- Interpret absolute value as distance from 0. - “Less than” a positive distance describes points between two boundary values. - Decide whether the boundary points themselves are included.

The inequality asks for numbers whose distance from 0 is less than 4. Therefore, \(-4<x<4\), which is the interval \((-4, 4)\).

\(-4<x<4\), or \((-4, 4)\). These are the numbers less than 4 units from 0.

- Try placing the three candidates on a number line around zero. - Compare their distances, not whether the numbers themselves are positive or negative.

1. The absolute values are \(3\), \(3\), and \(4\). 2. The sign \(\le\) includes a distance of exactly \(3\), so \(-3\) and \(3\) work, but \(4\) does not.

\(-3\) and \(3\); the boundary distance is included.

- Which number is the center of the distance? - How far is zero from that center?

1. Substitute \(x=0\): \(|0-5|=5\). 2. Since \(5<6\), \(x=0\) is a solution: it lies fewer than \(6\) units from \(5\).

Yes. \(|0-5|=5<6\), so \(0\) is only \(5\) units from \(5\), within the allowed distance.

- Which two points are exactly \(5\) units from zero? - Consider a number far to the left and another far to the right. Must both conditions hold at once?

1. Distance at least \(5\) from zero includes points on either side, including the boundary points. 2. Therefore \(x\le-5\) or \(x\ge5\). Use “or” because no number can be both at most \(-5\) and at least \(5\).

\(x\le-5\) or \(x\ge5\). Use “or” because either side of zero can be far enough away.

- 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 \(2\). - Locate the two boundary points exactly \(4\) units from \(2\). - Because the inequality includes equality, decide what that means for the endpoints.

1. The expression \(|x-2|\) is the distance from \(x\) to \(2\). 2. The condition says this distance is at most \(4\), so \(-4\le x-2\le4\). 3. Add \(2\) to all three parts: \(-2\le x\le6\). 4. Therefore, every real number in \([-2,6]\) is a solution.

\([-2,6]\)

- Find the midpoint of the two interval endpoints. - Compare the midpoint's distance to either endpoint. - Use the brackets to decide whether the boundary distance is included.

The center of the interval is the midpoint of \(-2\) and 6: \(a=\frac{-2+6}{2}=2\). The distance from the center to either endpoint is 4, so \(b=4\). Because both endpoints are included, the equivalent inequality is \(\lvert x-2\rvert\le4\).

\(\lvert x-2\rvert\le4\), with center \(a=2\) and distance \(b=4\).

- What is the smallest possible value of an absolute value? - Compare that minimum with the right side \(-3\). - Decide whether any real x-value could make the inequality false.

1. Absolute value is always nonnegative, so \(|2x-5|\ge0\) for every real \(x\). 2. Every nonnegative number is greater than or equal to \(-3\). 3. Therefore, every real number satisfies the inequality. There are no boundary points because the right side is negative.

All real numbers, \((-\infty,\infty)\).

- Think about points on both sides of zero, not just positive points. - Test whether your general rule would accept a point exactly \(2\) units away.

1. The distances are \(|-3|=3\) and \(|1|=1\). 2. Since \(3>2\), \(x=-3\) works, but \(x=1\) does not. 3. Every point more than \(2\) units from zero satisfies \(|x|>2\).

a) \(x=-3\) works; \(x=1\) does not. Their distances are \(3\) and \(1\). b) \(|x|>2\)

- Which axis numbers are printed, and how many ticks separate them? - What distance from a single center reaches both marked ends? - How does a filled endpoint affect the inequality sign?

1. Each tick is \(1\) unit. The left endpoint is \(-2\), not \(-3\). Both endpoints are filled, so the interval is \([-2,4]\). 2. The center is \(h=\frac{-2+4}{2}=1\), and each endpoint is \(3\) units from it. 3. The inequality is \(|x-1|\le3\).

The left endpoint is \(-2\); interval \([-2,4]\); \(|x-1|\le3\).

- Compare the distances of a number near \(4\) and one far from \(4\). - Test values just below and just above your proposed boundary points. - Check your two intervals against the original inequality.

1. The distance from \(4\) must exceed \(2\), so \(x-4<-2\) or \(x-4>2\). 2. Add \(4\): \(x<2\) or \(x>6\). 3. The intervals are \((-\infty,2)\cup(6,\infty)\). Numbers between \(2\) and \(6\) are at most \(2\) units from \(4\).

\(x<2\) or \(x>6\); \((-\infty,2)\cup(6,\infty)\). Values from \(2\) through \(6\) are no farther than \(2\) units from \(4\), so they do not work.

- 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.

a) The condition \(|x| < 7\) means the distance from \(x\) to \(0\) is less than \(7\), so \(-7 < x < 7\). b) \(|x| \le 3.5\) means \(-3.5 \le x \le 3.5\), so the interval is \([-3.5, 3.5]\). c) The symmetric open interval \((-9.2, 9.2)\) is represented by \(|x| < 9.2\). d) Absolute value represents distance and is always nonnegative. It cannot be less than \(-1\), so there is no solution.

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 and look for a counterexample if one interval is not contained in the other.

1. Ethan's statement is \(|z|\le6\), which is equivalent to \(-6\le z\le6\). 2. Maya's statement is \(|z-4|\le3\), which is equivalent to \(1\le z\le7\). 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.

- In each part, isolate the absolute-value expression before translating it into ordinary inequalities. - A strict “less than” absolute-value inequality describes an inside interval. - An “at least” absolute-value inequality describes two outside regions.

1. In (a), rewrite as \(-10<2x+6<10\). Subtract \(6\): \(-16<2x<4\). Divide by \(2\): \(-8<x<2\). 2. In (b), add \(5\): \(4|x-1|\ge12\). Divide by \(4\): \(|x-1|\ge3\). 3. A distance of at least \(3\) from \(1\) means \(x-1\le-3\) or \(x-1\ge3\). 4. Therefore, \(x\le-2\) or \(x\ge4\).

a) \((-8,2)\) b) \((-\infty,-2]\cup[4,\infty)\)

- How far from \(-1\) must a solution lie? - Are values at the center or far away supposed to work? - Test a number well to the left of \(-1\) and another well to the right in the original inequality.

1. Aino used an inside interval, which applies when the absolute value is less than \(5\). Here the distance from \(-1\) must be greater than \(5\), so the solutions lie outside the boundary points. 2. Write \(x+1<-5\) or \(x+1>5\). Subtract \(1\): \(x<-6\) or \(x>4\). 3. The interval form is \((-\infty,-6)\cup(4,\infty)\).

a) Aino used an inside/AND condition instead of the outside/OR condition. b) \(x<-6\) or \(x>4\); \((-\infty,-6)\cup(4,\infty)\)

- Translate “within” into a statement about distance from the target fill amount. - The phrase “including the tolerance limits” determines whether the boundary values are included. - After removing the absolute value, keep both lower and upper bounds together as one compound inequality.

1. “Within \(0.3\)” of \(16\) means the distance from \(v\) to \(16\) is at most \(0.3\): \(|v-16|\le0.3\). 2. Rewrite as \(-0.3\le v-16\le0.3\). 3. Add \(16\) to all three parts: \(15.7\le v\le16.3\). 4. The endpoints are included, so the graph has closed points at \(15.7\) and \(16.3\) with the segment between them shaded.

a) \(|v-16|\le0.3\) b) \([15.7,16.3]\,\text{fl oz}\) c) Closed points at \(15.7\) and \(16.3\), shaded between them.