<PRE><B>Question 582681</B>
No luck involved.


Translate into an algebra problem (this requires practice) where x is the number


*[tex \LARGE 3(x+4) = \frac{1}{2}x - 8]


*[tex \LARGE 3x + 12 = \frac{1}{2}x - 8]


Subtract x/2 from both sides


*[tex \LARGE \frac{5}{2}x + 12 = -8]


Subtract 12 from both sides


*[tex \LARGE \frac{5}{2}x = -20 \Rightarrow x = (-20)(\frac{2}{5}) = -8]</PRE>