Three sets of real numbers are defined by A = {x : |2x - 3| < 7}, B = {x : x² - 9x + 8 ≤ 0}, C = {x…

The three intervals. $-7 < 2x - 3 < 7$ gives $A = \left(-2, 5\right)$. $x^2 - 9x + 8 = \left(x - 8\right) \left(x - 1\right) \le 0$ gives $B = \left[1, 8\right]$. $2^x > 4 = 2^{2}$ gives $C = \left(2, \infty\right)$.

(a) $A \cap B = \left[1, 5\right)$, whose integers are $1, 2, 3, 4$: 4 of them.

(b) $A \cup B = \left(-2, 8\right]$, of length $8 - (-2) = 10$. (The two overlap, so the lengths $7$ and $7$ do not simply add.)

(c) $A - C = \left(-2, 2\right]$: the integers $-1, 0, 1, 2$, so 4. The end $2$ is in, because $C$ starts just after $2$.

(d) $A \cap B \cap C = \left(2, 5\right)$, whose integers are $3, 4$. So $S$ has $2$ elements and $2^{2} = 4$ subsets (counting the empty set and $S$ itself).

(e) $B$ has the integers $1$ to $8$, $8$ in all. $A \cap C = \left(2, 5\right)$ has the integers $3, 4$. So $8 - 2 = 6$.