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Mathematics 17 Online
OpenStudy (anonymous):

|2x + 10| > 26

OpenStudy (agent47):

|2x|>16 |x|>8

OpenStudy (agent47):

x<-8 or x>8

OpenStudy (zzr0ck3r):

no try -10 for x that does not work 2x+10>26 2x+10 < -26 2x> 16 x > 8 2x < -36 x<-18 or x > 8

OpenStudy (agent47):

oh zz is right, ok this means that i'm off for today..

OpenStudy (anonymous):

x < -18

OpenStudy (zzr0ck3r):

try to break these up into two equations |x| > y is the same as x > y and x < -y

OpenStudy (anonymous):

3|2x + 4| – 1 = 11

OpenStudy (zzr0ck3r):

ok first get it down to just the absolut value on the left |2x+4| = 6 right?

OpenStudy (zzr0ck3r):

now we have 2x+4 = 6 and 2x+4 = -6 solve for x in both equations

OpenStudy (anonymous):

12/3 = 4

OpenStudy (zzr0ck3r):

sorry thought that was a 2 on the outside

OpenStudy (zzr0ck3r):

3|2x + 4| – 1 = 11 3|2x+4| = 12 |2x+4| = 4 2x+4 = 4 so x = 0 2x+4 = -4 so x = -4

OpenStudy (zzr0ck3r):

agent I think you might want to review solving with absolutes

OpenStudy (agent47):

ok nvm me, im just bad at absolute values.. but this time my mistake was dividing 12/3 lol

OpenStudy (agent47):

apparently i think that 12/3=6, i shld be sent back to 1st grade

OpenStudy (zzr0ck3r):

well and you are not breaking them up into two equations, you need to do that once you have isolated the abolute values signs

OpenStudy (anonymous):

|2x – 1| = 3

OpenStudy (zzr0ck3r):

lets let him do this one...

OpenStudy (agent47):

no i like to simplify it down all the way first

OpenStudy (agent47):

OK I GOT THIS!

OpenStudy (zzr0ck3r):

what do you want to do first ternvilla?

OpenStudy (anonymous):

Hey, don't mind me! I'm getting so hungry watching you guys solving so fast!

OpenStudy (zzr0ck3r):

no not you you agent, the dude who posts it

OpenStudy (agent47):

ok deleted

OpenStudy (zzr0ck3r):

I trust in you:)

OpenStudy (zzr0ck3r):

@trenvilla what do you think the first step should be?

OpenStudy (anonymous):

x+2 or x + -1

OpenStudy (zzr0ck3r):

when we try to solve for something on the inside of a absolute value sign we do this |a| = b so a = b and a = -b so if we had |2x+6| = 5 we take the inside and set it equal to the outside 2x+6 = 5 AND we take the inside and set it equal to the negative of the outside 2x+6 = -5 then we solve for x in both equations |2x-1| = 3 so 2x-1 = 3 and 2x-1 = -3 solve for both of these, you should end up with two x = something at the end.

OpenStudy (anonymous):

cool ty

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