1. ## word problem

a train covered 120 miles at a certian speed. Had the train had been able to travel 10 miles per hour faster, the trip would have been 2 hours shorter. how fast did the train go

2. $
\frac{{120}}
{{V + 10}} + 2 = \frac{{120}}
{V}
$

...

3. why are u adding 2 woundt be -2 because the trip would be two hours shorter?

4. i would be sooo lost lol

5. I think you're right, should be -2

because 120/v is the longer time (with the velocity it has)
120/(v+10) is the shorter time (faster velocity),
so you need to put -2 to the longer time, in order to equal the shorter time.

6. ok but i am still lost what do i do next?

ok but i am still lost what do i do next?
What Peritus did was use a simple $Time=\frac{Distance}{Velocity}$ forumula

Distance=D=120 in miles
Veolicty=V in miles/hours
Time=T hours

So for the first train
$T=\frac{120}{V}$
For the second train Velocity is increased by 10 and time is decreased by 2 hours. So
$T-2=\frac{120}{V+10}$
$T=\frac{120}{V+10}+2=\frac{120}{V}$

Then just solve for V.
It will be a quadratic giving you -30 and 20 but -30 is inadmisable because the train would be traveling backwards so your answer is V=20miles per hours

a train covered 120 miles at a certian speed. Had the train had been able to travel 10 miles per hour faster, the trip would have been 2 hours shorter. how fast did the train go
case 1
LEt the distance covered = 120

let x be the speed of the train and let y be the time taken

Speed=distance/time
x=120/y

so xy = 120 ------------------> equation 1

case 2

If the train traveled 10 mph faster then time is reduced by 2 hours

so x+10 = 120/y-2

(x+10)(y-2) = 120

multiplying we get

xy -2x +10y -20 = 120
from equation 1,

120 - 2x + 10y - 20 = 120
subtracting 120 on both sides

-2x + 10y -20 = 0

so x = (10y - 20)/2

x = 5y - 10

substituting this in equation 1 we get

(5y-10)y = 120

5y^2 -10y - 120 = 0

so now its a quadratic equation, we see that 5 is common on both sides so we divede 5 on both sides

y^2 -2y - 24 = 0

factorizing we get

(y+4)(y-6) = 0

so either y= -4 or y = 6

but time can't be negative so we conclude that the time taken is 6 hours, substituting this in equation 1 we get x = 120/6

x = 20 mph

so the train was traveling at 20 mph, it's so slow, you can use the bus

Regards,
Ice Sync