
now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then
![\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2016%7D%7D%7By%20%3D%2016%20-%2028%7D%5Cimplies%20y%20%3D%20-12%20%5C%5C%5C%5C%5C%5C%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2012%7D%7D%7By%20%3D%2012%20-%2028%7D%5Cimplies%20y%20%3D%20-16%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7Bx%7D%7B16%7D~~%2C~~%5Cstackrel%7By%7D%7B-12%7D%29%5Cqquad%2C%5Cqquad%20%28%5Cstackrel%7Bx%7D%7B12%7D~~%2C~~%5Cstackrel%7By%7D%7B-16%7D%29~%5Chfill)
Given:
Nick has x toy race cars. This is one-third of what Bob has.
So, the number of Nick toys = one-third of what Bob has.
Let the Bob has y toys
So, we can write the following expression:

Solve for y, multiply both sides by (3):

So, the answer will be:
Bob has (3x) toy race cars.
An area That rises above surrounding landforms and has a rounded top; lower and usually less steep than a mountain. ... A high, flat-topped landform with cliff-like sides; larger than a butte.
Hope this helps!
For this case, the first thing we must do is define a variable.
We have then:
p: rate in miles per hour for the last 1.5 hours
We now write the equation that models the problem.
We have then:

Rewriting we have:

From here, we clear the value of p.
We have then:
Answer:
DeAngelo's rate for the last 1.5 hours of his run is 7 miles per hour.
Slope is -3 and y intercept is 2