problem using the method of elimination.

I am trying to figure this problem out.

Two planes start from the same airport and fly in opposite directions. The second plane starts 1/2 hour after the first plane, but its speed is 33 miles per hour faster. Find the airspeed of each plane if, 2 hours after the first plane departs, the planes are 3000 miles apart.




STEP 1:

Let x be the speed of the plane that leaves first and let y be the speed of the plane that leaves second. Complete the following system of equations which describes this situation.


33 = y - x
3000 = 2x + 3 (y/2)

STEP 2:

Find the solution to the system of equations in Step 1. Enter your answer in the form (x, y).


STEP 3:

Rewrite your result from Step 2 in the context of the problem.
The first plane that leaves the airport travels at 843 mph.
The second plane that leaves the airport travels at _____ mph.

Can anyone help me figure out the mph for the second plane. 10 points to the best answer!! Thank you so much.

-x + y = 33
2x + (3/2)y = 3000

multiply first equation by 2 and add to the second equation
-2x + 2y = 66
2x + (3/2)y = 3000
---------------------------
(7/2)y = 3066

y = 3066(2/7) = 876 miles per hour

-x + 876 = 33
x = 843 miles per hour

(x, y) = (876, 843)

Let speed of plane A = x mph
After 1/2 h travels x/2 miles
Speed of plane B = x + 33 mph

After 2h plane A has travelled 2x miles
After 2 hours plane B has travelled (3/2)(x + 33) miles

2x + (3/2)(x + 33) = 300
4x + 3x + 99 = 6000
7x = 5901
x = 843

Speed of first plane = 843 mph
Speed of second plane = 876 mph

y=x+33
=843+33
=876 mph