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Algebra 12 Online
OpenStudy (anonymous):

|x/3+1|=6 A. {-7, 15} B. {-21, 15} C. {7, 15} D. {-15, 21}

terenzreignz (terenzreignz):

\[\large \left|\frac{x}3+1\right|=6\]

terenzreignz (terenzreignz):

All this really means is that you have two equations to solve. \[\Large \frac{x}3+1 = 6\] and \[\Large \frac{x}3+1 = -6\]

OpenStudy (anonymous):

I just don't know how to do it with fractions exactly... @terenzreignz

terenzreignz (terenzreignz):

Well then, you can always multiply everything by 3, thereby cancelling out the denominator. \[\Large x + 3 = 18\] and \[\Large x+3 = -18\]

OpenStudy (anonymous):

oh, right right. After that would I divide?

OpenStudy (anonymous):

@terenzreignz

terenzreignz (terenzreignz):

Not yet, you solve for x, on both equations.

OpenStudy (anonymous):

would it be {15, -21}

terenzreignz (terenzreignz):

Yes, indeed :)

OpenStudy (anonymous):

YES! Thank you so much! :)

OpenStudy (anonymous):

\[\left| \frac{ x }{3 }+1 \right|=6\] \[\left| \frac{ x+3 }{3 } \right|=6,\left| x+3 \right|=18,x+3=\pm18\] now you can solve the two equations.

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