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Algebra 13 Online
OpenStudy (genny7):

Which of the following values of x is in the solution set of 2|x + 4| > 14? A. x = 0 B. x = 1 C. x = –3 D. x = –12

OpenStudy (jdoe0001):

\(\large {2|x + 4| > 14\implies |x + 4| \cfrac{\cancel{ 14 }}{\cancel{ 2 }} \\ \quad \\ |x+4|>7\to \begin{cases} +(x+4)>7 \\ \quad \\ -(x+4)>7 \end{cases} }\)

OpenStudy (anonymous):

thats wrong too ._.

OpenStudy (anonymous):

lol we both messed up XD

OpenStudy (genny7):

so you were wrong chucho?

OpenStudy (anonymous):

mhmm :C

OpenStudy (anonymous):

still working on it tho

OpenStudy (genny7):

So how do you solve it then!

OpenStudy (jdoe0001):

you'd solve each case or scenario, thus 2 scenarios, thus 2 values for "x" I don't see anything wrong above though

OpenStudy (anonymous):

one sec almost done

OpenStudy (anonymous):

Got it! :D

OpenStudy (anonymous):

you ready for my awesome explanation? Replace X with one of the numbers, and check if its true!

OpenStudy (anonymous):

2[-12+4>14 2*8 16>14

OpenStudy (genny7):

Uhhh I'm lost Chucho!

OpenStudy (anonymous):

kk lemme explain slower

OpenStudy (anonymous):

|dw:1405720698631:dw|

OpenStudy (anonymous):

does this help? just go from top to bottom :3

OpenStudy (genny7):

Yeah, but thats not one of the answers???

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