Absolute Value Equations and Inequalities

19 questions

Question 1Question →

For all real numbers yy that satisfy the inequality ∣∣y−3∣−6∣≤2||y - 3| - 6| \leq 2, what is the sum of all possible integer values of yy?

Show answer & explanation

Answer: 3030

Answer

The sum of all possible integer values of yy is 3030.
The compound inequality 4≤∣y−3∣≤84 \leq |y - 3| \leq 8 splits into two parts: ∣y−3∣≤8|y - 3| \leq 8 (which gives −5≤y≤11-5 \leq y \leq 11) and ∣y−3∣≥4|y - 3| \geq 4 (which gives y≤−1y \leq -1 or y≥7y \geq 7). The intersection of these intervals is [−5,−1]∪[7,11][-5, -1] \cup [7, 11]. Summing all the integers in these intervals gives (−5+−4+−3+−2+−1)+(7+8+9+10+11)=−15+45=30(-5 + -4 + -3 + -2 + -1) + (7 + 8 + 9 + 10 + 11) = -15 + 45 = 30.

Step-by-Step Solution

1
Set up the compound inequality representing the outer absolute value.
−2≤∣y−3∣−6≤2-2 \leq |y - 3| - 6 \leq 2
By definition, ∣u∣≤a|u| \leq a is equivalent to −a≤u≤a-a \leq u \leq a for a≥0a \geq 0.
2
Isolate the inner absolute value term by adding 66 to all parts of the inequality.
4≤∣y−3∣≤84 \leq |y - 3| \leq 8
Isolating the absolute value allows us to split the compound inequality into two separate cases.
3
Split the compound inequality into two separate inequalities and solve each one.
∣y−3∣≤8|y - 3| \leq 8 and ∣y−3∣≥4|y - 3| \geq 4
The expression ∣y−3∣|y - 3| must be simultaneously less than or equal to 88 and greater than or equal to 44.
4
Solve the first inequality, ∣y−3∣≤8|y - 3| \leq 8.
−5≤y≤11-5 \leq y \leq 11
Rewriting the inequality gives −8≤y−3≤8-8 \leq y - 3 \leq 8, and adding 33 to all parts yields the interval [−5,11][-5, 11].
5
Solve the second inequality, ∣y−3∣≥4|y - 3| \geq 4.
y≤−1y \leq -1 or y≥7y \geq 7
By definition, ∣u∣≥a|u| \geq a is equivalent to u≥au \geq a or u≤−au \leq -a. Thus, y−3≥4  ⟹  y≥7y - 3 \geq 4 \implies y \geq 7, and y−3≤−4  ⟹  y≤−1y - 3 \leq -4 \implies y \leq -1.
6
Determine the intersection of the two solution sets.
y∈[−5,−1]∪[7,11]y \in [-5, -1] \cup [7, 11]
The values of yy must lie within [−5,11][-5, 11] and also satisfy y≤−1y \leq -1 or y≥7y \geq 7.
7
Identify the integers within the final intervals and calculate their sum.
Sum = 3030
The integers in [−5,−1][-5, -1] are −5,−4,−3,−2,−1-5, -4, -3, -2, -1 (sum = −15-15). The integers in [7,11][7, 11] are 7,8,9,10,117, 8, 9, 10, 11 (sum = 4545). The total sum is −15+45=30-15 + 45 = 30.

Key Concept

Solving compound and nested absolute value inequalities

Alternative Method

Instead of solving the inequality algebraically, one can test the integer values around the critical points. Since yy must satisfy 4≤∣y−3∣≤84 \leq |y - 3| \leq 8, we can see that the distance of yy from 33 must be between 44 and 88 units. The integers at distance 4,5,6,7,84, 5, 6, 7, 8 to the right of 33 are 7,8,9,10,117, 8, 9, 10, 11. The integers at distance 4,5,6,7,84, 5, 6, 7, 8 to the left of 33 are −1,−2,−3,−4,−5-1, -2, -3, -4, -5. Summing these ten integers yields 3030.
Estimated Time:3m 0s
Question 2Question →

For all real values of xx where the expression is defined, which of the following inequality expressions represents the complete set of values that satisfy the inequality below?

∣x−2∣−3∣x+1∣−2≤0\frac{|x - 2| - 3}{|x + 1| - 2} \leq 0
Show answer & explanation

Answer: −3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5

Answer

−3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5
The correct answer represents the union of the two valid cases where the numerator and denominator have opposite signs. Specifically, the numerator ∣x−2∣−3|x - 2| - 3 is non-positive when −1≤x≤5-1 \leq x \leq 5, and the denominator ∣x+1∣−2|x + 1| - 2 is positive when x<−3x < -3 or x>1x > 1. Their intersection is the interval 1<x≤51 < x \leq 5. Alternatively, the numerator is non-negative when x≤−1x \leq -1 or x≥5x \geq 5, and the denominator is negative when −3<x<1-3 < x < 1. Their intersection is the interval −3<x≤−1-3 < x \leq -1. Combining these yields the solution set −3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5.

Step-by-Step Solution

1
Analyze the conditions required for a rational expression to be non-positive.
The fraction ND≤0\frac{N}{D} \leq 0 is satisfied when the numerator NN and denominator DD have opposite signs, while ensuring the denominator is not equal to zero (D≠0D \neq 0). Thus, we must evaluate two cases: Case 1 where N≤0N \leq 0 and D>0D > 0, and Case 2 where N≥0N \geq 0 and D<0D < 0.
This establishes the logical framework for solving the rational inequality.
2
Determine the critical intervals for the numerator, N=∣x−2∣−3N = |x - 2| - 3.
Solve ∣x−2∣−3≤0  ⟹  ∣x−2∣≤3  ⟹  −3≤x−2≤3  ⟹  −1≤x≤5|x - 2| - 3 \leq 0 \implies |x - 2| \leq 3 \implies -3 \leq x - 2 \leq 3 \implies -1 \leq x \leq 5. Thus, the numerator is non-positive on [−1,5][-1, 5] and non-negative on (−∞,−1]∪[5,∞)(-\infty, -1] \cup [5, \infty).
This identifies the values of xx where the numerator changes sign.
3
Determine the critical intervals for the denominator, D=∣x+1∣−2D = |x + 1| - 2, and identify restrictions.
Solve ∣x+1∣−2<0  ⟹  ∣x+1∣<2  ⟹  −2<x+1<2  ⟹  −3<x<1|x + 1| - 2 < 0 \implies |x + 1| < 2 \implies -2 < x + 1 < 2 \implies -3 < x < 1. Thus, the denominator is negative on (−3,1)(-3, 1) and positive on (−∞,−3)∪(1,∞)(-\infty, -3) \cup (1, \infty). The boundary points x=−3x = -3 and x=1x = 1 make the denominator zero and must be excluded from the domain.
This identifies the values of xx where the denominator changes sign or makes the expression undefined.
4
Evaluate Case 1 and Case 2, then combine the solutions.
For Case 1 (N≤0N \leq 0 and D>0D > 0), find the intersection of [−1,5][-1, 5] and (−∞,−3)∪(1,∞)(-\infty, -3) \cup (1, \infty), which yields 1<x≤51 < x \leq 5. For Case 2 (N≥0N \geq 0 and D<0D < 0), find the intersection of (−∞,−1]∪[5,∞)(-\infty, -1] \cup [5, \infty) and (−3,1)(-3, 1), which yields −3<x≤−1-3 < x \leq -1. Combining these intervals gives the final solution set: −3<x≤−1-3 < x \leq -1 or 1<x≤51 < x \leq 5.
This determines the overall solution set that satisfies the starting inequality.

Key Concept

Solving rational inequalities containing absolute value expressions by finding critical intervals.
Estimated Time:3m 0s
Question 3Question →

Which of the following is the solution set for the inequality ∣5−2x∣<9|5 - 2x| < 9?

Show answer & explanation

Answer: −2<x<7-2 < x < 7

Answer

−2<x<7-2 < x < 7
The correct solution is found by setting up the compound inequality −9<5−2x<9-9 < 5 - 2x < 9. Subtracting 55 from all parts gives −14<−2x<4-14 < -2x < 4. Finally, dividing by −2-2 and reversing the inequality signs yields the interval −2<x<7-2 < x < 7.

Step-by-Step Solution

1
Rewrite the absolute value inequality as a compound inequality.
−9<5−2x<9-9 < 5 - 2x < 9
An absolute value inequality of the form ∣u∣<c|u| < c represents all points within distance cc from 00, which translates to −c<u<c-c < u < c.
2
Subtract 55 from all three parts of the inequality.
−14<−2x<4-14 < -2x < 4
To isolate the variable term, we perform the inverse operation of adding 55, which is subtracting 55.
3
Divide all three parts by −2-2 and reverse the inequality signs.
−2<x<7-2 < x < 7
Dividing by a negative number reverses the inequality direction. Doing so gives 7>x>−27 > x > -2, which is conventionally written as −2<x<7-2 < x < 7.

Key Concept

Solving absolute value inequalities of the form ∣ax+b∣<c|ax + b| < c
Question 4Question →

Which of the following inequality expressions represents the complete set of real values of pp that satisfy the inequality ∣7−2p∣<13|7 - 2p| < 13?

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Answer: −3<p<10-3 < p < 10

Answer

−3<p<10-3 < p < 10
To solve ∣7−2p∣<13|7 - 2p| < 13, write it as the compound inequality −13<7−2p<13-13 < 7 - 2p < 13. Subtracting 77 from all parts gives −20<−2p<6-20 < -2p < 6. Dividing by −2-2 and reversing the inequality signs yields −3<p<10-3 < p < 10.

Step-by-Step Solution

1
Write the absolute value inequality as a compound inequality.
−13<7−2p<13-13 < 7 - 2p < 13
An absolute value inequality of the form ∣x∣<c|x| < c is equivalent to −c<x<c-c < x < c.
2
Subtract 77 from all three parts of the compound inequality.
−20<−2p<6-20 < -2p < 6
To isolate the variable term, we perform the inverse operation of adding 77 by subtracting 77 from each part.
3
Divide all parts by −2-2 and reverse the inequality signs.
10>p>−310 > p > -3, which simplifies to −3<p<10-3 < p < 10
Dividing an inequality by a negative number requires reversing the direction of the inequality signs to preserve the truth of the statement.

Key Concept

Solving absolute value inequalities of the form ∣ax+b∣<c|ax + b| < c
Question 5Question →

For what values of the real number pp is the inequality −2∣p−4∣≥−10-2|p - 4| \geq -10 true?

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Answer: −1≤p≤9-1 \leq p \leq 9

Answer

−1≤p≤9-1 \leq p \leq 9
The correct answer is the inequality showing that pp is between −1-1 and 99, inclusive. First, divide both sides of the inequality −2∣p−4∣≥−10-2|p - 4| \geq -10 by −2-2 and reverse the inequality sign to obtain ∣p−4∣≤5|p - 4| \leq 5. Next, set up the compound inequality −5≤p−4≤5-5 \leq p - 4 \leq 5. Finally, add 44 to all parts of the inequality to isolate pp, which yields −1≤p≤9-1 \leq p \leq 9.

Step-by-Step Solution

1
Divide both sides of the inequality −2∣p−4∣≥−10-2|p - 4| \geq -10 by −2-2.
∣p−4∣≤5|p - 4| \leq 5
Dividing by a negative number reverses the direction of the inequality sign.
2
Rewrite the absolute value inequality ∣p−4∣≤5|p - 4| \leq 5 as a compound inequality.
−5≤p−4≤5-5 \leq p - 4 \leq 5
An inequality of the form ∣x∣≤a|x| \leq a for a≥0a \geq 0 is equivalent to −a≤x≤a-a \leq x \leq a.
3
Add 44 to all parts of the compound inequality to isolate pp.
−1≤p≤9-1 \leq p \leq 9
Adding a constant to all parts of an inequality preserves the inequality relationships and isolates the variable.

Key Concept

Solving absolute value inequalities involving multiplication or division by a negative number.
Question 6Question →

For all real values of ww that satisfy the inequality 4−∣3−2w∣≤−14 - |3 - 2w| \leq -1, which of the following expressions represents the complete set of possible values of ww?

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Answer: w≤−1w \leq -1 or w≥4w \geq 4

Answer

The complete set of possible values is represented by the inequality w≤−1w \leq -1 or w≥4w \geq 4.
The correct answer is the solution set representing w≤−1w \leq -1 or w≥4w \geq 4. Isolating the absolute value expression yields ∣3−2w∣≥5|3 - 2w| \geq 5. This splits into two cases: 3−2w≥53 - 2w \geq 5 (which solves to w≤−1w \leq -1 after dividing by −2-2 and reversing the inequality sign) and 3−2w≤−53 - 2w \leq -5 (which solves to w≥4w \geq 4 after dividing by −2-2 and reversing the inequality sign). Combining these two cases gives the union w≤−1w \leq -1 or w≥4w \geq 4.

Step-by-Step Solution

1
Isolate the absolute value expression on one side of the inequality.
−∣3−2w∣≤−5-|3 - 2w| \leq -5, which simplifies to ∣3−2w∣≥5|3 - 2w| \geq 5 after multiplying by −1-1 and reversing the inequality sign.
Before splitting an absolute value inequality, the absolute value term must be isolated.
2
Split the absolute value inequality ∣3−2w∣≥5|3 - 2w| \geq 5 into two separate compound inequalities.
3−2w≥53 - 2w \geq 5 or 3−2w≤−53 - 2w \leq -5
An absolute value inequality of the form ∣u∣≥c|u| \geq c (where c>0c > 0) is equivalent to u≥cu \geq c or u≤−cu \leq -c.
3
Solve the first inequality: 3−2w≥53 - 2w \geq 5.
Subtracting 33 from both sides gives −2w≥2-2w \geq 2. Dividing both sides by −2-2 and reversing the inequality sign gives w≤−1w \leq -1.
Dividing or multiplying an inequality by a negative number requires reversing the direction of the inequality sign.
4
Solve the second inequality: 3−2w≤−53 - 2w \leq -5.
Subtracting 33 from both sides gives −2w≤−8-2w \leq -8. Dividing both sides by −2-2 and reversing the inequality sign gives w≥4w \geq 4.
Dividing or multiplying an inequality by a negative number requires reversing the direction of the inequality sign.
5
Combine the individual solutions to find the total solution set.
w≤−1w \leq -1 or w≥4w \geq 4
The solution to a 'greater than or equal to' absolute value inequality is the union of the solutions of the two split cases.

Key Concept

Solving absolute value inequalities with negative variable coefficients by isolating the absolute value term, splitting into cases, and reversing inequality signs when multiplying/dividing by a negative number.
Estimated Time:2m 0s
Question 7Question →

A manufacturer of precision components produces cylindrical rods. The target diameter of the rods is 1.20 centimeters1.20\text{ centimeters}. A rod is classified as Grade A if its actual diameter, dd centimeters, satisfies the inequality ∣3−2.5d∣≤0.15|3 - 2.5d| \leq 0.15. To be used in a specific high-stress assembly, the rod's diameter must also satisfy the tolerance inequality ∣d−1.22∣<0.04|d - 1.22| < 0.04. Which of the following inequality expressions represents the complete set of all possible diameters, in centimeters, of rods that qualify as Grade A and are suitable for the assembly?

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Answer: 1.18<d<1.261.18 < d < 1.26

Answer

The set of diameters satisfying both conditions is 1.18<d<1.261.18 < d < 1.26.
To satisfy both conditions, a rod's diameter must meet the Grade A requirement of 1.14≤d≤1.261.14 \le d \le 1.26 and the assembly requirement of 1.18<d<1.261.18 < d < 1.26. The intersection of these two intervals is the more restrictive range, which is 1.18<d<1.261.18 < d < 1.26. This ensures both inequalities are simultaneously true.

Step-by-Step Solution

1
Solve the first inequality representing Grade A rods: ∣3−2.5d∣≤0.15|3 - 2.5d| \le 0.15.
1.14≤d≤1.261.14 \le d \le 1.26
Rewrite the absolute value inequality as a compound inequality: −0.15≤3−2.5d≤0.15-0.15 \le 3 - 2.5d \le 0.15. Subtract 33 from all parts to get −3.15≤−2.5d≤−2.85-3.15 \le -2.5d \le -2.85. Divide all parts by −2.5-2.5, reversing the direction of the inequality signs: 1.26≥d≥1.141.26 \ge d \ge 1.14, which simplifies to 1.14≤d≤1.261.14 \le d \le 1.26.
2
Solve the second inequality representing suitability for the assembly: ∣d−1.22∣<0.04|d - 1.22| < 0.04.
1.18<d<1.261.18 < d < 1.26
Rewrite the absolute value inequality as a compound inequality: −0.04<d−1.22<0.04-0.04 < d - 1.22 < 0.04. Add 1.221.22 to all parts to isolate dd: 1.18<d<1.261.18 < d < 1.26.
3
Find the intersection of the two solution sets: [1.14,1.26]∩(1.18,1.26)[1.14, 1.26] \cap (1.18, 1.26).
1.18<d<1.261.18 < d < 1.26
For a rod to qualify for Grade A and be suitable for the assembly, its diameter must satisfy both conditions. The overlapping range is bounded below by the stricter lower bound of 1.181.18 (exclusive) and above by the stricter upper bound of 1.261.26 (exclusive).

Key Concept

Absolute Value Equations and Inequalities
Question 8Question →

For all real values of tt that satisfy the inequality 12−4∣3−2t∣≤−812 - 4|3 - 2t| \leq -8, which of the following inequalities represents the complete set of possible values of tt?

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Answer: t≤−1t \leq -1 or t≥4t \geq 4

Answer

t≤−1t \leq -1 or t≥4t \geq 4
The correct answer is t≤−1t \leq -1 or t≥4t \geq 4. To solve the inequality, we first subtract 12 from both sides to get −4∣3−2t∣≤−20-4|3 - 2t| \leq -20. Next, dividing both sides by −4-4 and reversing the inequality sign gives ∣3−2t∣≥5|3 - 2t| \geq 5. This absolute value inequality splits into two cases: 3−2t≥53 - 2t \geq 5 or 3−2t≤−53 - 2t \leq -5. Solving the first case, we subtract 3 to get −2t≥2-2t \geq 2, and dividing by −2-2 while reversing the inequality sign yields t≤−1t \leq -1. Solving the second case, we subtract 3 to get −2t≤−8-2t \leq -8, and dividing by −2-2 while reversing the inequality sign yields t≥4t \geq 4. Combining these, we obtain the solution set t≤−1t \leq -1 or t≥4t \geq 4.

Step-by-Step Solution

1
Isolate the absolute value term by subtracting 12 from both sides of the inequality.
−4∣3−2t∣≤−20-4|3 - 2t| \leq -20
Before splitting an absolute value inequality, the absolute value expression must be isolated on one side.
2
Divide both sides of the inequality by −4-4 and reverse the inequality sign.
∣3−2t∣≥5|3 - 2t| \geq 5
Dividing or multiplying an inequality by a negative number requires reversing the direction of the inequality sign.
3
Split the absolute value inequality ∣u∣≥c|u| \geq c (where c>0c > 0) into two separate inequalities: u≥cu \geq c or u≤−cu \leq -c.
3−2t≥53 - 2t \geq 5 or 3−2t≤−53 - 2t \leq -5
An absolute value inequality of the form ∣u∣≥c|u| \geq c represents values that are at least cc units away from zero, which lie in two disjoint intervals.
4
Solve the first inequality: 3−2t≥53 - 2t \geq 5.
−2t≥2  ⟹  t≤−1-2t \geq 2 \implies t \leq -1
Subtracting 3 from both sides gives −2t≥2-2t \geq 2. Dividing by −2-2 and reversing the inequality sign yields t≤−1t \leq -1.
5
Solve the second inequality: 3−2t≤−53 - 2t \leq -5.
−2t≤−8  ⟹  t≥4-2t \leq -8 \implies t \geq 4
Subtracting 3 from both sides gives −2t≤−8-2t \leq -8. Dividing by −2-2 and reversing the inequality sign yields t≥4t \geq 4.
6
Combine the solutions from both cases to express the complete solution set.
t≤−1t \leq -1 or t≥4t \geq 4
The complete solution set is the union of the solutions to both cases.

Key Concept

Solving absolute value inequalities of the form ∣ax+b∣≥c|ax + b| \geq c by translating them into compound inequalities and carefully reversing the inequality sign when multiplying or dividing by negative numbers.
Question 9Question →

What is the set of all real numbers yy that make the inequality 2−3∣1−2y∣≥−102 - 3|1 - 2y| \ge -10 a true statement?

Show answer & explanation

Answer: −1.5≤y≤2.5-1.5 \le y \le 2.5

Answer

−1.5≤y≤2.5-1.5 \le y \le 2.5
To solve the inequality 2−3∣1−2y∣≥−102 - 3|1 - 2y| \ge -10, we first isolate the absolute value term by subtracting 2 from both sides to get −3∣1−2y∣≥−12-3|1 - 2y| \ge -12, and then dividing both sides by −3-3. Since we divide by a negative number, the inequality sign reverses, giving ∣1−2y∣≤4|1 - 2y| \le 4. We rewrite this as the compound inequality −4≤1−2y≤4-4 \le 1 - 2y \le 4. Subtracting 1 from all parts yields −5≤−2y≤3-5 \le -2y \le 3. Finally, dividing by −2-2 and reversing the inequality signs gives 2.5≥y≥−1.52.5 \ge y \ge -1.5, which is rewritten from least to greatest as −1.5≤y≤2.5-1.5 \le y \le 2.5.

Step-by-Step Solution

1
Subtract 2 from both sides of the inequality to begin isolating the absolute value term.
−3∣1−2y∣≥−12-3|1 - 2y| \ge -12
Isolating the absolute value expression allows us to rewrite it as a standard inequality.
2
Divide both sides by −3-3 and reverse the inequality sign because of division by a negative number.
∣1−2y∣≤4|1 - 2y| \le 4
Dividing by a negative value requires reversing the inequality direction to preserve the truth of the statement.
3
Rewrite the absolute value inequality as a compound inequality.
−4≤1−2y≤4-4 \le 1 - 2y \le 4
An inequality of the form ∣u∣≤c|u| \le c (where c>0c > 0) is equivalent to the compound inequality −c≤u≤c-c \le u \le c.
4
Subtract 1 from all parts of the compound inequality.
−5≤−2y≤3-5 \le -2y \le 3
This is the next step to isolate the variable yy in the middle.
5
Divide all parts of the compound inequality by −2-2 and reverse the inequality signs.
2.5≥y≥−1.52.5 \ge y \ge -1.5, which is equivalent to −1.5≤y≤2.5-1.5 \le y \le 2.5
Dividing by the negative coefficient −2-2 requires reversing the direction of all inequality signs.

Key Concept

Solving absolute value inequalities involving negative coefficients by isolating the absolute value and reversing inequality signs when multiplying or dividing by negative numbers.
Estimated Time:1m 30s
Question 10Question →

In the inequality 7−3∣2z+1∣≥−87 - 3|2z + 1| \ge -8, which of the following inequality expressions represents the complete solution set for zz?

Show answer & explanation

Answer: −3≤z≤2-3 \le z \le 2

Answer

−3≤z≤2-3 \le z \le 2
Subtracting 7 from both sides of 7−3∣2z+1∣≥−87 - 3|2z + 1| \ge -8 yields −3∣2z+1∣≥−15-3|2z + 1| \ge -15. Dividing both sides by −3-3 and reversing the inequality sign results in ∣2z+1∣≤5|2z + 1| \le 5. Writing this as the compound inequality −5≤2z+1≤5-5 \le 2z + 1 \le 5, then subtracting 1 and dividing by 2 yields the correct solution interval −3≤z≤2-3 \le z \le 2.

Step-by-Step Solution

1
Subtract 7 from both sides to begin isolating the absolute value term.
−3∣2z+1∣≥−15-3|2z + 1| \ge -15
To solve an absolute value inequality, we must first isolate the absolute value term on one side.
2
Divide both sides by -3 and reverse the direction of the inequality sign.
∣2z+1∣≤5|2z + 1| \le 5
Dividing an inequality by a negative number reverses the direction of the inequality sign.
3
Rewrite the absolute value inequality as a compound inequality.
−5≤2z+1≤5-5 \le 2z + 1 \le 5
An inequality of the form ∣u∣≤a|u| \le a (where a≥0a \ge 0) is equivalent to the compound inequality −a≤u≤a-a \le u \le a.
4
Subtract 1 from all three parts of the compound inequality.
−6≤2z≤4-6 \le 2z \le 4
This is the first step to isolate the variable zz in the middle.
5
Divide all three parts by 2.
−3≤z≤2-3 \le z \le 2
This fully isolates zz, giving the final solution interval.

Key Concept

Solving multi-step absolute value inequalities, including isolating the absolute value expression, reversing the inequality sign when dividing by a negative number, and expressing the solution as a compound inequality.
Question 11Question →

If yy is a real number such that ∣3−12y∣=5|3 - \frac{1}{2}y| = 5, what is the sum of all possible values of yy?

Show answer & explanation

Answer: 12

Answer

12
Solving the equation ∣3−12y∣=5|3 - \frac{1}{2}y| = 5 requires setting up two cases: 3−12y=53 - \frac{1}{2}y = 5 and 3−12y=−53 - \frac{1}{2}y = -5. Solving the first case gives y=−4y = -4, and solving the second case gives y=16y = 16. The sum of these two solutions is −4+16=12-4 + 16 = 12.

Step-by-Step Solution

1
Set up the two equations represented by the absolute value expression.
3−12y=53 - \frac{1}{2}y = 5 and 3−12y=−53 - \frac{1}{2}y = -5
An absolute value equation ∣u∣=C|u| = C (where C≥0C \ge 0) splits into two cases: u=Cu = C and u=−Cu = -C.
2
Solve the first equation 3−12y=53 - \frac{1}{2}y = 5.
y=−4y = -4
Subtracting 3 from both sides yields −12y=2-\frac{1}{2}y = 2. Multiplying both sides by −2-2 isolates yy.
3
Solve the second equation 3−12y=−53 - \frac{1}{2}y = -5.
y=16y = 16
Subtracting 3 from both sides yields −12y=−8-\frac{1}{2}y = -8. Multiplying both sides by −2-2 isolates yy.
4
Calculate the sum of all possible values of yy.
1212
Add the two solutions together: −4+16=12-4 + 16 = 12.

Key Concept

Absolute Value Equations
Estimated Time:1m 30s
Question 12Question →

For all real values of qq that satisfy the inequality 8−2∣3q−1∣≥−68 - 2|3q - 1| \ge -6, which of the following inequality expressions represents the complete solution set for qq?

Show answer & explanation

Answer: $−2≤q≤83-2 \le q \le \frac{8}{3}

Answer

−2≤q≤83-2 \le q \le \frac{8}{3}
The correct answer is the interval containing all real numbers between −2-2 and 83\frac{8}{3} inclusive. This is determined by subtracting 8 from both sides of the inequality, dividing by −2-2 (which reverses the inequality sign to yield ∣3q−1∣≤7|3q - 1| \le 7), expressing this as the compound inequality −7≤3q−1≤7-7 \le 3q - 1 \le 7, and isolating qq.

Step-by-Step Solution

1
Subtract 8 from both sides of the inequality to isolate the absolute value term.
−2∣3q−1∣≥−14-2|3q - 1| \ge -14
To solve for the variable, we must first isolate the term containing the absolute value by performing inverse operations.
2
Divide both sides by −2-2 and reverse the inequality sign.
∣3q−1∣≤7|3q - 1| \le 7
Dividing both sides of an inequality by a negative number requires reversing the direction of the inequality sign.
3
Rewrite the absolute value inequality as a compound inequality.
−7≤3q−1≤7-7 \le 3q - 1 \le 7
An inequality of the form ∣x∣≤d|x| \le d (where d≥0d \ge 0) is equivalent to the compound inequality −d≤x≤d-d \le x \le d.
4
Add 1 to all three parts of the compound inequality.
−6≤3q≤8-6 \le 3q \le 8
Adding 1 eliminates the constant term from the middle section of the inequality.
5
Divide all three parts by 3 to solve for qq.
−2≤q≤83-2 \le q \le \frac{8}{3}
Dividing by 3 isolates the variable qq, yielding the complete solution set.

Key Concept

Solving multi-step absolute value inequalities, including reversing the inequality sign when dividing by a negative number and setting up a compound inequality to represent both positive and negative cases.
Question 13Question →

What is the complete solution set for the inequality ∣4−3v∣≥7|4 - 3v| \ge 7?

Show answer & explanation

Answer: v≤−1v \le -1 or v≥113v \ge \frac{11}{3}

Answer

The solution set is the union of two open-ended intervals, representing all values of vv such that vv is less than or equal to −1-1 or vv is greater than or equal to 113\frac{11}{3}.
To solve the absolute value inequality ∣4−3v∣≥7|4 - 3v| \ge 7, we rewrite it as two separate inequalities: 4−3v≥74 - 3v \ge 7 or 4−3v≤−74 - 3v \le -7. For the first case, subtracting 4 from both sides of 4−3v≥74 - 3v \ge 7 results in −3v≥3-3v \ge 3. Dividing by −3-3 and reversing the inequality sign gives v≤−1v \le -1. For the second case, subtracting 4 from both sides of 4−3v≤−74 - 3v \le -7 results in −3v≤−11-3v \le -11. Dividing by −3-3 and reversing the inequality sign gives v≥113v \ge \frac{11}{3}. Combining these gives the correct solution set: v≤−1v \le -1 or v≥113v \ge \frac{11}{3}.

Step-by-Step Solution

1
Set up two separate linear inequalities based on the absolute value inequality template.
4−3v≥74 - 3v \ge 7 or 4−3v≤−74 - 3v \le -7
An absolute value inequality of the form ∣A∣≥B|A| \ge B where B>0B > 0 is equivalent to the compound statement A≥BA \ge B or A≤−BA \le -B.
2
Solve the first inequality: 4−3v≥74 - 3v \ge 7. Subtract 4 from both sides, then divide by −3-3 and reverse the inequality sign.
v≤−1v \le -1
Subtracting 4 yields −3v≥3-3v \ge 3. Dividing by the negative number −3-3 requires reversing the direction of the inequality sign.
3
Solve the second inequality: 4−3v≤−74 - 3v \le -7. Subtract 4 from both sides, then divide by −3-3 and reverse the inequality sign.
v≥113v \ge \frac{11}{3}
Subtracting 4 yields −3v≤−11-3v \le -11. Dividing by the negative number −3-3 requires reversing the direction of the inequality sign.
4
Combine the individual solutions into a single compound statement.
v≤−1v \le -1 or v≥113v \ge \frac{11}{3}
The solution to a 'greater than or equal to' absolute value inequality is the union of the individual solutions.

Key Concept

Solving absolute value inequalities by splitting them into two linear cases and correctly reversing the inequality sign when dividing by a negative number.

Alternative Method

Alternatively, test test-values from each interval. For instance, choosing v=0v = 0 (which is in the middle interval) yields ∣4−0∣=4≥7|4 - 0| = 4 \ge 7, which is false. Choosing v=−2v = -2 (which is in the left interval) yields ∣4−3(−2)∣=∣10∣=10≥7|4 - 3(-2)| = |10| = 10 \ge 7, which is true. Choosing v=4v = 4 (which is in the right interval since 4>1134 > \frac{11}{3}) yields ∣4−12∣=∣−8∣=8≥7|4 - 12| = |-8| = 8 \ge 7, which is true. This confirms the solution set must cover v≤−1v \le -1 and v≥113v \ge \frac{11}{3}.
Estimated Time:1m 30s
Question 14Question →

A thermometer is considered accurate if its temperature reading, TT degrees Fahrenheit, differs from the actual temperature by less than 1.5∘F1.5^\circ\text{F}. If the actual temperature is 72.0∘F72.0^\circ\text{F}, which of the following inequalities represents the range of reading temperatures, TT, that are NOT considered accurate?

Show answer & explanation

Answer: ∣T−72.0∣≥1.5|T - 72.0| \ge 1.5

Answer

∣T−72.0∣≥1.5|T - 72.0| \ge 1.5
The difference between the temperature reading, TT, and the actual temperature of 72.0∘F72.0^\circ\text{F} is represented by ∣T−72.0∣|T - 72.0|. A thermometer is accurate when this difference is less than 1.5∘F1.5^\circ\text{F} (∣T−72.0∣<1.5|T - 72.0| < 1.5). Therefore, the thermometer is not accurate when the difference is greater than or equal to 1.5∘F1.5^\circ\text{F}, which is written as ∣T−72.0∣≥1.5|T - 72.0| \ge 1.5.

Step-by-Step Solution

1
Represent the difference between the reading and the actual temperature.
∣T−72.0∣|T - 72.0|
The difference between the reading temperature and the actual temperature of 72.0∘F72.0^\circ\text{F} is given by the absolute value expression, representing the distance between the two temperatures on a thermometer.
2
Formulate the condition for an accurate reading.
∣T−72.0∣<1.5|T - 72.0| < 1.5
An accurate reading differs from the actual temperature by less than 1.5∘F1.5^\circ\text{F}.
3
Determine the complement condition for a reading that is NOT accurate.
∣T−72.0∣≥1.5|T - 72.0| \ge 1.5
To find the range of readings that are NOT accurate, we take the complement of the accurate condition. The opposite of 'less than 1.51.5' is 'greater than or equal to 1.51.5'.

Key Concept

Absolute Value Inequalities in Real-World Contexts
Estimated Time:1m 0s
Question 15Question →

If ww is a real number such that ∣3−2w∣≤9|3 - 2w| \le 9, what is the complete range of possible values for ww?

Show answer & explanation

Answer: −3≤w≤6-3 \le w \le 6

Answer

−3≤w≤6-3 \le w \le 6
The inequality ∣3−2w∣≤9|3 - 2w| \le 9 is equivalent to the compound inequality −9≤3−2w≤9-9 \le 3 - 2w \le 9. Subtracting 33 from all parts yields −12≤−2w≤6-12 \le -2w \le 6. Dividing all parts by −2-2 and reversing the inequality signs results in 6≥w≥−36 \ge w \ge -3, which can be rewritten as −3≤w≤6-3 \le w \le 6.

Step-by-Step Solution

1
Set up the compound inequality representing the absolute value inequality.
−9≤3−2w≤9-9 \le 3 - 2w \le 9
An absolute value inequality of the form ∣f(x)∣≤c|f(x)| \le c is equivalent to −c≤f(x)≤c-c \le f(x) \le c.
2
Subtract 33 from all three parts of the inequality to isolate the term containing ww.
−12≤−2w≤6-12 \le -2w \le 6
To solve for ww, we must isolate the variable term by performing inverse operations on all parts of the inequality.
3
Divide all three parts of the inequality by −2-2 and reverse the inequality signs.
6≥w≥−36 \ge w \ge -3, which is equivalent to −3≤w≤6-3 \le w \le 6
Dividing an inequality by a negative number requires reversing the direction of the inequality signs to preserve the truth of the statement.

Key Concept

Solving absolute value inequalities of the form ∣ax+b∣≤c|ax + b| \le c by converting them into compound inequalities and solving for the variable while reversing the inequality signs when dividing by a negative number.
Estimated Time:1m 30s
Question 16Question →

Which of the following inequality expressions represents the complete solution set for pp in the inequality 15−3∣2p+5∣<615 - 3|2p + 5| < 6?

Show answer & explanation

Answer: p<−4p < -4 or p>−1p > -1

Answer

p<−4p < -4 or p>−1p > -1
The correct solution is obtained by first isolating the absolute value expression. Subtracting 15 from both sides of the inequality 15−3∣2p+5∣<615 - 3|2p + 5| < 6 yields −3∣2p+5∣<−9-3|2p + 5| < -9. Dividing both sides by −3-3 and reversing the inequality sign gives ∣2p+5∣>3|2p + 5| > 3. This absolute value inequality splits into two cases: 2p+5>32p + 5 > 3 (which simplifies to p>−1p > -1) or 2p+5<−32p + 5 < -3 (which simplifies to p<−4p < -4). Combining these yields the complete solution set p<−4p < -4 or p>−1p > -1.

Step-by-Step Solution

1
Subtract 15 from both sides of the inequality 15−3∣2p+5∣<615 - 3|2p + 5| < 6.
−3∣2p+5∣<−9-3|2p + 5| < -9
To isolate the absolute value term, first subtract the constant term from both sides.
2
Divide both sides of −3∣2p+5∣<−9-3|2p + 5| < -9 by −3-3 and reverse the inequality sign.
∣2p+5∣>3|2p + 5| > 3
Dividing an inequality by a negative number requires reversing the direction of the inequality sign.
3
Solve the absolute value inequality ∣2p+5∣>3|2p + 5| > 3 by setting up two separate inequalities.
2p+5>32p + 5 > 3 or 2p+5<−32p + 5 < -3
An absolute value inequality of the form ∣u∣>c|u| > c splits into u>cu > c or u<−cu < -c.
4
Solve each linear inequality for pp.
p>−1p > -1 or p<−4p < -4
Subtract 5 from both sides and then divide by 2 for both inequalities to isolate pp.

Key Concept

Solving absolute value inequalities involving algebraic manipulation and reversing the inequality sign when multiplying or dividing by a negative number.
Question 17Question →

What is the complete solution set for the inequality ∣7−3n∣<11|7 - 3n| < 11?

Show answer & explanation

Answer: −43<n<6-\frac{4}{3} < n < 6

Answer

−43<n<6-\frac{4}{3} < n < 6
To solve ∣7−3n∣<11|7 - 3n| < 11, rewrite it as the compound inequality −11<7−3n<11-11 < 7 - 3n < 11. Subtracting 77 from all parts yields −18<−3n<4-18 < -3n < 4. Finally, dividing by −3-3 and reversing the inequality signs gives 6>n>−436 > n > -\frac{4}{3}, which simplifies to the interval −43<n<6-\frac{4}{3} < n < 6. This matches the correct option.

Step-by-Step Solution

1
Express the absolute value inequality as a compound inequality.
−11<7−3n<11-11 < 7 - 3n < 11
An absolute value inequality of the form ∣u∣<c|u| < c is equivalent to the compound inequality −c<u<c-c < u < c.
2
Subtract 77 from all three parts of the inequality.
−18<−3n<4-18 < -3n < 4
To isolate the term containing nn, we subtract 77 from all parts of the inequality.
3
Divide all three parts by −3-3 and reverse the inequality signs.
6>n>−436 > n > -\frac{4}{3}, which is equivalent to −43<n<6-\frac{4}{3} < n < 6
Dividing an inequality by a negative number requires reversing the direction of the inequality signs to maintain a true statement.

Key Concept

Absolute Value Equations and Inequalities
Question 18Question →

Which of the following is the set of all real values of pp for which the inequality ∣7−3p∣≥14|7 - 3p| \ge 14 is true?

Show answer & explanation

Answer: p≤−73 or p≥7p \le -\frac{7}{3} \text{ or } p \ge 7

Answer

p≤−73 or p≥7p \le -\frac{7}{3} \text{ or } p \ge 7
To solve the inequality ∣7−3p∣≥14|7 - 3p| \ge 14, we split it into two separate inequalities: 7−3p≥147 - 3p \ge 14 or 7−3p≤−147 - 3p \le -14. Solving the first inequality gives −3p≥7-3p \ge 7, which simplifies to p≤−73p \le -\frac{7}{3} after dividing by −3-3 and reversing the inequality sign. Solving the second inequality gives −3p≤−21-3p \le -21, which simplifies to p≥7p \ge 7 after dividing by −3-3 and reversing the inequality sign. Combining these two cases yields the set of values p≤−73p \le -\frac{7}{3} or p≥7p \ge 7.

Step-by-Step Solution

1
Set up the two compound inequalities representing the absolute value inequality ∣7−3p∣≥14|7 - 3p| \ge 14.
7−3p≥147 - 3p \ge 14 or 7−3p≤−147 - 3p \le -14
An absolute value inequality of the form ∣u∣≥c|u| \ge c (where c>0c > 0) is equivalent to the union of u≥cu \ge c or u≤−cu \le -c.
2
Solve the first inequality 7−3p≥147 - 3p \ge 14.
p≤−73p \le -\frac{7}{3}
Subtract 77 from both sides to get −3p≥7-3p \ge 7. Then, divide by −3-3 and reverse the inequality sign because of division by a negative number.
3
Solve the second inequality 7−3p≤−147 - 3p \le -14.
p≥7p \ge 7
Subtract 77 from both sides to get −3p≤−21-3p \le -21. Then, divide by −3-3 and reverse the inequality sign because of division by a negative number.
4
Combine the two case solutions to state the final solution set.
p≤−73 or p≥7p \le -\frac{7}{3} \text{ or } p \ge 7
The complete solution set is the union of the solutions from both individual cases.

Key Concept

Solving absolute value inequalities of the form ∣ax+b∣≥c|ax + b| \ge c
Question 19Question →

What is the complete solution set, expressed in interval notation, for the inequality ∣2−3t4∣>2\left| \frac{2-3t}{4} \right| > 2?

Show answer & explanation

Answer: (−∞,−2)∪(103,∞)(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)

Answer

The complete solution set is (−∞,−2)∪(103,∞)(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)
The correct answer shows (−∞,−2)∪(103,∞)(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right) because solving the absolute value inequality ∣2−3t∣>8|2-3t| > 8 requires splitting it into two inequalities: 2−3t>82-3t > 8 and 2−3t<−82-3t < -8. Solving the first yields t<−2t < -2 after reversing the inequality sign when dividing by −3-3. Solving the second yields t>103t > \frac{10}{3} after also reversing the inequality sign. The union of these two intervals is (−∞,−2)∪(103,∞)(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right).

Step-by-Step Solution

1
Multiply both sides of the inequality by 44 to isolate the absolute value term.
∣2−3t∣>8|2-3t| > 8
Eliminating the denominator simplifies the absolute value expression.
2
Split the absolute value inequality into two separate linear inequalities representing the positive and negative cases.
2−3t>82-3t > 8 or 2−3t<−82-3t < -8
An absolute value greater than a positive number cc is equivalent to the expression being greater than cc or less than −c-c.
3
Solve the first inequality: subtract 22 from both sides, then divide by −3-3 and reverse the inequality sign.
−3t>6  ⟹  t<−2-3t > 6 \implies t < -2
Dividing an inequality by a negative number requires flipping the inequality sign.
4
Solve the second inequality: subtract 22 from both sides, then divide by −3-3 and reverse the inequality sign.
−3t<−10  ⟹  t>103-3t < -10 \implies t > \frac{10}{3}
Dividing an inequality by a negative number requires flipping the inequality sign.
5
Combine the two solutions using interval notation.
(−∞,−2)∪(103,∞)(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)
The union of the two intervals represents the complete set of values that satisfy either inequality.

Key Concept

Solving absolute value inequalities of the form ∣ax+b∣>c|ax+b| > c by splitting them into two cases and reversing the inequality sign when dividing by a negative number.
Estimated Time:1m 30s
Absolute Value Equations and Inequalities Practice Questions — ACT | Examkin