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Please help if you're fairly good in algebra 2 medals will b awarded :) whats the solution set to the given inequality? |x + 3| < 12
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yess hold on
(-infinity, -15) U (9, infin)
(-infinity, -5) upside down U (9, infinity)
[-15,9]
(-15,9)
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a,b,c,d those are the options
|a| < b gives you a< b or a>-b (or equivalently -b<a<b) so what about |x+3| < 12 ?
so is correct?
b
(=infin, -15) upsidedown U (9, infin)
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why upsiddown U ?
oh wait it a union right?
so reg U
no, its not the first choice either
x+3<12 gives you x<9 right ?
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ugh :(
yess.
good, and x+3 > -12 gives you x>-15 this ?
agree
so x is between -15 and 9, right ? so (-15,9)
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ohhhh.. i think im starting to understand it more :D
i am glad to hear that! :)
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