Solve the compound inequality. Express the answer in interval notation.
(B) $2x-3\le 1$ or $-x+4\le 10$

• A. $\left(-\infty,-14\right\rbrack\cup\left\lbrack\left.-1,\infty\right)\right.$
• B. $\left\lbrack-6,2\right\rbrack$
• C. $\left(-\infty,-6\right\rbrack\cup\left\lbrack\left.2,\infty\right)\right.$
• D. $\left(-\infty,\infty\right)$

Answer: (D)

Verified step by step guidance

1. Start by writing down the two inequalities separately: 1) \$2x - 3 \leq 1\( and 2) \)-x + 4 \leq 10$.
2. Solve the first inequality for \(x\): Add 3 to both sides to get \$2x \leq 4\(, then divide both sides by 2 to isolate \)x\(, resulting in \)x \leq 2$.
3. Solve the second inequality for \(x\): Subtract 4 from both sides to get \(-x \leq 6\), then multiply both sides by \(-1\) (remember to reverse the inequality sign) to get \(x \geq -6\).
4. Since the compound inequality uses "or", the solution is the union of the two solution sets: \(x \leq 2\) or \(x \geq -6\).
5. Analyze the union of these intervals: \((-\infty, 2] \cup [-6, \infty)\), which covers all real numbers, so the solution in interval notation is \((-\infty, \infty)\).