Help for a medal, anyone?
I'm rusty on inequalities and need some help on a few problems to boost my memory.
Here's the first one...
-9 < 4x + 8 < 10

- anonymous

- schrodinger

See more answers at brainly.com

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- amistre64

what are your thoughts ...

- anonymous

If I remember correctly, you're supposed to simplify the equation by adding, divining..etc terms on either side, but I'm not sure how to do that with an inequality.

- anonymous

dividing*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

the math doesnt change ... what you do to one part, you do to all the parts to keep them "equal" right?

- amistre64

in this case, the equation is in the middle instead of to the left or right ... what would you think we should do first?

- amistre64

just make a guess at what you think we should do to start with ... what comes to mind?
4x+8 is what we want to manipulate, how do we work about getting 'x' all by itself?

- anonymous

Would you solve one equality at a time like
-9 < 4x + 8 or 4x + 8 < 10 ?

- amistre64

you can if you want, but i just work it all at once to avoid the extra work of writing it all out twice

- amistre64

-9 < 4x + 8 < 10
-8 -8 -8
----------------
-17< 4x < 2

- anonymous

Ah, I see. Then you would divide all 3 parts by 4 to get x by itself.

- amistre64

correct, the only thing to watch out for is when we have to mulitply or divide by a negative number ... which they really should include the rule when doing equalities
the sign reverses ... but we do not have to worry about that in this particular case do we

- amistre64

if -a < b, then a > -b
if -a > b, then a < -b
if -a = b, then a = -b
... the reversing of an equals sign just produces an equals sign, but it helps to make the rule consistent for all 3 equality signs.

- anonymous

So would
-17/4 < x < 1/2
be the final answer or is there another step?

- amistre64

thats the answer for me, but im not grading it

- amistre64

17/4 cant be reduced, but it can be written as 4 and 1/4 if need be

- anonymous

Okay, thank you. :)
Could you help with one other? It looks a bit different from the first.

- amistre64

i spose

- anonymous

\[\left|\frac{ x+4 }{ 5 } \right| < 6\]

- amistre64

absolute value, what does that mean to you?

- anonymous

I know that nothing inside of it can be negative (or at least that I recall)

- amistre64

first off |5| = 5 so we can go ahead and multiply thru by 5
|x+4| < 30

- amistre64

it can be negative inside of it, but the bars cancel out the sign

- amistre64

|-30| = 30
and
|30| = 30
when does x+4 = 30? or -30?

- amistre64

as a side note:
|a| < k can be written as:
-k < a < k

- amistre64

|x+4| < 30 can be written as
-30 < x+4 < 30

- anonymous

Then you just subtract 4 from both sides...
-34 < x < 26
Also does the |a| < k or -k < a < k
rule always apply?

- amistre64

it is a rule, so it always applies yes. its just my brain that doesnt always remember it in time lol

- amistre64

\[|\frac{x+4}{5}|<6\]
\[-6<\frac{x+4}{5}<6\]
\[-306

- amistre64

not -306, but yeah

- anonymous

Oh, okay.
So -34 < x < 26 would be the answer correct?

- amistre64

looks good to me

- anonymous

Thank you so much for your help. :)

- amistre64

youre welcome, and good luck :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.