alegbraic problem, can you solve?

Question

and explain how you did?

A group of students went on a trip. During the first day of their trip they walked with a constant speed of 6 km/h. On the next day they walked with a constant speed of 5 km/h. The time of walking on the second day was longer by 3 hours than the time of walking on the first day. They distance they walked on the second day was longer by 8km than the distance they walked on the second day. Find the total distance they walked during the two days.

Answer 1

The answer is 92 kms.

I'll explain.

The speed is given. We know that speed=distanct/time=x/y

on the first day speed = 6km/h

ie x/y=6 re writing we get x=6y and

x-6y=0------(1)

on the second day,they travel distance greater by 8km ie (x+8) and time 3 hours longer ie (y+3)

so we have (x+8)/(y+3)=5

rearrangin we get the second equation as x+8=5y+15 or

x-5y=7------(2)

Solving (1) & (2) we have x=42 and y=7
now this x and y denotes just the distance and time on the first day.
So first day's distance covered is 42kms. Second day is 8km more than that. Hence second day is 50kms.

Therefore the entire distance is 42+50=92kms

oof..!

Answer 2

ok...hope i can make this clear....

for day one, speed is 6kmh. let the total time be x hrs.
so the total distance walked for day one is 6x (speed x time) km.

for day two, speed is 5 kmh. total time is (x + 3 )hrs
so the total distance for the second day is ( 5x + 15) km.

second days distance is longer than first days distance by 8 km, so :
( 5x +15) - 8 = 6x
therefore, x = 7

so total distance walked is :

6x + ( 5x + 15) =
6(7) + [ 5(7) + 15] = 92 km.

hope i'm right !