Graph and Solve 4-2x>8 & 4-2x<0: Interval Notation (-2,2)

[elmosworld403]

l 4-2x l >8
4-2x>0 4-2x>8
-2x>-4 -2x>4
x>2 x>-2

4-2x<0 4-2x<8
-2x<-4 -2x<4
X<2 X<-2

-2< X < 2

ANSWER- (-2,2) interval notation

Do I have to keep doing the 4-2x>0 4-2x<0 Even tho it doesn't say it in the question?
It is also asking me to graph it, just graph it on a number line?

 

[jbunniii]

l 4-2x l >8



4-2x>0 4-2x>8
-2x>-4 -2x>4
x>2 x>-2

4-2x<0 4-2x<8
-2x<-4 -2x<4
X<2 X<-2




-2< X < 2

ANSWER- (-2,2)

Do I have to keep doing the 4-2x>0 4-2x<0 Even tho it doesn't say it in the question?

You can see that the answer can't be (-2,2) by taking a number in that interval such as 0 and noticing that it doesn't satisfy the inequality.

I'm not sure why you wrote 4 - 2x > 0 and 4 - 2x < 0. These are mutually exclusive (they can't both be true), and in any case neither one of them follows from the original inequality.

|4-2x| > 8 means precisely that 4-2x > 8 or 4-2x < -8.

 

[elmosworld403]

Kk so i don't have to do the zeros. I redid the question.

-(4-2x)>8
-4+2X>8
2x>12
X>6

4-2x>8
-2x>4
x>-2

Interval notation

(-2,∞)

 

[jbunniii]

Kk so i don't have to do the zeros. I redid the question.

-(4-2x)>8
-4+2X>8
2x>12
X>6

OK

4-2x>8
-2x>4
x>-2

No, when you divide by sides by -2 you have to reverse the inequality, so the last line should be x < -2.

 

[pierce15]

Are you sure that this should be in the calculus section?