a) All real numbers \( x \) less than 4 units from 8 . b) All real numbers \( x \) more than 7 units from 0 c) All real numbers \( x \) at least 9 units from 7 . (d) All real numbers \( x \) at most 4 units from 3 .

Let's analyze each part of the problem and express the conditions in mathematical notation. a) **All real numbers \( x \) less than 4 units from 8.** This can be expressed as: \[ |x - 8| < 4 \] To solve this inequality, we can break it down: \[ -4 < x - 8 < 4 \] Adding 8 to all parts: \[ 4 < x < 12 \] So, the solution is: \[ x \in (4, 12) \] b) **All real numbers \( x \) more than 7 units from 0.** This can be expressed as: \[ |x| > 7 \] This inequality can be split into two cases: 1. \( x > 7 \) 2. \( x < -7 \) Thus, the solution is: \[ x \in (-\infty, -7) \cup (7, \infty) \] c) **All real numbers \( x \) at least 9 units from 7.** This can be expressed as: \[ |x - 7| \geq 9 \] This inequality can also be split into two cases: 1. \( x - 7 \geq 9 \) which simplifies to \( x \geq 16 \) 2. \( x - 7 \leq -9 \) which simplifies to \( x \leq -2 \) Thus, the solution is: \[ x \in (-\infty, -2] \cup [16, \infty) \] d) **All real numbers \( x \) at most 4 units from 3.** This can be expressed as: \[ |x - 3| \leq 4 \] Breaking this down gives: \[ -4 \leq x - 3 \leq 4 \] Adding 3 to all parts: \[ -1 \leq x \leq 7 \] So, the solution is: \[ x \in [-1, 7] \] In summary, the solutions are: a) \( x \in (4, 12) \) b) \( x \in (-\infty, -7) \cup (7, \infty) \) c) \( x \in (-\infty, -2] \cup [16, \infty) \) d) \( x \in [-1, 7] \)