Free math worksheets

- 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.

- 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, \((-\infty,\infty)\).

- 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) \((-\infty,-2]\cup[4,\infty)\)

- Interpret the inequality as a statement about distance from the center \(-1\). - Decide whether “greater than 5” describes points between the boundaries or outside them. - For an outside solution set, consider the two separate directions from the center.

a) The student used an inside interval, which corresponds to an absolute value being less than 5. The symbol \(>\) asks for points whose distance from \(-1\) is greater than 5, so the solutions lie outside the two boundary points and require an OR statement. b) Write \(x+1<-5\) or \(x+1>5\). Subtract 1 in each inequality: \(x<-6\) or \(x>4\). In interval notation, the solution is \((-\infty, -6)\cup(4, \infty)\).

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

- Use the V's vertex to identify its horizontal shift. - The horizontal line's y-value gives the comparison level. - For the algebraic check, translate the absolute-value inequality into a compound inequality.

1. The V-shaped graph has vertex \((1,0)\), so \(f(x)=|x-1|\). The horizontal graph is at \(y=3\), so \(g(x)=3\). 2. The graphs intersect at \(x=-2\) and \(x=4\). Between these values, including the intersections, \(f\) is at or below \(g\). Thus, the interval is \([-2,4]\). 3. The corresponding inequality is \(|x-1|\le3\). 4. Rewrite it as \(-3\le x-1\le3\). Add \(1\) to all three parts: \(-2\le x\le4\), confirming \([-2,4]\).

a) \(f(x)=|x-1|\) and \(g(x)=3\) b) \([-2,4]\) c) \(|x-1|\le3\), which gives \(-2\le x\le4\)

- 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.