Topic review (newest first)
- 2012-08-10 06:03:44
I'll check into it and probably some free ones too.
- 2012-08-10 04:15:55
That's exactly what I needed! Thanks a million.
- bob bundy
- 2012-08-10 03:57:14
It's called Geometer's Sketchpad.
It's not free like Geogebra, but has some excellent features.
- 2012-08-10 01:35:05
Nice graphs! What program are you using to get those?
- bob bundy
- 2012-08-09 17:29:04
Arrrhhh! At last the question makes sense.
Q2 shown in red.
Q3 shown in blue
Q5. shown in green.
You have asked about x + y = 3 before because I remember making a diagram for this.
You just need the x and y numbers to add up to a total of 3. eg. (3,0) (2,1) (1,2) (0,3)
Q6 shown in purple.
I plotted the b point (0,-1) and then used the gradient ... 3 across 1 up, to make more points.
- 2012-08-09 14:28:54
Oh, shame on me. You guys were right (of course).
Okay, I found the other points, "the given points" are points I have to find. I made an error in thinking I already had them all it it was confusing me greatly.
HOPEFULLY! This works better. Sorry about my screw up guys..
2. (0, -4) (-4/3, 0) and (1, -1) (2, -2)
3. (0, -5)(- 15/4, 0) and
y = x + 2
m = 1
b = 2
5. (2, 4) (3, 6) and (2, 1) (I can't find the other point for some reason, grr..) The equation the point needs to be found from is x + y = 3.
6. I found m = 1/3 b =-1 from x = 3 - 3y
(0, -2) and (-2, 0)
Sorry for my mistakes. I hope this works better! Thanks for all the help!
- 2012-08-09 07:46:35
The quote says "Graph the following using the given points using the system of linear equations."
This instruction is poorly worded. I've taught algebra for over 40 years but I can't understand what
this is asking. Perhaps if it comes from a book and the book has had several examples using this
terminology one could figure out what is being asked.
As mentioned in the previous posts it take TWO points to determine a straight line. Then it takes
TWO such lines to determine a system of equations and a point of crossing (if they cross. Distinct
parallel lines do not cross.)
I would think that since the question says "using the system of linear equations" that one would have
to first get the equations for the two lines (like #2 and #3 together). Then one could solve for the
exact point where they cross. Often it is difficult to get better than just an approximation to the
point where they cross since the point may involve fractions or decimals: difficult to figure out from
a drawn graph. If using a graphing calculator then one can zoom in to get better approximations.
Also one doesn't need the system of equations to just graph the lines. The system is needed to
get an accurate point of intersection.
May you be much blessed in your mathematical endeavors!
- bob bundy
- 2012-08-08 17:12:34
Well one point is not enough to fix a line. (see latest diagram)
So what I did in post 2 is the best I can offer for this. Sorry.
- 2012-08-08 09:43:55
It says, "Graph the following using the given points using the system of linear equations."
The ones I can't get are:
2. (0, -4) and (-4/3, 0)
3. (0, -5) and (- 15/4, 0)
5. (2, 4) and (2, 6)
6. (0, -2) and (-2, 0)
And under each problem would be a graph where I draw my lines.
- bob bundy
- 2012-08-08 07:38:53
I understand the theory of systems of linear equations.
What I want is the exact wording of the questions you have been set.
It is not possible to draw a single line where only one point is given.
When you know two points, you can join them with a line.
Then if you have another two points, you can join them with another line.
Then you can look to see where the lines cross.
I have decided to assume that Q2 and Q3 are meant to be taken as such a system.
The diagram below then shows what answer you would get.
If that is not what the questions ask, you will have to give the exact wording of the questions.
- 2012-08-08 04:26:34
the graph needs to lines, one line from (0, -4) and another from (-4/3, 0)
Takes two points to determine a line. There are bajillions of lines that go through (0,-4).
- 2012-08-08 03:21:03
Okay, I'm going to try to explain it better.
My lesson is about liner equations, which states, "A system of linear equations is two or more linear equations that are solved at the same time. To solve a system of linear equations by graphing, we graph the equations on one coordinate plane. The solution is the point where the lines cross."
So, the line doesn't need to join them together. The graph needs two lines, not one line joining the coordinates. So, using the coordinates from #2; the graph needs to lines, one line from (0, -4) and another from (-4/3, 0). Not one line made from joining both coordinates.
Hopefully, this clarifies. By the way, I'm not sure if the lines will cross each other on every graph. It's okay if they're not..
- bob bundy
- 2012-08-07 17:42:34
I'm confused about what you want.
That is two points so I showed them in red on my diagram with a line joining them.
That's another two points, shown this time in blue, with a line joining them.
Perhaps you need to post the actual question wording that went with these coordinates.
- 2012-08-07 05:52:25
There needs to be two lines on each graph, once graphed. *Each 2 point sets* are for one line. So for #2, (0, -4) is for one line and the other point set is for another line. That's one graph.
- bob bundy
- 2012-08-07 03:42:57
Why are they all on one graph?
Strictly speaking, my diagram shows four graphs on a single set of axes.
And I did this to save time.
If you need four separate answers, just make 4 separate diagrams.
But sometimes, you need to put two or more lines onto a single set of axes, so you can see where they cross.