Free Math Worksheets

- Rewrite the absolute value inequality as a compound inequality. - Represent the second condition on the same number line. - Find the overlap of the two solution sets.

1. The condition \(|x| < 2.5\) is equivalent to \(-2.5 < x < 2.5\). 2. The second condition requires \(x > -1\). 3. The intersection of the two conditions is \(-1 < x < 2.5\).

\(-1 < x < 2.5\)

- Interpret \(|x - 2|\) as distance from \(2\). - Identify the points exactly \(4\) units from \(2\). - Because the inequality includes equality, include both endpoints. - Rewrite the condition as a compound inequality.

1. The expression \(|x - 2|\) is the distance from \(x\) to \(2\). 2. The condition says this distance is at most \(4\). 3. Rewrite the absolute value inequality as \(-4 \le x - 2 \le 4\). 4. Add \(2\) to all three parts: \(-2 \le x \le 6\). 5. Therefore, the solution set is every rational number in \([-2, 6]\).

\(S = \{x \in \mathbb{Q} \mid -2 \le x \le 6\}\)

- List the integers satisfying the first condition. - Interpret absolute value as distance from \(0\). - Keep only values whose distance is strictly greater than \(2\).

1. The integers satisfying \(-3 < z < 5\) are \(-2, -1, 0, 1, 2, 3, 4\). 2. The condition \(|z| > 2\) means that the distance from \(0\) is greater than \(2\). 3. Among the listed integers, only \(3\) and \(4\) have absolute value greater than \(2\).

\(S = \{3, 4\}\)

- Interpret absolute value as distance from \(0\). - A distance less than a positive number creates a symmetric interval. - Decide whether each endpoint is included. - A distance cannot be negative.

1. The condition \(|x| < 7\) means that the distance from \(x\) to \(0\) is less than \(7\), so \(-7 < x < 7\). 2. The condition \(|x| \le 3.5\) means \(-3.5 \le x \le 3.5\), so the interval is \([-3.5, 3.5]\). 3. The symmetric open interval \((-9.2, 9.2)\) is represented by \(|x| < 9.2\). 4. Absolute value represents distance and is always nonnegative. It can never be less than \(-1\), so the solution set is empty.

a) \(-7 < x < 7\) b) \([-3.5, 3.5]\) c) \(|x| < 9.2\) d) No solution; \(|x|\) cannot be negative.

- Rewrite each absolute value statement as an interval. - Interpret \(|z - 4|\) as distance from \(4\). - Compare the endpoints of the two intervals. - Look for a counterexample in Maya's interval but outside Ethan's.

1. Ethan's statement is \(|z| \le 6\), which is equivalent to \(-6 \le z \le 6\). 2. Maya's statement is \(|z - 4| \le 3\), which is equivalent to \(1 \le z \le 7\). 3. Maya's interval extends to \(7\), while Ethan's interval ends at \(6\). 4. The number \(7\) is a counterexample: it is \(3\) units from \(4\), but \(|7| = 7 > 6\). 5. Therefore, not every number satisfying Maya's statement satisfies Ethan's statement.

No. Maya''s solution set is \([1, 7]\), while Ethan''s is \([-6, 6]\). For example, \(z = 7\) satisfies Maya''s condition but not Ethan''s.

- Isolate the absolute value expression first. - An inequality of the form \(|u| < a\) describes values between two boundaries. - An inequality of the form \(|u| \ge a\) describes two outer regions. - Solve both cases in part b).

1. For a), rewrite the inequality as \(-10 < 2x + 6 < 10\). 2. Subtract \(6\): \(-16 < 2x < 4\). Divide by \(2\): \(-8 < x < 2\). 3. For b), add \(5\): \(4|x - 1| \ge 12\). Divide by \(4\): \(|x - 1| \ge 3\). 4. A distance of at least \(3\) from \(1\) means \(x - 1 \ge 3\) or \(x - 1 \le -3\). 5. Solve the two cases: \(x \ge 4\) or \(x \le -2\).

a) \(S = \{x \in \mathbb{Q} \mid -8 < x < 2\}\) b) \(S = \{x \in \mathbb{Q} \mid x \le -2 \text{ or } x \ge 4\}\)