Thank you for responding to previous question. Here is another I can't seem to get.?

Question

The homework question is: for the absolute value 3x-6y less than/equal to 9, is the box in the area defined by the inequality, out of the area, or has both in and out of the area defined by the inequality. The coordinates in the box are[(1,1), (2,1), (2,2), (1,2)].
When I graph the inequality, I get points in and out of the area. I am not sure if this is right. Any help is appreciated. Thanks.

Answer 1

3x - 6y ≤ 9

y = 0

3x - 6(0) ≤ 9

3x - 0 ≤ 9

3x ≤ 9

3x/3 ≤9/3

x ≤ 3

- - - - - - -

The solution set { 3, 0 }

- - - - - - - - - - -

x = 0

3x - 6y ≤ 9

3(0) - 6y ≤ 9

0 - 6y ≤ 9

- 6y ≤ 9

- 6y / - 6 ≥ 9 / - 6. . .Negative number inequality sign changes.

y = ≥ - 3/2

The solution set is { 0, - 3/2 }

- - - - - - - - -s-

Answer 2

|3x-6y| <= 9
|x-2y| <= 3
x-2y <= 3
-2y <= -x+3
y>= x/2 -3/2 This is a line passing through (0,-3/2) and (3/2,0)
x-2y <= -3
-2y <= -3-x
y >= x/2+3/2 This is a line passing through (0,3/2) and (-3/2,0)

Only the point (1,2) satisfies the two lines. The rest of the box is totally outside the range.

You could think of the allowable area to be on or above the area above both of the two lines which make a "V" with the point of the V at (0,3/2).

Answer 3

I think you're right; some in the box and some out of the box.