# Thread: solution set/ interval notation

1. ## solution set/ interval notation

I tried the latex thing and it still won't work...

[tex]\frac{\pi^2}{6}[tex]

The problem I need help with is this, all steps please so I can follow along and see how you got to the answer.

1)
6-x/12 > 1/4 or 6-x/12 < -1/6

2) Absolute Problem

lwl+8=2

For this I am thinking it has one answer because isn't it true if the number on the right side of the equal sign is negative the negative equation is false?

w+8=2 w+8=-2
w=6 w=-10

Now seems both do not work... is this correct?

2. 1) the "math" in the brackets should be capitol and you need a space after pi so it can register that that's a separate call.

and you need to close it with [\MATH ] (minus the space)

2) I don't know what you mean with the "or" but here goes:

$6 - \frac{x}{12} > \frac{1}{4}
\Rightarrow 72 - x > 3
\Rightarrow -x > -69
\Rightarrow x < 69$

$6 - \frac{x}{12} > -\frac{1}{6}
\Rightarrow 72 - x > -2
\Rightarrow -x > -74
\Rightarrow x < 74$

For the last question there is no way for that to be true. Because if you move things around you get |w| = -6 and an absolute value's not going to be negative.

But.. hmm..

(case 1: w > 0 : fails, by above
case 2: $w < 0 : -w + 8 = 2 \Rightarrow w = 6$)

I'm not sure about what I wrote in there, I think that was from graphing and when you had to find the intervals for which the equation worked or something. I don't know if it still applies but you can take it and show it to someone who knows what they're doing, unlike me obviously. :P

3. The "Or" meant that it is a solution set and either equality is the answer of both equalities.

4. Originally Posted by pippers
The "Or" meant that it is a solution set and either equality is the answer of both equalities.
oh, so.. then... x < 69, because all numbers < 69 are in both sets (as long as it's less than 69 then both equations work)?