## Task 2 - 120922

Determine all real numbers $x$ for which the quotient $\frac{8-3 x}{7 x-2}$ is negative!
It is $\frac{8-3 x}{7 x-2}<0$.

ii) It is $7 x-2<0$.

Regarding i): From $8-3 x<0$ follows $x>\frac{8}{3}$ and from $7 x-2>0$ follows $x>\frac{2}{7}$. Since $\frac{8}{3}>\frac{2}{7}$, $x>\frac{2}{7}$ is automatically satisfied if $x>\frac{8}{3}$ is satisfied. The first case thus occurs exactly when $x \in\left(\frac{8}{3}, \infty\right)$.

Regarding ii): From $7 x-2<0$ follows $x<\frac{2}{7}$. Since $\frac{2}{7}<\frac{8}{3}$, $x<\frac{8}{3}$ is automatically satisfied if $x<\frac{2}{7}$ is satisfied. The second case thus occurs exactly when $x \in\left(-\infty, \frac{2}{7}\right)$.

In total, the quotient $\frac{8-3 x}{7 x-2}$ is exactly negative when $x \in\left(-\infty, \frac{2}{7}\right) \cup\left(\frac{8}{3}, \infty\right)$.