Math - Word Problems?

Stumped yet agian !! :(?
If one-half of one integer is subtracted from three-fifths of the next consecutive integer, the difference is 3. What are the two integers?

I would guess that I would set up this equation as 1/2 - 3/5 = 3
But I am lost after that , and I am not even sure thats how I should set it up. Help !

Also I need help with this problem. It is the well know "a train taveling a such and such speed equation" THANK YOU

A passenger train can travel 325 miles in the same time a freight train takes to travel 200 miles. If the speed of the passenger train is 25 mi/hr faster than the speed of the freight train, find the speed of each.

First one:

1/2 of an integer:

1/2x

is subtracted from 3/5 of the next consecutive integer (which is x + 1):

3/5(x + 1) - 1/2x

And the difference is 3:

3/5x + 3/5 - 1/2x = 3

Multiply through by 10 to make it easier:

10(3/5)x + 10(3/5) - 10(1/2)x = 10(3)
6x + 6 - 5x = 30
x = 24

So the integers are 24 and 25.

Check:

3/5(25) - 1/2(24) = 15 - 12 = 3 check!

Next one:

distance (d) = rate (r) times time (t), so

time (t) = distance (d) over rate (r), or

t = d/r

The problem tells us two things:

1) The passenger train travels 325 miles (at a rate of p miles per hour) :

t = 325/p

in the same amount of time that the freight train travels 200 miles (at a rate of r miles per hour):

t = 200/r

Since both times are equal, you can set the equations equal too:

325/p = 200/r

Cross-multiplying:

325r = 200p

And dividing both sides by 25 to make it look nicer:

13r = 8p

2) The speed of the passenger train (p) is 25 miles per hour faster than the speed of the freight train (r):

r + 25 = p

So now you have your 2 equations:

r + 25 = p
13r = 8p

Substitute in for p in the second equation using the first equation:

13r = 8(r + 25) = 8r + 200
5r = 200
r = 40

Then plug that back into the equation for p:
p = r + 25 = 40 + 25 = 65

So the passenger train goes 65 miles per hour and the freight train goes 40 miles per hour.

Check:

The passenger train would travel 325 miles in 325 miles/65 miles per hour = 5 hours.

The freight train would travel 200 miles in 200 miles/40 miles per hour = 5 hours.

It works!

(1) 3/5*(i+1)-i/2=3
=> i=24
_______________
325/Vp=200/Vf
Vp=Vf+25

=> Vf=40, Vp=65
1. 3/5(x+1)-1/2x=3
3/5x-1/2x=3-3/5
1/10x=12/5
x=24 the integers are 24 & 25

2. a=ime the trains run, bmiles/hour
a x b = 200
a x (b+25) =325

200/a = (325/a) - 25
a=5, b=40
passenger train -- 65 miles/h, freight -- 40 miles/h
1. 3/5(x+1)-1/2x=3
3/5x-1/2x=3-3/5
1/10x=12/5
x=24 the integers are 24 & 25

2. a=ime the trains run, bmiles/hour
a x b = 200
a x (b+25) =325

200/a = (325/a) - 25
a=5, b=40
passenger train -- 65 miles/h, freight -- 40 miles/h
Ok, your first question
Let's X = the first integer
Based on your information, let X+1= second integer

3/5(X+1) - 1/2X=3
Multiply everything by 10 since least common denominator
6(x+1 - 5X= 30
6X+6 -5X= 30
X+6=30
X= 24
So X+1=25
So our integers are 24, and 25

Now Distance=Rate * time
Let the time both trains take is Y
Let X= speed of the freight train
Let X+25= speed of passenger train
freight train: 200= x*y
passenger train 325= (x+25) y
Solve for Y on both equation
200/x=y
325/(X+25)=y
Now since they both equal y, set them equal
200/x=325/(x+25)
cross multiply
200(x+25)=325x
200x + 5000 = 325x
5000= 125x
40 = x
65 = x+25
So speed of freight train is 40 mph
and passenger train is 65 mph