SOLUTION: Here is my problem: Sarah went to Jen house at 10:00 with 40 mph.The distance from her house to Jen house is 221 miles. Later, at 12:00, Jen went to Sarah house at a speech of 54 m

Question 265640: Here is my problem: Sarah went to Jen house at 10:00 with 40 mph.The distance from her house to Jen house is 221 miles. Later, at 12:00, Jen went to Sarah house at a speech of 54 mph.How many miles will Sarah meet Jen?
Not: This is all i have remember.This is one of problems on the test.
I just want to make sure I did it right.
Here is what i have done so far
R*T=D
Sarah 40 t 40t
Jen 54 2-t 54(2-t)
40t=54(2-t)
40t=100-54t
+54t +54t
94t=100
t=1/3/47
40*1/3/47=42/26/47
it was a multiple choice test. i put 140 miles.im not sure,i guess on this one
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
It looks like Sarah got a headstart of 2 hrs.
How far did she get in 2 hrs?

mi
Now, if I had a stopwatch I'd start it so that
they are both heading towards eachother and will
meet somewhere between the houses.
I'll write equations for both:
(1)
(2)
The time is the same for both because I'll
stop the stopwatch when they meet.
The houses are 221 mi apart, but now Sarah and Jen
are mi apart, so I know

From (1) and (2) above,

Multiply both sides by

By substitution:

And, since

Jen has to drive 81 mi and Sarah has to drive 60 mi
check answer:
(1)

hr
and
(2)

hr
The times are the same -OK

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