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- anonymous

graph/ solve

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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- anonymous

graph/ solve

- katieb

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- anonymous

|dw:1439060839044:dw|

- phi

we have negative numbers with a relation
so *we have to be careful*
the "fast way" is multiply both sides by -1 and *switch the relation operator*
the slow (but easy to do) way is add +3 to both sides
then add +3 | x-5| to both sides
either way, what do you get?

- anonymous

0

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- anonymous

/x-5/<1

- phi

do it in steps
-3 | x-5| < -3
which way do you want to do it? fast way or slow way?

- anonymous

fast way of course!

- phi

fast way:
multiply both sides by -1
*flip the relation operator*

- anonymous

would it be 3/x-5/<3 ??

- phi

you did step 1 ok.
but step 2 means you change < to >
or if you had > you would change it to <
or \(\le\) would be changed to \( \ge\)
that is the "flip" part

- anonymous

3/x-5/>3

- phi

yes. the problem with the fast way is you have to remember to flip the operator... if you don't have a good memory for "details", the slow way is safer (one less rule to remember)
so now you have
3 | x-5| > 3
it is safe to divide both sides by 3. you get
| x-5 | > 1

- phi

with absolute value problems we *always* have two conditions
either x-5 is positive , in which case , drop the | |
or x-5 is negative, in which case we replace the | | with -1*(x-5)

- phi

let's do the x-5 is positive condition.
we have
x-5 > 1
now add +5 to both sides

- anonymous

x>6

- phi

now do the x-5 is negative case. can you do that one?

- anonymous

yes(hopefully)!
So i got x<4

- phi

looks good.
x<4 or x>6
are the solutions
put an open circle at 4. the open (not solid , filled in) means 4 is not part of the answer, but we can get as close to 4 as we want (without being exactly 4)
and x smaller than 4. all numbers smaller (to the left) of 4 are part of the solution
as it would take a long time to draw a line showing all of the numbers, we use an arrow tip to show the answer includes all numbers the arrow points towards.

- phi

and then do the same for
x> 6
open circle at 6
an arrow pointing which way ??

- anonymous

gotcha:) So, the arrows will be pointing in opposite direction
x-------->6
and x<----------4
the arrows is showing thee way it'll go:)

- phi

yes. one last point: put the answers on the same number line

- anonymous

kk. Hold on...|dw:1439061932632:dw|

- phi

yes. looks good. for your teacher, make sure the "tick mark" for the 4 is clearly marked
(in your graph it is not clear which tick mark is the 4)

- anonymous

Will do:)! Thanks again phi<3<3

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