Question

$$4x+5 \geq 5y3x-7 \leq -1$$

Solve for x

$\left\{\begin{matrix}x\leq \frac{3}{2}\text{, }&y_{3}=\frac{4}{5}\\x\leq \frac{6}{5y_{3}}\text{, }&y_{3}>\frac{4}{5}\\x\geq -3\text{, }&y_{3}=0\\x\geq -\frac{12}{4-5y_{3}}\text{, }&y_{3}\geq -\frac{4}{5}\text{ and }y_{3}<0\\x\geq \frac{6}{5y_{3}}\text{, }&y_{3}<-\frac{4}{5}\\x\in \begin{bmatrix}-\frac{12}{4-5y_{3}},\frac{6}{5y_{3}}\end{bmatrix}\text{, }&y_{3}>0\text{ and }y_{3}<\frac{4}{5}\\x=\frac{6}{5y_{3}}\text{, }&|y_{3}|\geq \frac{4}{5}\text{ or }y_{3}\geq 0\\x=-\frac{12}{4-5y_{3}}\text{, }&y_{3}\geq -\frac{4}{5}\text{ and }y_{3}\leq \frac{4}{5}\end{matrix}\right.$

Solve for y_3

$\left\{\begin{matrix}y_{3}\in \mathrm{R}\text{, }&x=0\\y_{3}=\frac{6}{5x}\text{, }&x\geq -\frac{3}{2}\\y_{3}\leq \frac{6}{5x}\text{, }&x>0\\y_{3}\geq \frac{4}{5}+\frac{12}{5x}\text{, }&x\leq -\frac{3}{2}\\y_{3}\geq \frac{6}{5x}\text{, }&x>-\frac{3}{2}\text{ and }x<0\end{matrix}\right.$