3+1/2x-8 < 1
[tex]\it 3+\dfrac{1}{2}x -8 \ \textless \ 1 \Leftrightarrow \dfrac{1}{2}x \ \textless \ 1+8-3 \Leftrightarrow \dfrac{1}{2}x \ \textless \ 6|_{\cdot2} \Leftrightarrow x \ \textless \ 12 \Leftrightarrow
\\\;\\
\Leftrightarrow x \in (-\infty, \ 12)[/tex]
Prin urmare, mulțimea soluțiilor este :
S = (∞, 12).
