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

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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

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1. x> -6
correct :)

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Other answers:

how would i graph this
|dw:1435336629065:dw|
You could use: https://www.desmos.com/calculator
i mean on a number line sorry
it would be an open circle?
yes
Do you have a website for the number line?
essentially, you can use the website and only look at the x-axis
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.
ok so closed
|dw:1435336897914:dw| this has a dashed line so it means open circle
I drew an orange box around the dashed line.
alright i can graph the rest i just need help solving the rest
-2x +1 < -11 \[-2x +1 \le -11\]
for graphing it on a number line: |dw:1435337002162:dw|
For #2, we would first take the +1 to the other side. So we have to subtract that on both sides.
so -12
-2x less than or equal to -12
now we have \(\sf -2x\le -12\)
now devide -2?
divide*
Do you know that when you divide by a negative number on both sides that the sign flips
so it goes to >?
or this
For example: \(\sf\Large -x \le 5\) then \(\sf\Large x \ge 5\) another example: \(\sf\Large -2y > 6\) then \(\sf\Large y<-3\)
does it ever switch from \[\le \to \]
not showign up
without the line under it
no it can't
but this answer is x=6
\(\color{blue}{\text{Originally Posted by}}\) @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?
x= 6
oh x > 6
no, it would be \(\sf\Large x\ge6\)
yeah thats i meant my b
could i also say greater than or equal to?
remember these are the opposites: \(\sf \le\) and \(\sf \ge\) are opposites < and > are opposites
and this circle would be open?
nope closed
http://prntscr.com/7ln08a from: http://www.studyzone.org/mtestprep/math8/a/graphing_inequalities7l.cfm
|dw:1435337698545:dw|
correct :)
10 < -3x + 1
-1 both sides
9 < -3x
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
oh alright
+3x both sides 3x + 10 < 1 3x < -9 x< -3
|dw:1435338193021:dw|
i might have made a mistake
Should i have flipped the sign?
splendid work, you made no mistakes :)
we only flip the sign when we divide by a negative number
oh alright
2(x + 5) > 8x – 8
2(x + 5) > 8x - 8
First we have to take care of 2(x+5) distribute the 2 inside
2x + 10 > 8x - 8
-8x both sides -6x + 10 > -8 -10 both sides -6x> -18 divide -6 both sides x < 3
|dw:1435338649893:dw|
wow, amazing work once again! :D
Thank you :)
-2(x - 3) > 5 - (x +3)
–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
-2x + 6 > 5 -x - 3 @AbdullahM ?
amazing work! :D
no we can add the 2x to both sides and work from there :)
why te 2x?
+x both sides -x + 6 > 5-3 -6 both sides -x > -4 divide by -1 x < 4
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
not really a rtick, more like a rule xD
yeah :p i thought it' be easier
great work again :)
|dw:1435339061662:dw|
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)
\(\color{blue}{\text{Originally Posted by}}\) @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
alright and i think i did my last graph wrong
|dw:1435339286981:dw|
open circle, not closed. and everything else is correct
@AbdullahM Thank you for the help! You are a great teacher :)
anytime, that's why I'm on OpenStudy to help.
That's all i need for today but i'll hit ya up if i need help another time, Have a good afternoon!
sure :)

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