The slope of the line is -10.5.
let's recall that d = rt, distance = rate * time.
we know that Steve is twice as fast as Jill, so say if Jill has a speed or rate of "r", then Steve is traveling at 2r fast, now we know they both in opposite directions have covered a total of 120 miles, so if Jill covered "d" miles then Steve covered 120 -d, check the picture below.
![\begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Jill&d&r&2.5\\ Steve&120-d&2r&2.5 \end{array}~\hfill \begin{cases} d=2.5r\\[2em] 120-d=5r \end{cases} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{120-2.5r=5r\implies 120=7.5r}\implies \cfrac{120}{7.5}=r\implies \stackrel{Jill's}{16=r}~\hfill \stackrel{Steve's}{32}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Jill%26d%26r%262.5%5C%5C%20Steve%26120-d%262r%262.5%20%5Cend%7Barray%7D~%5Chfill%20%5Cbegin%7Bcases%7D%20d%3D2.5r%5C%5C%5B2em%5D%20120-d%3D5r%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B120-2.5r%3D5r%5Cimplies%20120%3D7.5r%7D%5Cimplies%20%5Ccfrac%7B120%7D%7B7.5%7D%3Dr%5Cimplies%20%5Cstackrel%7BJill%27s%7D%7B16%3Dr%7D~%5Chfill%20%5Cstackrel%7BSteve%27s%7D%7B32%7D)
(24 m)*(4 m) = (5 m)*height . . . . . area is the same no matter which base you use
height = (4/5)*(24 m) = 19.2 m
Y=4 when X=1
So it mean y=4(1)
Y=4
X equal 1 when they don't give you the number