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Topic review (newest first)
- bob bundy
- 2012-08-02 16:47:23
You're welcome.
Bob
- SlowlyFading
- 2012-08-02 04:15:41
Thanks!
- bob bundy
- 2012-08-01 03:36:02
see below
Bob
- SlowlyFading
- 2012-08-01 02:17:09
And, how exactly do you graph points like that?
- SlowlyFading
- 2012-07-31 02:29:30
Okay! Thanks. <3
- debjit625
- 2012-07-31 02:23:23
SlowlyFading wrote:
The (3, -1) is the point.
This point can't be from the equation
,let substitute the values of x and y in the equation- debjit625
- 2012-07-31 02:17:52
I think you have one equation and solving it will be like this
Let y be 2
Subtract 2 from both the sides
So your point will be
Now let x be 2
Subtract 2 from both the sides
So your point will be
Normally we use intercept to find any two points of a straight line i.e... take x as 0 and solve for the equation and then take y as 0 and solve the equation.
If you take x as 0 then y will be 3 and if you take y as 0 then x will be 3
Good Luck
- bobbym
- 2012-07-31 02:15:01
Hi;
It is certainly not on x + y = 3.
- SlowlyFading
- 2012-07-31 02:11:26
The (3, -1) is the point.
- bobbym
- 2012-07-31 01:42:19
Hi;
What is (3, -1)?
- SlowlyFading
- 2012-07-31 01:20:52
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..
- debjit625
- 2012-07-28 18:33:58
SlowlyFading wrote:
15/4
I got 3.75, so I rounded that up to 4.
(4, 0)
These coordinates or points are from cartesian coordinate system,they all have different meanings,you just cant change it a bit ,if you do the whole system will go wrong...
In short never round up any point of any coordinate,if the answer is in fraction keep it in fraction
So in this case your coordinate will be
SlowlyFading wrote:
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.
Like you picked 2 for x ,just pick any other number for x you will get another point.
Good Luck
- bob bundy
- 2012-07-28 03:55:50
hi SlowlyFading
Your working to get (2,4) and (2,1) is all correct. 
If you want more points, just choose another value for x, and work out the y values as you have done.
(0,-5) is correct for the first question.
(4,0) is not correct.
Every thing was correct up to
4x = -15
Then it went wrong. 
The x coordinate should have been negative.
And I would not have rounded it at all. When you come to draw the graph, the line will not be in the right place.
Bob
- SlowlyFading
- 2012-07-28 03:12:47
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?
- SlowlyFading
- 2012-07-28 03:06:28
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.