Rectangular Coordinate System

[LadyWater]

Alright, I can't seem to understand what they're doing on this problem.

The book (Saxon Algebra 1, second ed.) tells me to graph the following equations on a rectangular coordinate system.

Problem #11: x= -11/2
Problem #13 : y= 11/3x - 2

I understand #11 perfectly, but #13 has me totally confused. I can't seem to get the image in here, so here's the link to the solution, if someone could explain it to me.

 

[Denis]

Problem #13 : y= 11/3x - 2

Hmmm...hard on the peepers...
looks like 1^(1/3) times (x-2);
since 1^(anything) = 1, then that's simply x - 2 ...I think...

 

[LadyWater]

Thanks for the quick reply Denis!

Have you ever had one of those days? Apparently I'm experiencing one.

Somehow I seem to have turned my negative sign into a one. The real problem is: y = -1/3x - 2

If you could find it in your heart to assist me again, it would be greatly appreciated.

 

[Gene]

Ya gotta stop in at Wal-Mart and pick up a bunch of ()s. Then use them so we know if you mean
y=-1/(3x) -2 or
-1/(3x-2) or
(-1/3)*x -2 or ???
We could use PEMDAS (that would be the last one) but so many times you people don't that we wold appreciate your using them even if they aren't necessary.

 

[Denis]

Probably is: y = (-1/3)x - 2 ; y = mx + b
(since graphing seems to be involved)
If that's the case, you got your slope (-1/3) and y-intercept (-2)....