$$lx+1\le\frac{3x+10}{4}$$
$\left\{\begin{matrix}x\geq \frac{6}{4l-3}\text{, }&l<\frac{3}{4}\\x\leq \frac{6}{4l-3}\text{, }&l>\frac{3}{4}\\x\in \mathrm{R}\text{, }&l=\frac{3}{4}\\x=\frac{6}{4l-3}\text{, }&l\neq \frac{3}{4}\end{matrix}\right.$
$\left\{\begin{matrix}l\geq \frac{3}{4}+\frac{3}{2x}\text{, }&x<0\\l\leq \frac{3}{4}+\frac{3}{2x}\text{, }&x>0\\l\in \mathrm{R}\text{, }&x=0\\l=\frac{3}{4}+\frac{3}{2x}\text{, }&x\neq 0\end{matrix}\right.$