algebra word problem. pleeeease help.

[Cassandra123]

Ive been working on this problem for an hour and i cannot solve this.

A fire truck is called to a scene. Three minutes later, a second truck is called. The first truck averages only 30mi/h, but the the second averages 60mi/h. The trucks travel a total of 12 miles and arrive at the same time.
How long from the first call did the trucks take to arrive?
How far did each travel?

Please show me how you got the answer so i can understand it. Thank You!

 

[mmm4444bot]

Let x = the second truck's time

The distance driven by the second truck (I'll call it d2) can be written as rate (60) times time (x).

d2 = 60x

The first truck's time is 3/60 hours longer than the second truck's time. Its distance is also rate times time.

d1 = 30(x + 3/60)

The total distance driven by both trucks is 12 miles.

d1 + d2 = 12

30(x + 3) + 60x = 12

I set up an equation for you. Can you solve it for x, and then answer the two questions?

 

[BigGlenntheHeavy]

\ \ \ \ \ \ \ \ \(\displaystyle Rate \ \ \ \ X \ \ \ \ Time \ \ \ \ = \ \ \ \ Distance\)

\(\displaystyle 1st \ Truck \ \ \ \ 30 \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30x\)

\(\displaystyle 2nd \ Truck \ \ \ \ 60 \ \ \ \ \ \ \ \ \ x-1/20 \ \ \ \ \ \ \ \ 60x-3\)

\(\displaystyle Hence, \ 30x+ 60x-3 \ = \ 12, \ \implies \ x \ = \ \frac{1}{6} \ hrs \ = \ 10 \ min.\)

\(\displaystyle Therefore, \ first \ truck \ drove \ 5 \ miles \ and \ second \ truck \ drove \ 7 \ miles,\)

\(\displaystyle and \it \ took \ 7 \ min. \ from \ the \ first \ call \ for \ the \ trucks \ to \ arrive.\)