Absolute Value Inequalities - College Algebra

A delivery hub is at mile marker $12$ ; drivers stay in-zone if $|x-12|\le 5$ . What mile markers qualify?
Answer: $x\in[7,17]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 12| ≤ 5 represents a range of acceptable mile markers around 12, bounded by a distance of 5 miles. The correct answer choice correctly identifies the solution set as the closed interval [7,17], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A bottle fill volume targets $500$ mL; regulation requires $|x-500|<8$ . What is the solution set?
Answer: $x\in(492,508)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 500| < 8 represents a range of acceptable volumes around 500 mL, bounded by a distance of 8 mL. The correct answer choice correctly identifies the solution set as the open interval (492,508), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A circuit board thickness targets $1.6$ mm; specification is $|x-1.6|\le 0.1$ . What values satisfy this?
Answer: $x\in[1.5,1.7]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 1.6| ≤ 0.1 represents a range of acceptable thicknesses around 1.6 mm, bounded by a distance of 0.1 mm. The correct answer choice correctly identifies the solution set as the closed interval [1.5,1.7], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
Consider the compound inequality $|2x - 3| \geq 5$ AND $|x + 1| < 4$ . The solution set contains exactly how many integers?
Answer: $4$ integers We solve each inequality separately, then find their intersection. For $|2x - 3| \geq 5$ : This means $2x - 3 \geq 5$ or $2x - 3 \leq -5$ . From the first: $2x \geq 8$ , so $x \geq 4$ . From the second: $2x \leq -2$ , so $x \leq -1$ . Therefore, $|2x - 3| \geq 5$ when $x \in (-\infty, -1] \cup [4, \infty)$ . For $|x + 1| < 4$ : This means $-4 < x + 1 < 4$ , so $-5 < x < 3$ . Therefore, $|x + 1| < 4$ when $x \in (-5, 3)$ . The intersection is: $((-\infty, -1] \cup [4, \infty)) \cap (-5, 3) = (-5, -1] \cup \emptyset = (-5, -1]$ . The integers in the interval $(-5, -1]$ are $-4, -3, -2, -1$ . Let's verify: For $x = -4$ : $|2(-4) - 3| = |-11| = 11 \geq 5$ ✓ and $|-4 + 1| = 3 < 4$ ✓. For $x = -3$ : $|2(-3) - 3| = |-9| = 9 \geq 5$ ✓ and $|-3 + 1| = 2 < 4$ ✓. For $x = -2$ : $|2(-2) - 3| = |-7| = 7 \geq 5$ ✓ and $|-2 + 1| = 1 < 4$ ✓. For $x = -1$ : $|2(-1) - 3| = |-5| = 5 \geq 5$ ✓ and $|-1 + 1| = 0 < 4$ ✓. For $x = 0$ : $|2(0) - 3| = 3 \not\geq 5$ ✗. Therefore, there are exactly 4 integers in the solution set.
A monthly budget targets $1200$ dollars; to stay flexible, $|x-1200|\le 150$ . What spending range is allowed?
Answer: $x\in[1050,1350]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 1200| ≤ 150 represents a range of acceptable spending around 1200 dollars, bounded by a distance of 150 dollars. The correct answer choice correctly identifies the solution set as the closed interval [1050,1350], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A delivery zone is centered at $x=40$ on a straight highway map; service is offered when $|x-40|\le 9$ . What is the range?
Answer: $x\in[31,49]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 40| ≤ 9 represents a range of acceptable positions around 40, bounded by a distance of 9 units. The correct answer choice correctly identifies the solution set as the closed interval [31,49], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A bolt diameter targets $8$ mm; rejects occur unless $|x-8|<0.05$ . Which interval represents accepted diameters?
Answer: $x\in(7.95,8.05)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 8| < 0.05 represents a range of acceptable diameters around 8 mm, bounded by a distance of 0.05 mm. The correct answer choice correctly identifies the solution set as the open interval (7.95,8.05), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A lab incubator is set to $37^\circ\text{C}$ and must satisfy $|T-37|<0.8$ for valid trials. Which interval fits?
Answer: $T\in(36.2,37.8)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |T - 37| < 0.8 represents a range of acceptable temperatures around 37°C, bounded by a distance of 0.8°C. The correct answer choice correctly identifies the solution set as the open interval (36.2,37.8), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
If $|x - a| + |x - b| \leq c$ has no solution, where $a < b$ and $c > 0$ , then which condition must be true?
Answer: $c < b - a$ regardless of the value of $a + b$ The function $f(x) = |x - a| + |x - b|$ represents the sum of distances from $x$ to points $a$ and $b$ on the number line. The minimum value of this function occurs when $x$ is between $a$ and $b$ (i.e., $a \leq x \leq b$ ), and this minimum value is $b - a$ . Specifically, for any $x \in [a, b]$ , we have $f(x) = |x - a| + |x - b| = (x - a) + (b - x) = b - a$ . For $x < a$ , we have $f(x) = (a - x) + (b - x) = a + b - 2x > a + b - 2a = b - a$ . For $x > b$ , we have $f(x) = (x - a) + (x - b) = 2x - a - b > 2b - a - b = b - a$ . Therefore, the minimum value of $|x - a| + |x - b|$ is exactly $b - a$ , achieved on the interval $[a, b]$ . For the inequality $|x - a| + |x - b| \leq c$ to have no solution, we need $c$ to be strictly less than the minimum possible value of $|x - a| + |x - b|$ . Since this minimum value is $b - a$ , we need $c < b - a$ . The value of $a + b$ is irrelevant to this condition - what matters is only the distance $b - a$ between the two points and how it compares to $c$ . Therefore, the condition is $c < b - a$ regardless of the value of $a + b$ .
A lens curvature measurement targets $12$ units; rework is needed when $|x-12|\ge 0.6$ . Which set needs rework?
Answer: $x\in(-\infty,11.4]\cup[12.6,\infty)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 12| ≥ 0.6 represents measurements outside a range around 12 units, bounded by a distance of 0.6 units, requiring rework. The correct answer choice correctly identifies the solution set as the union of intervals (-∞,11.4] ∪ [12.6,∞), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A machined gear targets $x=32$ mm; tolerance is $|x-32|\le 0.4$ to ensure fit. What is the solution set?
Answer: $x\in[31.6,32.4]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 32| ≤ 0.4 represents a range of acceptable gear sizes around 32 mm, bounded by a distance of 0.4 mm. The correct answer choice correctly identifies the solution set as the closed interval [31.6,32.4], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A factory machines rods with target length $50$ mm; quality requires $|x-50|\le 0.2$ . What is the acceptable range?
Answer: $x\in[49.8,50.2]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 50| ≤ 0.2 represents a range of acceptable lengths around 50 mm, bounded by a distance of 0.2 mm. The correct answer choice correctly identifies the solution set as the closed interval [49.8,50.2], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A courier operates near downtown at $x=30$ on a linear map; policy is $|x-30|<6$ . Which interval applies?
Answer: $x\in(24,36)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 30| < 6 represents a range of acceptable positions around 30, bounded by a distance of 6 units. The correct answer choice correctly identifies the solution set as the open interval (24,36), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A travel budget centers at $950$ dollars; it stays on-plan if $|x-950|<75$ . Which interval represents acceptable $x$ ?
Answer: $x\in(875,1025)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 950| < 75 represents a range of acceptable budgets around 950 dollars, bounded by a distance of 75 dollars. The correct answer choice correctly identifies the solution set as the open interval (875,1025), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A rescue boat must stay at least $4$ nautical miles from reef at $x=18$ ; rule is $|x-18|\ge 4$ . What is the solution set?
Answer: $x\in(-\infty,14]\cup[22,\infty)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 18| ≥ 4 represents positions outside a range around 18, bounded by a distance of 4 nautical miles. The correct answer choice correctly identifies the solution set as the union of intervals (-∞,14] ∪ [22,∞), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A thermostat is calibrated at $21^\circ\text{C}$ ; comfort requires $|T-21|<3$ . Which option gives the correct range?
Answer: $T\in(18,24)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |T - 21| < 3 represents a range of acceptable temperatures around 21°C, bounded by a distance of 3°C. The correct answer choice correctly identifies the solution set as the open interval (18,24), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A conference budget centers at $15{,}000$ dollars; it must satisfy $|x-15000|<1200$ . What is the solution set?
Answer: $x\in(13800,16200)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 15000| < 1200 represents a range of acceptable budgets around 15000 dollars, bounded by a distance of 1200 dollars. The correct answer choice correctly identifies the solution set as the open interval (13800,16200), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
The solution set of $|3x + 2| - |x - 4| > 2x + 6$ includes which of the following intervals?
Answer: $(-\infty, -4)$ is entirely contained in the solution set We solve $|3x + 2| - |x - 4| > 2x + 6$ by considering cases based on the critical points $x = -\frac{2}{3}$ (where $3x + 2 = 0$ ) and $x = 4$ (where $x - 4 = 0$ ). Case 1: $x < -\frac{2}{3}$ . Then $|3x + 2| = -(3x + 2) = -3x - 2$ and $|x - 4| = -(x - 4) = 4 - x$ . The inequality becomes $(-3x - 2) - (4 - x) > 2x + 6$ , which simplifies to $-3x - 2 - 4 + x > 2x + 6$ , giving $-2x - 6 > 2x + 6$ . This leads to $-4x > 12$ , so $x < -3$ . Since we need $x < -\frac{2}{3} \approx -0.67$ , the solution in this case is $x < -3$ . Case 2: $-\frac{2}{3} \leq x < 4$ . Then $|3x + 2| = 3x + 2$ and $|x - 4| = 4 - x$ . The inequality becomes $(3x + 2) - (4 - x) > 2x + 6$ , which simplifies to $3x + 2 - 4 + x > 2x + 6$ , giving $4x - 2 > 2x + 6$ . This leads to $2x > 8$ , so $x > 4$ . But this contradicts our case assumption $x < 4$ , so there are no solutions in this interval. Case 3: $x \geq 4$ . Then $|3x + 2| = 3x + 2$ and $|x - 4| = x - 4$ . The inequality becomes $(3x + 2) - (x - 4) > 2x + 6$ , which simplifies to $3x + 2 - x + 4 > 2x + 6$ , giving $2x + 6 > 2x + 6$ . This simplifies to $0 > 0$ , which is false, so there are no solutions for $x \geq 4$ . Therefore, the solution set is $(-\infty, -3)$ , which means $(-\infty, -4)$ is entirely contained in the solution set.
A club expects $450$ dollars in dues; it remains solvent only if $|x-450|<40$ . Which interval fits $x$ ?
Answer: $x\in(410,490)$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 450| < 40 represents a range of acceptable dues around 450 dollars, bounded by a distance of 40 dollars. The correct answer choice correctly identifies the solution set as the open interval (410,490), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.
A project estimates $8000$ dollars; approval requires $|x-8000|\le 600$ . What is the permitted cost range?
Answer: $x\in[7400,8600]$ This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 8000| ≤ 600 represents a range of acceptable costs around 8000 dollars, bounded by a distance of 600 dollars. The correct answer choice correctly identifies the solution set as the closed interval [7400,8600], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

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