Sally and ron are taking a trip. sally left at 6 p. m. and traveled an average of 40 miles per hour. ron left at 11 am and traveled an average of 65 miles per hour. at what time are they at the same time? write a system of equation to represent thus situation. how many miles awat from home will they be at that time? _

Explanation:

Assuming Ron takes off at 11 am the next day , so that he left (6+11)=17 hours later

Let the time that they meet be x hours after Ron left, then

40*(17+x) = 65*(x)

solve for x

680+40x = 65x

680 = (65-40)x = 25x

x = 680/25 = 27.2 hours (after Ron left)

Time he catches up with her  = 27.2 + 11 = 38.2 = (38.2-24)= 14.2 hours on the third day, or 2:12 pm on the third day.

Solving by a system of equations:

Let x = time required for sally since she started

let y = time required for Ron since he started.

Since there is a lag of 17 hours, we have

x = y + 17 (1)

For them to travel the same distance,

40x = 65y (2)

Substitute (1) in (2)

40(y+17) = 65y

40y + 680 = 65y

680 = (65-40) y = 25y

y = 680/25 = 27.2 hours

So they would meet 27.2 hours after Ron started, or one day and 3 hours 12 minutes after 11:00 am, the next day.

So they would meet at 2:12 pm on the third day.

answer; enforce the rules;

x=3

step-by-step explanation:

this is really simple i saw that there was only one term (5x) and thought if that was 15 what would i have to multiply everything else to get it there? well if 5x became 15 then x would have to be 3 so plug the 3 in the x for the other equations and you get 15 for the other ones. so there you go. : )