CLST 103 — Liberty University Math Assessment · Part 2 65-Question Comprehensive Exam
CLST 103 — Liberty University Mathematics
Assessment
Part 2 — Comprehensive Algebra & Functions Evaluation
Liberty University • College of General Studies
Total Questions: 65 • Sections: 9 • Format: Multiple Choice (A–D) • Cognitive Mix: 35% Recall · 45% Application · 20%
Analysis
Instructions: Read each question carefully. Select the single best answer from choices A–D. Each question
provides a step-by-step rationale showing the correct procedure and explaining why each distractor is incorrect.
Distractors are designed to identify specific mathematical misconceptions such as sign errors, order-of-operations
mistakes, incorrect factoring, misapplied formulas, and failure to check for extraneous solutions.
Section 1: Linear Equations, Inequalities, and Absolute Value (Q1–Q8)
Q1. Solve for x: 3(x − 4) + 2 = 5x − 6
A. x = −2 [CORRECT]
B. x = −1
C. x = 1
D. x = 4
Correct Answer: A
Rationale: Distribute 3 to obtain 3x − 12 + 2 = 5x − 6, which simplifies to 3x − 10 = 5x − 6. Subtract 3x from both sides: −10
= 2x − 6, then add 6 to get −4 = 2x, so x = −2. Choice B reflects a sign error when moving the constant; Choice C results from
dropping the −12; Choice D comes from adding 12 instead of subtracting it.
Q2. Solve the inequality −2(x + 3) > 4 − x and express the solution in interval notation.
A. (−∞, −10) [CORRECT]
B. (−10, ∞)
C. (−∞, 10)
D. (10, ∞)
Correct Answer: A
Rationale: Distribute −2: −2x − 6 > 4 − x. Add x to both sides: −x − 6 > 4. Add 6: −x > 10. Divide by −1 and FLIP the
inequality: x < −10, which is (−∞, −10). Choice B forgets to flip the inequality when dividing by a negative; Choice C uses
+10 (sign error in the constant); Choice D reflects both errors.
Q3. Solve the absolute value equation |2x − 5| = 11.
A. x = −3 or x = 8 [CORRECT]
B. x = 3 or x = −8
C. x = 3 or x = 8
D. x = −3 or x = −8
Correct Answer: A
Rationale: Split into two equations: 2x − 5 = 11 ⇒ 2x = 16 ⇒ x = 8; and 2x − 5 = −11 ⇒ 2x = −6 ⇒ x = −3. Solutions are x
= −3 or x = 8. Choice B uses the wrong sign on −3; Choice C forgets the negative branch; Choice D applies the wrong sign to
both branches.
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, CLST 103 — Liberty University Math Assessment · Part 2 65-Question Comprehensive Exam
Q4. Solve |3x + 1| ≤ 7 and write the solution in interval notation.
A. [−8/3, 2] [CORRECT]
B. [−2, 8/3]
C. (−∞, −8/3] ∪ [2, ∞)
D. (−∞, −2] ∪ [8/3, ∞)
Correct Answer: A
Rationale: Rewrite as a compound inequality: −7 ≤ 3x + 1 ≤ 7. Subtract 1: −8 ≤ 3x ≤ 6. Divide by 3: −8/3 ≤ x ≤ 2, which is
[−8/3, 2]. Choice B swaps the endpoints (sign error); Choice C is the solution to |3x + 1| ≥ 7; Choice D is the “≥” solution with
endpoints swapped.
Q5. Solve the literal equation P = 2L + 2W for W.
A. W = (P − 2L) / 2 [CORRECT]
B. W = (P + 2L) / 2
C. W = P − L
D. W = 2P − 2L
Correct Answer: A
Rationale: Subtract 2L from both sides: P − 2L = 2W. Divide by 2: W = (P − 2L)/2. Choice B adds 2L instead of subtracting
(sign error); Choice C drops the factor of 2 in the denominator; Choice D multiplies both sides by 2 instead of dividing.
Q6. A taxi charges $3.00 plus $0.75 per mile. If the total fare was $12.75, how many miles was the ride?
A. 13 miles [CORRECT]
B. 17 miles
C. 11 miles
D. 21 miles
Correct Answer: A
Rationale: Set up 3 + 0.75m = 12.75. Subtract 3: 0.75m = 9.75. Divide by 0.75: m = 13 miles. Choice B (17) results from
adding 3 instead of subtracting; Choice C (11) comes from dividing 9.75 by 0.85 (incorrect divisor); Choice D (21) results
from multiplying by 0.75 instead of dividing.
Q7. Solve −3(x − 2) + 5x = 2(x + 3). Classify the solution.
A. All real numbers (identity) [CORRECT]
B. x = 0
C. No solution (contradiction)
D. x = 6
Correct Answer: A
Rationale: Distribute on both sides: −3x + 6 + 5x = 2x + 6, which simplifies to 2x + 6 = 2x + 6. Since this statement is true for
every value of x, the equation is an identity; the solution is all real numbers. Choice B identifies one solution but not the
complete set; Choice C applies to contradictions; Choice D is an arbitrary single value.
Q8. Solve the absolute value inequality |x − 4| > 3 and express the solution in interval notation.
A. (−∞, 1) ∪ (7, ∞) [CORRECT]
B. (1, 7)
C. (−∞, −7) ∪ (1, ∞)
D. (−7, 1)
Correct Answer: A
Page 2
CLST 103 — Liberty University Mathematics
Assessment
Part 2 — Comprehensive Algebra & Functions Evaluation
Liberty University • College of General Studies
Total Questions: 65 • Sections: 9 • Format: Multiple Choice (A–D) • Cognitive Mix: 35% Recall · 45% Application · 20%
Analysis
Instructions: Read each question carefully. Select the single best answer from choices A–D. Each question
provides a step-by-step rationale showing the correct procedure and explaining why each distractor is incorrect.
Distractors are designed to identify specific mathematical misconceptions such as sign errors, order-of-operations
mistakes, incorrect factoring, misapplied formulas, and failure to check for extraneous solutions.
Section 1: Linear Equations, Inequalities, and Absolute Value (Q1–Q8)
Q1. Solve for x: 3(x − 4) + 2 = 5x − 6
A. x = −2 [CORRECT]
B. x = −1
C. x = 1
D. x = 4
Correct Answer: A
Rationale: Distribute 3 to obtain 3x − 12 + 2 = 5x − 6, which simplifies to 3x − 10 = 5x − 6. Subtract 3x from both sides: −10
= 2x − 6, then add 6 to get −4 = 2x, so x = −2. Choice B reflects a sign error when moving the constant; Choice C results from
dropping the −12; Choice D comes from adding 12 instead of subtracting it.
Q2. Solve the inequality −2(x + 3) > 4 − x and express the solution in interval notation.
A. (−∞, −10) [CORRECT]
B. (−10, ∞)
C. (−∞, 10)
D. (10, ∞)
Correct Answer: A
Rationale: Distribute −2: −2x − 6 > 4 − x. Add x to both sides: −x − 6 > 4. Add 6: −x > 10. Divide by −1 and FLIP the
inequality: x < −10, which is (−∞, −10). Choice B forgets to flip the inequality when dividing by a negative; Choice C uses
+10 (sign error in the constant); Choice D reflects both errors.
Q3. Solve the absolute value equation |2x − 5| = 11.
A. x = −3 or x = 8 [CORRECT]
B. x = 3 or x = −8
C. x = 3 or x = 8
D. x = −3 or x = −8
Correct Answer: A
Rationale: Split into two equations: 2x − 5 = 11 ⇒ 2x = 16 ⇒ x = 8; and 2x − 5 = −11 ⇒ 2x = −6 ⇒ x = −3. Solutions are x
= −3 or x = 8. Choice B uses the wrong sign on −3; Choice C forgets the negative branch; Choice D applies the wrong sign to
both branches.
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, CLST 103 — Liberty University Math Assessment · Part 2 65-Question Comprehensive Exam
Q4. Solve |3x + 1| ≤ 7 and write the solution in interval notation.
A. [−8/3, 2] [CORRECT]
B. [−2, 8/3]
C. (−∞, −8/3] ∪ [2, ∞)
D. (−∞, −2] ∪ [8/3, ∞)
Correct Answer: A
Rationale: Rewrite as a compound inequality: −7 ≤ 3x + 1 ≤ 7. Subtract 1: −8 ≤ 3x ≤ 6. Divide by 3: −8/3 ≤ x ≤ 2, which is
[−8/3, 2]. Choice B swaps the endpoints (sign error); Choice C is the solution to |3x + 1| ≥ 7; Choice D is the “≥” solution with
endpoints swapped.
Q5. Solve the literal equation P = 2L + 2W for W.
A. W = (P − 2L) / 2 [CORRECT]
B. W = (P + 2L) / 2
C. W = P − L
D. W = 2P − 2L
Correct Answer: A
Rationale: Subtract 2L from both sides: P − 2L = 2W. Divide by 2: W = (P − 2L)/2. Choice B adds 2L instead of subtracting
(sign error); Choice C drops the factor of 2 in the denominator; Choice D multiplies both sides by 2 instead of dividing.
Q6. A taxi charges $3.00 plus $0.75 per mile. If the total fare was $12.75, how many miles was the ride?
A. 13 miles [CORRECT]
B. 17 miles
C. 11 miles
D. 21 miles
Correct Answer: A
Rationale: Set up 3 + 0.75m = 12.75. Subtract 3: 0.75m = 9.75. Divide by 0.75: m = 13 miles. Choice B (17) results from
adding 3 instead of subtracting; Choice C (11) comes from dividing 9.75 by 0.85 (incorrect divisor); Choice D (21) results
from multiplying by 0.75 instead of dividing.
Q7. Solve −3(x − 2) + 5x = 2(x + 3). Classify the solution.
A. All real numbers (identity) [CORRECT]
B. x = 0
C. No solution (contradiction)
D. x = 6
Correct Answer: A
Rationale: Distribute on both sides: −3x + 6 + 5x = 2x + 6, which simplifies to 2x + 6 = 2x + 6. Since this statement is true for
every value of x, the equation is an identity; the solution is all real numbers. Choice B identifies one solution but not the
complete set; Choice C applies to contradictions; Choice D is an arbitrary single value.
Q8. Solve the absolute value inequality |x − 4| > 3 and express the solution in interval notation.
A. (−∞, 1) ∪ (7, ∞) [CORRECT]
B. (1, 7)
C. (−∞, −7) ∪ (1, ∞)
D. (−7, 1)
Correct Answer: A
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