1. **State the problem:** We have 50 stereo systems in total. The lowest priced stereo system costs $800. The highest priced stereo system costs $3000. The total value of all stereo systems is $111000. We want to find the maximum number of $3000 stereo systems in the inventory. 2. **Define variables:** Let $x$ be the number of $3000 systems. Let $y$ be the number of $800 systems. 3. **Write the equations:** Total number of systems: $$x + y = 50$$ Total value of systems: $$3000x + 800y = 111000$$ 4. **Express $y$ in terms of $x$ from the first equation:** $$y = 50 - x$$ 5. **Substitute $y$ into the value equation:** $$3000x + 800(50 - x) = 111000$$ 6. **Simplify and solve for $x$:** $$3000x + 40000 - 800x = 111000$$ $$2200x + 40000 = 111000$$ $$2200x = 111000 - 40000$$ $$2200x = 71000$$ $$x = \frac{71000}{2200} = 32.2727...$$ 7. **Interpret the result:** Since $x$ must be an integer and cannot exceed the total inventory, the maximum number of $3000 systems is 32. **Final answer:** $$\boxed{32}$$ stereo systems priced at $3000 can be in the inventory at most.