\ANG{ 
L"ose:
\begin{enumerate}
\item      $7\cdot(8x+3)-(5x-3)\cdot 8=0$
\item      $\frac{1}{2}\cdot(4x+\frac{1}{3})-\frac{1}{4}\cdot(12x+1)=(9x-\frac{1}{4})\cdot\frac{1}{3}$
\item      $2\cdot(x+a)=6 - 3x +2a$
\end{enumerate} 
}

\LSG{
(a) $x=-2\frac{13}{16}$\qquad (b) $x=0$\qquad (c) $x=1\frac15$
}
