I have some math problems involving that you have to solve an inequality in interval notations and I don't know how to do it.

- anonymous

- jamiebookeater

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- anonymous

|dw:1438573641556:dw|

- jim_thompson5910

Rule:
\[\Large |x| > k\]
breaks down into
\[\Large x > k \ \ \text{or} \ \ x < -k\]
where k is some positive number

- jim_thompson5910

so in your case,
|x-2| > 13
turns into
x-2 > 13 or x-2 < -13

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## More answers

- anonymous

I don't get it. What would my answer be?

- anonymous

would it be 15??

- jim_thompson5910

I'm assuming you added 2 to both sides of x-2 > 13, right?

- anonymous

yes

- jim_thompson5910

|dw:1438574103833:dw|

- jim_thompson5910

you have the right idea
|dw:1438574120370:dw|

- jim_thompson5910

|dw:1438574134671:dw|

- jim_thompson5910

so `x-2 > 13` turns into `x > 15`

- jim_thompson5910

what does x-2 < -13 turn into when you solve for x ?

- anonymous

-11?

- jim_thompson5910

not just -11, but something else
is it x = -11 ?
x < -11 ?
x > -11 ?

- anonymous

x<-11

- jim_thompson5910

yes, `x-2 < -13` turns into `x < -11`

- jim_thompson5910

|x-2| > 13
turns into
x-2 > 13 or x-2 < -13
and then that becomes
x > 15 or x < -11

- anonymous

so the answer is x is greater than 15 but less than -11??

- anonymous

how would I write that?

- jim_thompson5910

in interval notation you mean?

- anonymous

yes

- jim_thompson5910

let's just focus on x > 15 for now

- jim_thompson5910

any ideas of what the interval notation of x > 15 is ?

- anonymous

i really don't know.

- jim_thompson5910

x > 15 means x is some number larger than 15
so if we describe an interval, we would write (15, infinity)
the parenthesis say "don't include the enpoints"

- jim_thompson5910

|dw:1438574679944:dw|

- jim_thompson5910

|dw:1438574694799:dw|

- anonymous

okay i get that part what about the -11?

- jim_thompson5910

you tell me, what does x < -11 translate to?

- anonymous

|dw:1438574720200:dw|

- jim_thompson5910

close

- anonymous

is that it?

- jim_thompson5910

you need a negative before the infinity

- anonymous

|dw:1438574761239:dw|

- jim_thompson5910

\[\Large x > 15 \rightarrow (15, \infty)\]
\[\Large x < -11 \rightarrow (-\infty, 11)\]

- jim_thompson5910

The "or" in the solution `x > 15 or x < -11` means we will have a union symbol U
Which is why this is the final answer
\[\Large (-\infty, 11) \cup (15, \infty)\]

- jim_thompson5910

oops I meant to write -11 instead of 11

- jim_thompson5910

so it should be \[\Large (-\infty, -11) \cup (15, \infty)\]

- anonymous

thank you very much for all of your help. Im probably going to struggle with the rest of these inequality problems. That seemed like an easy one compared to the others i have

- jim_thompson5910

with more practice, the rest should come easier too

- anonymous

can I message you for more help??

- anonymous

the step by step process really helped!

- jim_thompson5910

I'm glad it helped and yeah you can ask a few more questions

- anonymous

|dw:1438575156359:dw|

- anonymous

to get rid of the 3, do I multiply it by 8?

- jim_thompson5910

first you need to use the rule
|x| < k
becomes
-k < x < k

- jim_thompson5910

where again, k is some positive number

- jim_thompson5910

|dw:1438575356447:dw|

- jim_thompson5910

notice
|dw:1438575401382:dw|

- jim_thompson5910

what would be the next step?

- anonymous

8*3??

- jim_thompson5910

no

- jim_thompson5910

|dw:1438575499181:dw|

- anonymous

do the opposite which is multiply

- jim_thompson5910

yes, multiply ALL sides by 3
|dw:1438575563784:dw|

- jim_thompson5910

|dw:1438575578883:dw|

- jim_thompson5910

we are then left with this
\[\Large -36 \le x+8 \le 36\]

- jim_thompson5910

then what?

- anonymous

okay that helped. and next we have to get x by itself

- jim_thompson5910

yes so you have to somehow undo that +8

- jim_thompson5910

btw, we are following PEMDAS in reverse when it comes to isolating variables

- anonymous

so subtract 8 from each side?|dw:1438575726830:dw|

- jim_thompson5910

good

- anonymous

like that?

- jim_thompson5910

correct, which ends you up with what?

- anonymous

|dw:1438575822811:dw|

- jim_thompson5910

you nailed it

- jim_thompson5910

x is finally isolated so you're done

- anonymous

|dw:1438576160100:dw|

- anonymous

I tried putting in my answer all types of ways, but it still says it's incorrect.

- jim_thompson5910

|dw:1438576344123:dw|

- anonymous

what do you mean?

- jim_thompson5910

|dw:1438576379769:dw|

- jim_thompson5910

we include both endpoints so the interval notation would be [-44, 28]

- jim_thompson5910

the square brackets say "include endpoints

- jim_thompson5910

|dw:1438576439943:dw|

- anonymous

i just entered it and I didn't need the infinity signs! but it worked.

- jim_thompson5910

we don't need infinity signs because the interval starts at -44 and ends at 28

- anonymous

|dw:1438576618671:dw|

- jim_thompson5910

you only do infinity signs if you have something like x > 15 where it starts at 15 but doesn't stop

- anonymous

ohhh I understand

- anonymous

So first I would undo the cube root right?

- anonymous

which would be x-12??

- jim_thompson5910

factor out x first
|dw:1438576760465:dw|

- jim_thompson5910

then factor x^2 - 36 to get (x-6)(x+6)
|dw:1438576790668:dw|