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Algebra 18 Online
OpenStudy (kise_ryouta):

Solve for p. 3+|1+p|=123+|1+p|=12 a.p = 8 or p=−10p=−10 b.p = 14 or p=−16p=−16 c.p = 10 or p=−10p=−10 d.p = 8 or p=−8

OpenStudy (kise_ryouta):

3+|1+p|=12 sorry for the confusion

OpenStudy (benji):

Ok so when we have absolute values, what do you have to do first before touching anything inside the brackets?

OpenStudy (kise_ryouta):

um subtract 3 from 12 right

OpenStudy (benji):

Yes so we have \[\left| 1+p \right|=9\]

OpenStudy (benji):

And if \[\left| a \right|=a\] or \[\left| a \right|=-a\] What do you think we do next?

OpenStudy (kise_ryouta):

subtracting or adding

OpenStudy (benji):

\[\left| 1+p \right|=9\] becomes\[1+p=9\]and\[1-p=9\] right?

OpenStudy (kise_ryouta):

yea and i subtact 1 from 9 right

OpenStudy (benji):

Yes so p can equal 8 or?

OpenStudy (kise_ryouta):

-8 THANKS ALOT KOOKIE

OpenStudy (benji):

ya boi

OpenStudy (irishboy123):

wrong

OpenStudy (kise_ryouta):

thank you benji it was right

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