How do I solve this word problem?

Question

I was finishing up my AP Calculus I assignment and I came across a question that confused me. Anyone know how to solve it? Step-by-step instructions would be greatly appreciated :)!

Two planes start from the same airport and fly in opposite directions. The second plane starts one-half hour after the first plane, but its speed is 80 kilometers per hour faster. Find the air speed of each plane if 2 hours after the first plane departs the planes are 3200 kilometers apart.

Answer 1

Speed of the 1st plane

2x + ([1 - 1/2][x + 80]) = 3,200

2x + ([1 1/2][x + 80]) = 3,200

2x + (3/2[x + 80]) = 3,200

2x + 3/2x + 120 = 3,200

4/2x + 3/2x = 3,080

7/2x = 3,080

x = 880

Speed of the 2nd plane:

= 880 + 80

= 960

Answer: 1st plane, 880 kph; 2nd plane, 960 kph.

Proof (distance apart is 3,200 km):

= 2 hrs(880 kph) + ([2 - 1/2 hrs][960 kph])

= 1,760 km + ([1 1/2 hrs][960 kph])

= 1,760 km + (3/2 hrs[960 kph])

= 1,760 km + 1,440 kph

= 3,200 km

Answer 2

I solved this problem like this:

v = velocity @ 2hours travel time

v + 80 = velocity @ 1.5 hours travel time

s = distance for v

3200 - x = distance for v + 80

--------------------------------------...

2v = 3200 - s

1.5(v + 80) = s

-----------------------------------

2v + s = 3200

1.5v - s = 120

------------------------------------

3.5v = 3080

v = 880

v + 80 = 960