anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1438573641556:dw|
jim_thompson5910
  • jim_thompson5910
Rule: \[\Large |x| > k\] breaks down into \[\Large x > k \ \ \text{or} \ \ x < -k\] where k is some positive number
jim_thompson5910
  • jim_thompson5910
so in your case, |x-2| > 13 turns into x-2 > 13 or x-2 < -13

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anonymous
  • anonymous
I don't get it. What would my answer be?
anonymous
  • anonymous
would it be 15??
jim_thompson5910
  • jim_thompson5910
I'm assuming you added 2 to both sides of x-2 > 13, right?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
|dw:1438574103833:dw|
jim_thompson5910
  • jim_thompson5910
you have the right idea |dw:1438574120370:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438574134671:dw|
jim_thompson5910
  • jim_thompson5910
so `x-2 > 13` turns into `x > 15`
jim_thompson5910
  • jim_thompson5910
what does x-2 < -13 turn into when you solve for x ?
anonymous
  • anonymous
-11?
jim_thompson5910
  • jim_thompson5910
not just -11, but something else is it x = -11 ? x < -11 ? x > -11 ?
anonymous
  • anonymous
x<-11
jim_thompson5910
  • jim_thompson5910
yes, `x-2 < -13` turns into `x < -11`
jim_thompson5910
  • jim_thompson5910
|x-2| > 13 turns into x-2 > 13 or x-2 < -13 and then that becomes x > 15 or x < -11
anonymous
  • anonymous
so the answer is x is greater than 15 but less than -11??
anonymous
  • anonymous
how would I write that?
jim_thompson5910
  • jim_thompson5910
in interval notation you mean?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
let's just focus on x > 15 for now
jim_thompson5910
  • jim_thompson5910
any ideas of what the interval notation of x > 15 is ?
anonymous
  • anonymous
i really don't know.
jim_thompson5910
  • 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
  • jim_thompson5910
|dw:1438574679944:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438574694799:dw|
anonymous
  • anonymous
okay i get that part what about the -11?
jim_thompson5910
  • jim_thompson5910
you tell me, what does x < -11 translate to?
anonymous
  • anonymous
|dw:1438574720200:dw|
jim_thompson5910
  • jim_thompson5910
close
anonymous
  • anonymous
is that it?
jim_thompson5910
  • jim_thompson5910
you need a negative before the infinity
anonymous
  • anonymous
|dw:1438574761239:dw|
jim_thompson5910
  • jim_thompson5910
\[\Large x > 15 \rightarrow (15, \infty)\] \[\Large x < -11 \rightarrow (-\infty, 11)\]
jim_thompson5910
  • 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
  • jim_thompson5910
oops I meant to write -11 instead of 11
jim_thompson5910
  • jim_thompson5910
so it should be \[\Large (-\infty, -11) \cup (15, \infty)\]
anonymous
  • 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
  • jim_thompson5910
with more practice, the rest should come easier too
anonymous
  • anonymous
can I message you for more help??
anonymous
  • anonymous
the step by step process really helped!
jim_thompson5910
  • jim_thompson5910
I'm glad it helped and yeah you can ask a few more questions
anonymous
  • anonymous
|dw:1438575156359:dw|
anonymous
  • anonymous
to get rid of the 3, do I multiply it by 8?
jim_thompson5910
  • jim_thompson5910
first you need to use the rule |x| < k becomes -k < x < k
jim_thompson5910
  • jim_thompson5910
where again, k is some positive number
jim_thompson5910
  • jim_thompson5910
|dw:1438575356447:dw|
jim_thompson5910
  • jim_thompson5910
notice |dw:1438575401382:dw|
jim_thompson5910
  • jim_thompson5910
what would be the next step?
anonymous
  • anonymous
8*3??
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
|dw:1438575499181:dw|
anonymous
  • anonymous
do the opposite which is multiply
jim_thompson5910
  • jim_thompson5910
yes, multiply ALL sides by 3 |dw:1438575563784:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438575578883:dw|
jim_thompson5910
  • jim_thompson5910
we are then left with this \[\Large -36 \le x+8 \le 36\]
jim_thompson5910
  • jim_thompson5910
then what?
anonymous
  • anonymous
okay that helped. and next we have to get x by itself
jim_thompson5910
  • jim_thompson5910
yes so you have to somehow undo that +8
jim_thompson5910
  • jim_thompson5910
btw, we are following PEMDAS in reverse when it comes to isolating variables
anonymous
  • anonymous
so subtract 8 from each side?|dw:1438575726830:dw|
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
like that?
jim_thompson5910
  • jim_thompson5910
correct, which ends you up with what?
anonymous
  • anonymous
|dw:1438575822811:dw|
jim_thompson5910
  • jim_thompson5910
you nailed it
jim_thompson5910
  • jim_thompson5910
x is finally isolated so you're done
anonymous
  • anonymous
|dw:1438576160100:dw|
anonymous
  • anonymous
I tried putting in my answer all types of ways, but it still says it's incorrect.
jim_thompson5910
  • jim_thompson5910
|dw:1438576344123:dw|
anonymous
  • anonymous
what do you mean?
jim_thompson5910
  • jim_thompson5910
|dw:1438576379769:dw|
jim_thompson5910
  • jim_thompson5910
we include both endpoints so the interval notation would be [-44, 28]
jim_thompson5910
  • jim_thompson5910
the square brackets say "include endpoints
jim_thompson5910
  • jim_thompson5910
|dw:1438576439943:dw|
anonymous
  • anonymous
i just entered it and I didn't need the infinity signs! but it worked.
jim_thompson5910
  • jim_thompson5910
we don't need infinity signs because the interval starts at -44 and ends at 28
anonymous
  • anonymous
|dw:1438576618671:dw|
jim_thompson5910
  • jim_thompson5910
you only do infinity signs if you have something like x > 15 where it starts at 15 but doesn't stop
anonymous
  • anonymous
ohhh I understand
anonymous
  • anonymous
So first I would undo the cube root right?
anonymous
  • anonymous
which would be x-12??
jim_thompson5910
  • jim_thompson5910
factor out x first |dw:1438576760465:dw|
jim_thompson5910
  • jim_thompson5910
then factor x^2 - 36 to get (x-6)(x+6) |dw:1438576790668:dw|
jim_thompson5910
  • jim_thompson5910
has your teacher gone over sign charts?
anonymous
  • anonymous
I am a freshman about to start college. I didn't take a math class my senior year so I'm definitely a little rusty with my math.
anonymous
  • anonymous
I'm stuck on what to do next..
jim_thompson5910
  • jim_thompson5910
if you set x(x-6)(x+6) equal to 0, you'd get x(x-6)(x+6) = 0 then use the zero product property to break up x(x-6)(x+6)=0 into x = 0 or x-6 = 0 or x+6 = 0
jim_thompson5910
  • jim_thompson5910
what are the other two solutions?
anonymous
  • anonymous
x=6?
jim_thompson5910
  • jim_thompson5910
and what else
anonymous
  • anonymous
this is confusing. I'm about to give up..
anonymous
  • anonymous
x=0? i really don't know
anonymous
  • anonymous
or x=-6?
jim_thompson5910
  • jim_thompson5910
yes x+6 = 0 leads to x=-6
jim_thompson5910
  • jim_thompson5910
so we have these three roots: x = 0, x = 6, or x = -6
jim_thompson5910
  • jim_thompson5910
draw out a number line with those numbers on it |dw:1438577170206:dw|
jim_thompson5910
  • jim_thompson5910
notice we have these 4 regions |dw:1438577220408:dw|
jim_thompson5910
  • jim_thompson5910
what number is in region A?
anonymous
  • anonymous
everything less than -6
jim_thompson5910
  • jim_thompson5910
so let's pick a particular value, say x = -7 |dw:1438577304486:dw|
jim_thompson5910
  • jim_thompson5910
let me draw up a sign chart real quick first
jim_thompson5910
  • jim_thompson5910
|dw:1438577336529:dw|
jim_thompson5910
  • jim_t