1. **Stating the problem:** We are given two inequalities involving variables and the variable $x$: $$1 - \geq \lambda x \pi$$ $$0 \leq \xi + x b$$ 2. **Rewrite the inequalities for clarity:** The first inequality can be interpreted as: $$1 \geq \lambda x \pi$$ which means: $$1 \geq \lambda \pi x$$ The second inequality is: $$0 \leq \xi + b x$$ 3. **Solve the first inequality for $x$:** Start with: $$1 \geq \lambda \pi x$$ Divide both sides by $\lambda \pi$ (assuming $\lambda \pi \neq 0$): $$\frac{1}{\cancel{\lambda \pi}} \geq \cancel{\frac{\lambda \pi}{\lambda \pi}} x$$ which simplifies to: $$x \leq \frac{1}{\lambda \pi}$$ 4. **Solve the second inequality for $x$:** Start with: $$0 \leq \xi + b x$$ Subtract $\xi$ from both sides: $$-\xi \leq b x$$ Divide both sides by $b$ (assuming $b \neq 0$): $$\frac{-\xi}{\cancel{b}} \leq \cancel{\frac{b}{b}} x$$ which simplifies to: $$x \geq \frac{-\xi}{b}$$ 5. **Summary of solution:** The values of $x$ satisfy both inequalities simultaneously: $$\frac{-\xi}{b} \leq x \leq \frac{1}{\lambda \pi}$$ This means $x$ lies in the interval between $\frac{-\xi}{b}$ and $\frac{1}{\lambda \pi}$, assuming all denominators are nonzero. **Final answer:** $$\boxed{\frac{-\xi}{b} \leq x \leq \frac{1}{\lambda \pi}}$$