recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%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%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)

Answer:
Both answers are correct
Step-by-step explanation:
both y = 1 - 3/2x and y = 1 + -3/2x are correct answers, it is just that y= 1 - 3/2x is the simpler version. Adding a negative number is the same as subtracting a positive.
2 is greater then -5.
2 is a positive number unlike -5 it is a negative which is less then 2.
Subtract.
1 1/3 - 2/3
Convert 1 1/3 into an improper fraction:
4/3 - 2/3
Subtract the numerators together:
2/3
So he read 2/3 more time on Saturday than on Friday.
Y intercept when x =0
<span> y=-x+3
y = -0 +3
y = 3
</span><span> the slope is -1 </span>
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