HELP!!!!!!! linear inequalitys

[scontreras]

3x+11< or = 6x+8

 

[Deleted member 4993]

3x+11< or = 6x+8

Hint:

Isolate 'x' - like you would in case you are dealing with an "equality"(with only equal sign).

Please read the post titled "Read before Posting".

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[scontreras]

i have got this far
3x+11< or = 6x+8
-8 -8
3x+3< or = 6x
-6x -6x

-3x+3< or = 0

that is what my teacher told me to do. to get zero on one side

 

[Deleted member 4993]

3x >= 3

Depending on what the original quest was (which OP has not stated) - you need to go one more step!!

 

[JeffM]

i have got this far
3x+11< or = 6x+8
-8 -8
3x+3< or = 6x
-6x -6x

-3x+3< or = 0

that is what my teacher told me to do. to get zero on one side

You have not told us what you are supposed to do. if it is to simplify the inequality then you are right as far as you have gone but you can go further.

$\displaystyle 3x + 11 \le 6x + 8 \implies$

$\displaystyle 3x + 11 - 6x - 8 \le 6x + 8 - 6x - 8 \implies$

$\displaystyle - 3x + 3 \le 0.$ That is certainly sound, but it may be viewed as incomplete.

$\displaystyle 3x + 11 \le 6x + 8 \implies$

$\displaystyle 3x + 11 - 6x \le 6x + 8 - 6x \implies$

$\displaystyle - 3x + 11 \le 8 \implies.$

$\displaystyle - 3x + 11 - 11 \le 8 - 11 \implies$

$\displaystyle - 3x \le - 3 \implies$

$\displaystyle \left(- \dfrac{1}{3}\right) * (- 3x) \ge \left(- \dfrac{1}{3}\right) * - 3 \implies$

$\displaystyle x \ge 1.$

Questions?