
\begin{enumerate}
\item $\begin{aligned}[t]
|x - 7| = 3 & \iff x-7=3\qquad\text{ou}\qquad -(x-7)=3\\
& \iff x=3+7 \qquad\text{ou}\qquad x-7=-3\\
& \iff x=10 \qquad\text{ou}\qquad x=4
\end{aligned}$

D'où $\boxed{S=\{4;10\}}$

\item $\begin{aligned}[t]
|x - 4| \leqslant 8 & \iff x-4 \leqslant 8 \qquad\text{ou}\qquad -(x-4) \leqslant8\\
& \iff x \leqslant 8+4 \qquad\text{ou}\qquad x-4 \geqslant-8\\
& \iff x \leqslant 12 \qquad\text{ou}\qquad x \geqslant-4
\end{aligned}$

Donc $\boxed{S=\intervFF{-4}{12}}$

\item $\begin{aligned}[t]
|x + 7| \geqslant 3 & \iff x+7 \geqslant3 \qquad\text{ou}\qquad -(x+7) \geqslant3\\
& \iff x \geqslant3-7 \qquad\text{ou}\qquad x+7 \leqslant -3\\
& \iff x \geqslant-4 \qquad\text{ou}\qquad x \leqslant-10
\end{aligned}$

D'où $\boxed{S=\intervOF{-\infty}{-10} \cup \intervFO{-4}{+\infty}}$
\end{enumerate}
