 Solution

Find the salesperson identifiers satisfying each price condition using relational selection and relational projection:Intersect these identifier sets, relational join the result to the salesperson records, and apply relational projection to names:If intersection is not a primitive of the chosen relational algebra, use $L\cap H=L-(L-H)$ with relational difference. There is no need to access Car: the foreign key already ensures that every sale refers to a car.

Intersecting names instead of identifiers is wrong. Two different salespeople called Alex, one making only $10000$ sales and the other only $30000$ sales, would make Alex appear in both name sets although neither person satisfies the condition. Nor can one sale satisfy both price tests: the two distinct prices require distinct sale records.