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#51 Re: Help Me ! » Adding and Subtracting Polynomials » 2012-08-12 15:51:06
Okay, thanks guys.
I have a few more similar problems from my next lesson and I'm going to post them and my answers. I would appreciate it someone check them for me and tell me where/how/what the correct answer is.
I think I'm understanding it. It seems really simple actually but I need to get used to and get a grasp on it. It's the simplifying part I don't understand how to do, I don't understand that part crystal clear yet. Oh, and the moving terms part. For example; (8x^2 x + 2) Does the negative sign or positive sign belong to the x? I don't know things like that.. Never done this before.
#52 Re: Help Me ! » Adding and Subtracting Polynomials » 2012-08-11 07:31:15
Awesome!
For;
12. (4x^2 x 5) + (6x^2 + x)
4x^2 - 6x^2 - 5 + x
10x^2 + 5
#53 Re: Help Me ! » Adding and Subtracting Polynomials » 2012-08-10 05:56:32
I don't even know what latex is bobbym..
Now 11:
11. \\(9x^2 + 3x 5) + (8x^2 x + 2)\\
So far I have, \\8x^2 + 11x - 3\\ and \\2x^2 - 1x - 1\\
Is that correct? What's next or is it done..?
I used the method from post #4
#54 Re: Help Me ! » Adding and Subtracting Polynomials » 2012-08-09 07:14:50
For 6 - 10:
6. 5x2 3x 7
The degree of this one is 2
7. Was already done for me.
8. 5x6 + 6x5 4x3 + x2 45
This one is 6.
9. 3x 4
It's 1 maybe?
10. 8x5 7x3 + 4
this one is 5.
Are these correct?
#55 Re: Help Me ! » How Do You Graph These? » 2012-08-09 06:15:55
That's exactly what I needed! Thanks a million. 
#56 Re: Help Me ! » How Do You Graph These? » 2012-08-08 16:28:54
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
and
(0, -2) and (-2, 0)
Sorry for my mistakes. I hope this works better! Thanks for all the help!
#57 Help Me ! » Adding and Subtracting Polynomials » 2012-08-08 07:25:38
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These are the problems I got wrong on my newest lesson. I've tried a couple times to correct them and keep coming up with the wrong answer. If somebody could solve and show their so I can understand how to do it, that would be awesome!
The Questions I got wrong:
6. 5x2 3x 7
7. 4x4 3x2 + 7x 11
8. 5x6 + 6x5 4x3 + x2 45
9. 3x 4
10. 8x5 7x3 + 4
For 7 - 10 it says, " identify the degree of the polynomial."
11. (9x2 + 3x 5) + (8x2 x + 2)
12. (4x2 x 5) + (6x2 + x)
For those two it says, "add or subtract the given polynomials."
THANKS! 
#58 Re: Help Me ! » How Do You Graph These? » 2012-08-07 11:43:55
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.
#59 Re: Help Me ! » How Do You Graph These? » 2012-08-07 05:21:03
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..
Thanks!
#60 Re: Help Me ! » How Do You Graph These? » 2012-08-06 07:52:25
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.
#61 Re: Help Me ! » How Do You Graph These? » 2012-08-06 03:24:42
Why are they all on one graph?
#62 Help Me ! » How Do You Graph These? » 2012-08-05 14:12:49
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2. (0, -4) and (-4/3, 0)
3. (0, -5) (- 15/4, 0)
5. (2, 4) (2, 6)
6. (0, -2) (-2, 0)
What do these look like once they're graphed? I've tried a bunch and keep getting it wrong. Help me!!
By the way, there should be 2 lines once graphed. Each 2 point sets are for one line.
#63 Re: Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-08-01 06:15:41
Thanks!
#64 Re: Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-07-31 04:17:09
And, how exactly do you graph points like that?
#65 Re: Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-07-30 04:29:30
Okay! Thanks. <3
#66 Re: Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-07-30 04:11:26
The (3, -1) is the point.
#67 Re: Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-07-30 03:20:52
Thanks guys!
How does this work:
x + y = 3
x + 2 = 3
-2 -2
-1/1
(3, -1)
It's the only one I'm not sure about..
#68 Re: Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-07-27 05:12:47
y = 2x, x + y = 3
Actually, I'm not exactly sure what's wrong with the problem above. Is there even a problem with that one? Do I need more points? 
#69 Help Me ! » Help me Find Two Points from this Problem and Show your Work! » 2012-07-27 05:06:28
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I need help finding 2 points from these equations. I tried solving them myself (My work is below) but it was wrong. So I need the correct answers and please show your work. Thank you!
I need help with this one: 4x + 3y = -15
My work for it:
4x + 3y = -15
4(0) + 3y = -15
0 + 3y = -15
3y = -15
-15/3
(0, -5)
4x + 3y = -15
4x + 3(0) = -15
4x + 0 = -15
4x = -15
-15/4
I got 3.75, so I rounded that up to 4.
(4, 0)
The points are (0, -5) and (4, 0)
The Equations: y = 2x, x + y = 3
y = 2x
y = 2(2)
y = 4
(2, 4)
x + y = 3
First I picked 2 for x.
2 + y = 3
If we subtract 2 from both sides of the equation we will get y by itself.
2 - 2 + y = 3 - 2
2 - 2 = 0 so the left hand side becomes
0 + y = 3 - 2
Now y + 0 is the same as y so the left hand side becomes
y = 3 - 2
Now 3 - 2 = 1 so the right side becomes
y = 1
(2, 1)
I don't think any of the work is wrong but the problem is only one point was found for each equation. I need 2 points from each equation.
#70 Re: Help Me ! » Could you help me find points from x + y = 0 and 3x + y = -4 » 2012-07-27 04:42:50
Thank you!
#71 Help Me ! » Could you help me find points from x + y = 0 and 3x + y = -4 » 2012-07-25 12:17:49
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I need two points from each problem! If you could show your work that would be great.
x + y = 0 and 3x + y = -4
#72 Re: Help Me ! » Help me Find Two Points in the Problem x + y = 3? » 2012-07-20 04:09:18
Thank you so much for helping me with that.
Well, that was the only problem like that on my work.
#73 Re: Help Me ! » Help me Find Two Points in the Problem x + y = 3? » 2012-07-19 16:44:28
Yes, I know how to graph such problems, once I have the points.
I'm sorry, I don't know how to solve 2 + y = 3
I could solve it easily if it were, say like this; 2x + 3y = 6
But it isn't so I'm lost.
#74 Help Me ! » Help me Find Two Points in the Problem x + y = 3? » 2012-07-19 04:29:57
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I can't get the right answer for this one, simple because I'm not sure how. So if someone could give me the answer and show/explain how they got it that would be perfect. That way I can learn to do ones like this myself.
The problem is:
x + y = 3
#75 Re: Help Me ! » Solving Equations using the elimination method. Help! » 2012-07-05 08:37:22
Thank you cmowla your explanation with quite helpful!