Can someone help me out on this word problem?

Question

Two planes, which are 3780 miles apart, fly toward each other. Their speeds differ by 45MPH. If they pass each other in 4 hours, what is the speed of each?

Answer 1

Let A and B be the speeds of the planes... wlog let A>B so we have A = B+45. Now if 4 hours they travel all together 3780 miles so we also have A*4 + B*4 = 3780. Now with 2 equations and 2 unknows solve the system to get
A=495MPH
B=450MPH

Answer 2

With two planes flying toward each other their closing speed is the sum of their two speeds.
Let x = the speed of one plane
Let x+45 = the speed of the other plane
Distance = Velocity x Time, so
Velocity = Distance/Time

Now substitute the values from above:

x = (x=45) =3780 miles/4 hours

2x + 45 = 945
2x= 900
x=450 miles/hour for plane 1

x+45 = 450 + 45 = 495 miles/hour for plane 2

Answer 3

When the planes pass each other, the sum of the distance each had traveled will be 3780 miles. Call the rate of the faster plane 'A' and the rate of the slower plane 'B.'
We know that A = B + 45.
And distance = rate × time, so:
(B)(4 hrs) + (B + 45)(4 hrs) = 3780 miles.
Solve for 'B' to get the rate of the slower plane.

4B + 4B + 180 = 3780,
8B = 3600,
B = 400,
A = 445.

Don't forget to check my work!
Feel free to email me for any clarifications.

Answer 4

total distance is 3780 miles
speed of 1st plane = v
speed of second plane = v + 45
time = t = 4
distance first plane flew is vt =4v
distance second plane flew is (v+45)t = 4v +180
total distance is sum of the distance flown by each:

3780 = 4v + 4v + 180
3600 = 8v
v= 450 mph = first plane speed
v + 45 = 495 mph = second plane speed