Question

One student begins in Cleveland, Ohio, and walks towards Buffalo, New York, a distance of 200 miles. She walks at the rate of 2.5 miles per hour. Another student begins riding his bicycle in Cleveland and rides towards Buffalo at the rate of 16 miles per hour. The walking student begins her journey at 8:00 AM. The riding student starts 4 hours later.

Answer 1

Part A Write a system of two equations that could be used to find the distance each student is from Cleveland, Ohio, at any given time. Let x represent the distance that the walking student is from Cleveland, let y represent the distance that the riding student is from Cleveland, and let t be the time that has elapsed since the walking student left Cleveland.
Part B Solve the system of equations for x and y to determine how many hours elapse before the students meet.

Answer 2

What method is used for this?
What is the equation?

Answer 3

lot of rules for this one
i am sure we can do it

Answer 4

@satellite73 yeah have no idea what method to use or how the equation would even look like

Answer 5

i think mostly it is distance is rate times time

Answer 6

She walks at the rate of 2.5 miles per hour
her rate is \(2.5\) so if her time is \(t\) then her distance is \(2.5t\)

Answer 7

similarly, the riding students distance is \(16t\) BUT the walker has a four hour head start

Answer 8

she walked for four hours before the riding student started
in four hours she went \(2.5\times 4=10\) miles

Answer 9

so add on 10 miles to her distance to make it
\[2.5t+10\] and the rider is \[16t\] and evidently they meet when their distances are the same
set
\[2.5t+10=16t\] and solve for \(t\)

Answer 10

@satellite73 but doesn't there need to be an x and y?!

Answer 11

and two equations?

Answer 12

ok really i object because it makes it sound like god give us one way to solve a problem
i really don't know how to do it with an \(x\) and \(y\) but the above answer i assure you is correct
we can try to bend to their method, but i am not sure how to do it in a natural way

Answer 13

@satellite73 sorry i was just going with what the problem was asking

Answer 14

x represent the distance that the walking student is from Cleveland
ok lets put
\[x=2.5t+10\]

Answer 15

and put
\[y=16t\]

Answer 16

@satellite73 yeah that makes more sense now

Answer 17

and since they have to be equal (as they meet at the same distance) set
\[2.5t+10=16t\] and solve for \(t\)

Answer 18

you good from there?

Answer 19

@satellite73 i think so..? just trying to figure out how i would solve it

Answer 20

subtract \(2.5t\) from both sides

Answer 21

then divide

Answer 22

10/13.5t ?!!?

Answer 23

i.e.
\[t=\frac{10}{13.5}\]

Answer 24

that is what i get, yes

Answer 25

so t = 1.35 ?!

Answer 26

no

Answer 27

\[\frac{13.5}{10}=1.35\] this is upside down

Answer 28

\[\frac{10}{13.5}=\frac{100}{135}\] which i guess you can reduce or turn in to a decimal

Answer 29

2.5(0.740)+10 = 11.85 16(0.740) = 11.84

Answer 30

yeah i got a time of \(.74\) hours

Answer 31

@satellite73 thank you!!

Answer 32

yw