Question

Solve the following system of inequalities:

∣2x−3∣≤11 and ∣ x − 2 ∣ ≥ 3 ∣x−2∣≥3

Answer 1

x ∈ [-4, -1] ∪ [5, 7]

Answer 2

To solve the given system of inequalities, we will solve each inequality separately and then find the intersection of their solution sets.

Inequality 1: ∣2x−3∣≤11

This absolute value inequality can be rewritten as a compound inequality: $-11 \le 2x - 3 \le 11$

Add 3 to all parts of the inequality: $-11 + 3 \le 2x \le 11 + 3$ $-8 \le 2x \le 14$

Divide all parts by 2: $\frac{-8}{2} \le x \le \frac{14}{2}$ $-4 \le x \le 7$ The solution set for the first inequality is $x \in [-4, 7]$. Let's call this set $S_1$.

Inequality 2: ∣x−2∣≥3

This absolute value inequality can be rewritten as two separate inequalities: $x - 2 \ge 3 \quad \text{OR} \quad x - 2 \le -3$

Solve the first part: $x - 2 \ge 3$ $x \ge 3 + 2$ $x \ge 5$

Solve the second part: $x - 2 \le -3$ $x \le -3 + 2$ $x \le -1$

The solution set for the second inequality is $x \in (-\infty, -1] \cup [5, \infty)$. Let's call this set $S_2$.

Finding the Intersection of the Solution Sets

We need to find the values of $x$ that satisfy both inequalities, which means finding the intersection of $S_1$ and $S_2$: $S_1 \cap S_2 = [-4, 7] \cap ((-\infty, -1] \cup [5, \infty))$

To find this intersection, we can consider the intersection of $[-4, 7]$ with each part of $S_2$ and then take their union: $([-4, 7] \cap (-\infty, -1]) \cup ([-4, 7] \cap [5, \infty))$

1. Intersection of [-4, 7] and (-∞, -1]: The numbers that are greater than or equal to -4 AND less than or equal to -1 are in the interval $[-4, -1]$.

2. Intersection of [-4, 7] and [5, ∞): The numbers that are less than or equal to 7 AND greater than or equal to 5 are in the interval $[5, 7]$.

Combining these two resulting intervals with a union gives the final solution: $[-4, -1] \cup [5, 7]$

The solution represents all real numbers $x$ such that $x$ is between -4 and -1 (inclusive) or $x$ is between 5 and 7 (inclusive).