|2x + 10| > 26
|2x|>16 |x|>8
x<-8 or x>8
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
oh zz is right, ok this means that i'm off for today..
x < -18
try to break these up into two equations |x| > y is the same as x > y and x < -y
3|2x + 4| – 1 = 11
ok first get it down to just the absolut value on the left |2x+4| = 6 right?
now we have 2x+4 = 6 and 2x+4 = -6 solve for x in both equations
12/3 = 4
sorry thought that was a 2 on the outside
3|2x + 4| – 1 = 11 3|2x+4| = 12 |2x+4| = 4 2x+4 = 4 so x = 0 2x+4 = -4 so x = -4
agent I think you might want to review solving with absolutes
ok nvm me, im just bad at absolute values.. but this time my mistake was dividing 12/3 lol
apparently i think that 12/3=6, i shld be sent back to 1st grade
well and you are not breaking them up into two equations, you need to do that once you have isolated the abolute values signs
|2x – 1| = 3
lets let him do this one...
no i like to simplify it down all the way first
OK I GOT THIS!
what do you want to do first ternvilla?
Hey, don't mind me! I'm getting so hungry watching you guys solving so fast!
no not you you agent, the dude who posts it
ok deleted
I trust in you:)
@trenvilla what do you think the first step should be?
x+2 or x + -1
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.
cool ty
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