ແກ້ |q(x)|+|r(x)|leq1

$$| q ( x ) | + | r ( x ) | \leq 1$$

ປະເມີນ (complex solution)

$$|x|\left(|q|+|r|\right)\leq 1$$

ແກ້ສຳລັບ x

$$\left\{\begin{matrix}x\in \begin{bmatrix}\frac{1}{q-|r|},-\frac{1}{q-|r|}\end{bmatrix}\text{, }&\left(q>-|r|\text{ and }q\leq 0\right)\text{ or }\left(q\leq -|r|\text{ and }q<|r|\right)\text{ or }q<0\\x\in \begin{bmatrix}-\frac{1}{|r|+q},\frac{1}{|r|+q}\end{bmatrix}\text{, }&\left(q\geq 0\text{ and }q>-|r|\right)\text{ or }\left(q>0\text{ and }q<|r|\right)\\x\in \begin{bmatrix}-\frac{1}{|r|+q},\frac{1}{|r|+q}\end{bmatrix}\text{, }&\left(q\geq 0\text{ and }q>-|r|\right)\text{ or }\left(r\neq 0\text{ and }q<|r|\text{ and }q>0\right)\\x\in \mathrm{R}\text{, }&q=0\text{ and }r=0\end{matrix}\right,$$

ແກ້ສຳລັບ r

$$\left\{\begin{matrix}r\in \mathrm{R}\text{, }&x=0\\r\in \begin{bmatrix}|q|-\frac{1}{|x|},-|q|+\frac{1}{|x|}\end{bmatrix}\text{, }&x\neq 0\text{ and }|q|\leq \frac{1}{|x|}\end{matrix}\right,$$