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u can write |2x + 5| as2x + 5 or -(2x + 5) like that |4x + 7| as(4x + 7) or -(4x + 7) so totally there as 4 possibilities
Alright. I'm still a bit confused...would I do them all separately?
plugin into wolfram ... something interesting there
lol...it is a bit interesting...I need to know how to do these without wolfram though...
Is that really what I do?
i'm not sure .... though this technique seems pretty promising
But all that results in is the removal of the absolute value no?
yep ... take square and see |dw:1344019775207:dw|
I am so confused...not going to lie...if you take the square of both sides, wouldn't you end up with the second side: |dw:1344019964409:dw| and where did the absolute value come back in?
sorry .. my bad
looks like this is going to be quite difficult ...
Yeah...my teacher said that there was a very simple trick to this, but she never had the opportunity to teach it to me, so I'm just stuck on what to do...
for x is positive.
Well, turns out that sai at the beginning was correct...you just work out the 4 different cases...
I found the solution set...
Thanks for trying though...
for x is negative, |dw:1344020308613:dw|
check the value for a = 3, which is less than 30|dw:1344020674321:dw|
Yup. I get it now. We just test out all 4 possibilities and combine the inequalities. Thank you :)
well ... the best would be do see carefully how it takes the least effort ...