1. x – 3 > –9
2. –2x + 1 ≤ –11
3. 10 < –3x + 1
4. 2(x + 5) > 8x – 8
5. –2(x – 3) ≥ 5 – (x + 3)
Solve each and graph for each

- anonymous

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- anonymous

@AbdullahM

- anonymous

1. x> -6

- AbdullahM

correct :)

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## More answers

- anonymous

how would i graph this

- AbdullahM

|dw:1435336629065:dw|

- AbdullahM

You could use: https://www.desmos.com/calculator

- anonymous

i mean on a number line sorry

- anonymous

it would be an open circle?

- AbdullahM

yes

- anonymous

Do you have a website for the number line?

- AbdullahM

essentially, you can use the website and only look at the x-axis

- AbdullahM

and since you see the graph has a dashed line, that means an open circle.
When the line on the graph is solid, then that means a closed circle.

- anonymous

ok so closed

- AbdullahM

|dw:1435336897914:dw|
this has a dashed line so it means open circle

- AbdullahM

I drew an orange box around the dashed line.

- anonymous

alright i can graph the rest i just need help solving the rest

- anonymous

-2x +1 < -11 \[-2x +1 \le -11\]

- AbdullahM

for graphing it on a number line:
|dw:1435337002162:dw|

- AbdullahM

For #2, we would first take the +1 to the other side. So we have to subtract that on both sides.

- anonymous

so -12

- anonymous

-2x less than or equal to -12

- AbdullahM

now we have
\(\sf -2x\le -12\)

- anonymous

now devide -2?

- anonymous

divide*

- AbdullahM

Do you know that when you divide by a negative number on both sides that the sign flips

- anonymous

so it goes to
>?

- anonymous

or this

- AbdullahM

For example:
\(\sf\Large -x \le 5\)
then
\(\sf\Large x \ge 5\)
another example:
\(\sf\Large -2y > 6\)
then
\(\sf\Large y<-3\)

- anonymous

does it ever switch from \[\le \to \]

- anonymous

not showign up

- anonymous

without the line under it

- AbdullahM

no it can't

- anonymous

but this answer is x=6

- AbdullahM

\(\color{blue}{\text{Originally Posted by}}\) @AbdullahM
now we have
\(\sf -2x\le -12\)
\(\color{blue}{\text{End of Quote}}\)
if we divide -2, and also we have to change the sign, what do we get?

- anonymous

x= 6

- anonymous

oh x > 6

- AbdullahM

no, it would be \(\sf\Large x\ge6\)

- anonymous

yeah thats i meant my b

- anonymous

could i also say greater than or equal to?

- AbdullahM

remember
these are the opposites:
\(\sf \le\) and \(\sf \ge\) are opposites
< and > are opposites

- anonymous

and this circle would be open?

- AbdullahM

nope closed

- AbdullahM

http://prntscr.com/7ln08a
from: http://www.studyzone.org/mtestprep/math8/a/graphing_inequalities7l.cfm

- anonymous

|dw:1435337698545:dw|

- anonymous

@AbdullahM

- AbdullahM

correct :)

- anonymous

10 < -3x + 1

- anonymous

-1 both sides

- anonymous

9 < -3x

- AbdullahM

10 < –3x + 1
lets make our life easier and not have to do the switching signs. So lets take 3x to the other side by addition

- anonymous

oh alright

- anonymous

+3x both sides
3x + 10 < 1
3x < -9
x< -3

- anonymous

|dw:1435338193021:dw|

- anonymous

@AbdullahM

- anonymous

i might have made a mistake

- anonymous

Should i have flipped the sign?

- AbdullahM

splendid work, you made no mistakes :)

- AbdullahM

we only flip the sign when we divide by a negative number

- anonymous

oh alright

- AbdullahM

2(x + 5) > 8x – 8

- anonymous

2(x + 5) > 8x - 8

- AbdullahM

First we have to take care of
2(x+5)
distribute the 2 inside

- anonymous

2x + 10 > 8x - 8

- anonymous

-8x both sides
-6x + 10 > -8
-10 both sides
-6x> -18
divide -6 both sides
x < 3

- anonymous

|dw:1435338649893:dw|

- AbdullahM

wow, amazing work once again! :D

- anonymous

Thank you :)

- anonymous

-2(x - 3) > 5 - (x +3)

- AbdullahM

–2(x – 3) ≥ 5 – (x + 3)
this time we have both sides where we have to dsitribute. Be careful, they have a negative number outside that needs to be distributed

- anonymous

-2x + 6 > 5 -x - 3
@AbdullahM ?

- AbdullahM

amazing work! :D

- AbdullahM

no we can add the 2x to both sides and work from there :)

- anonymous

why te 2x?

- anonymous

+x both sides
-x + 6 > 5-3
-6 both sides
-x > -4
divide by -1
x < 4

- AbdullahM

so we don't have to divide by a negative number on both sides, but it seems like you have mastered that trick anyways :D

- AbdullahM

not really a rtick, more like a rule xD

- anonymous

yeah :p i thought it' be easier

- AbdullahM

great work again :)

- anonymous

|dw:1435339061662:dw|

- anonymous

@AbdullahM

- anonymous

Explain, in complete sentences, when you would use an open circle or a closed circle, and when you would shade to the right or left, to graph an inequality on the number line. (3 points)
7. Describe a real life scenario where inequalities are used and state the inequality. (2 points)

- AbdullahM

\(\color{blue}{\text{Originally Posted by}}\) @AbdullahM
http://prntscr.com/7ln08a
from:
http://www.studyzone.org/mtestprep/math8/a/graphing_inequalities7l.cfm
\(\color{blue}{\text{End of Quote}}\)
use this for the first question

- anonymous

alright and i think i did my last graph wrong

- anonymous

|dw:1435339286981:dw|

- AbdullahM

open circle, not closed. and everything else is correct

- AbdullahM

for the second question, you can use:
http://openstudy.com/study#/updates/4f188153e4b00328e4c52f91
http://openstudy.com/study#/updates/50fb45a6e4b010aceb32e484
http://openstudy.com/study#/updates/511e6d1fe4b06821731bf312
https://answers.yahoo.com/question/index?qid=20100719235617AABwik9

- anonymous

@AbdullahM Thank you for the help! You are a great teacher :)

- AbdullahM

anytime, that's why I'm on OpenStudy to help.

- anonymous

That's all i need for today but i'll hit ya up if i need help another time, Have a good afternoon!

- AbdullahM

sure :)

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