00:01
Hello, welcome to this lesson in this lesson.
00:04
We have a cyclist that rides from the car to the road for the distance of 40 miles and his return takes two hours longer because his speed decreased by 10 miles per hour so here the time we took from a to b is x hours and this was 40 miles in a time it took the return so when he was moving when he was leaving to rockford this was the time x hours when he was coming back this was a time x plus 2 all right, so the time delayed by two hours, all right.
00:50
So means he took two hours longer than he was going.
00:55
All right, so let's look at the speed in both ways so we have speed the speed one.
01:06
All right, which would be close to the 40 miles divided by x hours all right.
01:20
And now we have the speed 2 so let's have this as s and let's have this as t.
01:31
So his speed 2 would be the 40 miles divided by x plus 2 hours but we are given that his speed decreased by 10 miles per hour this means that we have x which is equals to t plus 10.
01:53
All right, and we have x equals to 40 on x we have t which is equal to 40 on x plus 2 so given this equation, this is equation 1 equation 2 and equation 3 we can find the speed that he went with and the speed that he returned with so let's substitute equation 2 and 3 into equation 1.
02:25
We'll have 40 on x which is equals 40 on x plus 2 plus 10 so here the lcm is x times x plus 2...