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OpenStudy (anonymous):
|x/3+1|=6
A. {-7, 15}
B. {-21, 15}
C. {7, 15}
D. {-15, 21}
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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?
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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! :)
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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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