Question

−9≤32z−3≤3
Write the solution using interval notation.

Answer 1

`z \in \left[-(3)/(16), (3)/(16) \right]`

Explanation:

`-9 \leq 32z-3 \leq 3 \\ \\ -6 \leq 32z \leq 6 \\ \\ -(3)/(16) \leq z \leq (3)/(16) \\ \\ \therefore z \in \left[-(3)/(16), (3)/(16) \right]`