Math question help please?

Question

two trains left a station at the same time. one traveled north at a certain time and other traveled south twice the speed. after 4 hours, the trains were 600 miles apart. how fast was each train traveling?

Answer 1

QUESTION 1:

Create the equation to explain what you want to solve.  Four times (the number of hours) times the speed of Train #1 (x) and Train #2 (2x) since it was travelling at twice the speed.

4(x + 2x) = 600

Expand...

4x + 8x = 600

12x = 600

x = 50

The basic speed of the train was 50 mph.

Proof:

Train #1 travelled at 50 mph for 4 hours = 200

Train #2 travelled at 100 mph for 4 hours = 400

Total distance travelled = 600.

QUESTION #2:

Follow the same system.

Mike's bus for 3 hours at 20 mph slower than the other bus...

3(x-20)

Adam's bus for 3 hours...

3x

Equation:

3(x-20) + 3x = 270

Solve by diving by 3:

(x-20) + x = 90

is the same as

x - 20 + x = 90

2x = 90 + 20

2x = 110

x = 55

Proof:

3 (55-20) + 3 (55) = 270

3 (35) + 3 (55) = 270

105 + 165 = 270

Answer 2

to get the distance a train travelled, we multiply the speed by the time.

so let's say the northbound train (the slower one) was travelling n m/h and it travelled for 4 hours. then the other train was going 2n m/h for 4 hours (twice the speed).

so their combined distance is n*4 + 2n*4 and we know that this is equal to 600 miles

so:

n*4 + 2n*4 = 600

4n + 8n = 600

12n = 600

n = 50

since n is the speed of the slower train, it was travelling 50 m/h and the other was going 100 m/h

problem 2)

again, distance = speed*time

let adams speed be n m/h and mike's speed be (n - 20) m/h.

adams distance is n*3

mike's distance is (n - 20)*3

together they travelled n*3 + (n - 20)*3

and we know that together they travelled 270 miles. so:

n*3 + (n - 20)*3 = 270

3n + 3n - 60 = 270

6n - 60 = 270

6n = 330

n = 55

so adam was going 55 m/h and mike was going 35 m/h