Yes, similarily you could plot your two equations  x+3*y=200  and  2*x + 2*y=300

and again use solve on your equations to find the intersect point.  The procedure is very similar.  Hopefully I am answering your question properly. 

If you mean to add the intersect point as a text inserted into the graph then that's a little different, you can use textplot for that.  If you want the intersect point on the inequal graph, assign it a variable and use it with display and the pointplot of your intersect

Okay, so now you have the x and y points of the intersection, and now you want to display those intersection points.

b := allvalues(solve({y = 3*x+5, y = 2*x^2-2}, {x, y}))

         b := {x = 3/4+(1/4)*sqrt(65), y = 29/4+3*sqrt(65)*(1/4)}, {x = 3/4-(1/4)*sqrt(65), y = 29/4-3*sqrt(65)*(1/4)}

c := [[rhs(b[1, 1]), rhs(b[1, 2])], [rhs(b[2, 1]), rhs(b[2, 2])]]

        c := [[3/4+(1/4)*sqrt(65), 29/4+3*sqrt(65)*(1/4)], [3/4-(1/4)*sqrt(65), 29/4-3*sqrt(65)*(1/4)]]

with(plots):

d:=pointplot(c,symbol=circle, symbolsize=30, color=blue)

f := plot({3*x+5, 2*x^2-2})

display(d,f)

solve( { your equation1 , your equation 2 } , { x , y } )

Then use allvalues

plot({3*x+5, 2*x^2-2})

implicitplot might be what you are looking for.

The first question here has been answered about pointsizes.

My second question is that I want the point symbols to remain in 2d when I transform them over to a 3d graph.  If Maple offered a disc as a symbol option it would simplify things greatly.  So this is what I have come up with.

a := [[2, 3], [4, 5], [7, 2], [6, 9]]
with(plots):
pointplot(a, symbol = circle, symbolsize = 40, view = [0 .. 10, 0 .. 10])

b := pointplot(a, symbol = circle, symbolsize = 40, view = [0 .. 10, 0 .. 10])
with(plottools);
to3d := transform(proc (x, y) options operator, arrow; [x, y, 0] end proc);
display(to3d(b), axes = boxed);

However when Maple transformed the points it made them into spheres which I do not want.  So it seems I must create them manually.

disc := (c1, c2, r) -> [c1+r*cos(t), c2+r*sin(t), t = 0 .. 2*Pi]:
discs := seq(disc(op(a[i]), .5), i = 1 .. nops(a)):

plot([discs], scaling = constrained, color = blue)

c := plot([discs], scaling = constrained, color = blue)
display(to3d(c), axes = boxed)

So now this is what I wanted here.  As I said earlier if Maple offered a sybol=disc option it would make this operation much easier. 

Is this the only way to do this?  or is there an easier way?

It is confusing.

For example how would I extract weather data in Waterloo from the environment Canada website located here http://www.climate.weatheroffice.gc.ca/climateData/dailydata_e.html?Prov=XX&timeframe=2&StationID=32008&Day=1&Month=4&Year=2010&cmdB1=Go and capture the data? 

Exactly what I was looking for.  Thanks!

As an animation, yes.  But viewpoint doesn't appear to be an option for Maple 12.

How can we do that in 12? 

There is something called Javaviewlib for Maple but I don't use that and I'd rather not.  I'd like it to work completely within Maple.

Nice.  Thanks.  Both ways are better.

I noticed you used a command similar to

seq(p[1..2],p=a)

I don't think I've actually ever come across that way, much more eficient than the ways I've been using it. 

Could you also do this

e := seq( pointplot( a[1..2], symbolsize=trunc(p[3]*10) ), a=a ):

or is that really a no-no to have the variable you're working equal to the same variable?

Both plots are invisible in Maple 12.  All that's visible is the t=0..1 at the top of the plot.

Thanks. 

It appears the myriad of combinations I tried did not include select with lowercase.  I was using Select and not select.

Seems there is a difference and I got confused when Select didn't work.  Thanks for clearing things up even more.

Yes exaclty what I wanted.  I'm only starting to get use map, it is a little confusing.  I got a little confused with the format, but I think this has cleared it up. 

Thanks! 

Now could I apply it that way if I wanted a negative at the 2nd element of each list? 
Again I came up with something rather bloated

a:=map(a->[a[1],-a[2],a[3],a[4]],a)

My method wouldn't work too well if I had a large list of 100 element lists. 

anything easier?

Just a small correction to avoid some confusion.

a:=[seq(  your sum  ,n=1..i),i=1..  number of points you want  )]

hope it helps