Algebra- distance equation. HELP!?

Question

Hi. I need to solve this problem, but I cannot figure out what formula to use to solve it.

John drove to Seattle from Tacoma at an average speed of 60mph. On the way back, he drove at 50mph. The total trip took 4.4 hours. What is the distance between the 2 cities?

Thanks for your help! :)

Answer 1

Let D be the distance.

D = 60 * t1
D = 50 * t2
t1 + t2 = 4.4

Subtract the first two equations:
0 = 60 * t1 - 50 * t2

Now substitute t1 = 4.4 - t2
60(4.4 - t2) - 50t2 = 0

Solve for t2:
264 - 60t2 - 50t2 = 0
264 = 110t2
t2 = 264/110
t2 = 2.4 hours

Solve for t1:
t1 = 4.4 - 2.4
t1 = 2.0 hours

Solve for D:
D = 2.0*60
D = 120 miles

Answer 2

X = distance between citis
T(1) = time of first trip
T(2) = time of second trip
V = velocity
X = V * T

x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x

X = 60 * T(1) => T(1) = X / 60
X = 50 * T(2) => T(2) = X / 50

T(1) + T(2) = 4.4

x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x

X / 60 + X / 50 = 4.4 => X = 120 mile

Answer 3

since the distance he traveled when going and coming bak are the same, u can set up this equation:
t = time he spent driving to tacoma
4.4-5 = time he spent coming bak
60(t) = 50(4.4-t)
60t = 220 - 50t
110t = 220
t = 2 hours

since 60t = distance
distance = 60mph(2 hours) = 120m

Answer 4

airplane A flew at x mph and lined (x*3) mi in 3h. airplane B flew at 2x mph and lined (2x*3) mi in 3h. entire distance between 2 planes is 2700 mi 2700 mi = (x*3) mi + (2x*3) mi 2700 = 3x + 6x 2700 = 9x x = 3 hundred mph hence, airplane A flew 3 hundred mph airplane B flew 3 hundred*2 = six hundred mph