Math 1 Quiz: Critiquing Mathematical Reasoning

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Critiquing Mathematical ReasoningQuestion 1 of 20

A student claims: "Since the function f(x) = x² - 4x + 3 has a positive leading coefficient, the function is always increasing."

What is the primary error in the student's reasoning?

The student confused the behavior of quadratic functions with linear functions and ignored the vertex.

The student incorrectly calculated the leading coefficient, which should be negative for this function.

The student failed to consider that quadratic functions are only defined for positive values of x.

The student should have factored the function first before making any claims about its behavior.

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Math 1 Quiz

Math 1 Quiz: Critiquing Mathematical Reasoning

Practice Critiquing Mathematical Reasoning in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Critiquing Mathematical Reasoning, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A student claims: "Since the function f(x) = x² - 4x + 3 has a positive leading coefficient, the function is always increasing."

What is the primary error in the student's reasoning?

The student confused the behavior of quadratic functions with linear functions and ignored the vertex. (correct answer)
The student incorrectly calculated the leading coefficient, which should be negative for this function.
The student failed to consider that quadratic functions are only defined for positive values of x.
The student should have factored the function first before making any claims about its behavior.

Explanation: The correct answer is A. While a positive leading coefficient does determine that a parabola opens upward, this doesn't mean the function is always increasing. The function decreases until it reaches its vertex at x = 2, then increases afterward. Choice B is wrong because the leading coefficient is indeed positive (1). Choice C is wrong because quadratic functions are defined for all real numbers. Choice D is wrong because factoring isn't necessary to analyze the increasing/decreasing behavior.

Question 2

A student claims: "Since $\sin(30°) = \frac{1}{2}$ and $\sin$ is a one-to-one function, there is exactly one angle between $0°$ and $360°$ whose sine equals $\frac{1}{2}$ ." Identify the error in this reasoning.

The student incorrectly calculated the value of $\sin(30°)$ , which should be $\frac{\sqrt{3}}{2}$ , not $\frac{1}{2}$ .
The student incorrectly assumed that sine is a one-to-one function over the interval $[0°, 360°]$ . (correct answer)
The student should have considered cosine instead of sine when making claims about unique angle measures.
The student failed to convert from degrees to radians before applying trigonometric function properties.

Explanation: The correct answer is B. The sine function is not one-to-one over [0°, 360°]; it's only one-to-one on restricted domains like [-90°, 90°]. In fact, sin(150°) = 1/2 as well, so there are two angles in [0°, 360°] with sine equal to 1/2. Choice A is wrong because sin(30°) = 1/2 is correct. Choice C is wrong because cosine has the same issue with being one-to-one. Choice D is wrong because the unit system doesn't affect the one-to-one property.

Question 3

A student claims: "The equation $\log_2(x-1) + \log_2(x+1) = 3$ can be solved by using the product property: $\log_2[(x-1)(x+1)] = 3$ , so $(x-1)(x+1) = 8$ . This gives $x^2 - 1 = 8$ , so $x^2 = 9$ and $x = ±3$ . Both solutions are valid." What error did the student make?

The student incorrectly applied the logarithm product property when combining the two logarithmic terms.
The student made an algebraic error when expanding $(x-1)(x+1)$ and should have gotten $x^2 + 1$ .
The student failed to verify that both solutions satisfy the domain restrictions of the original logarithmic equation. (correct answer)
The student should have used the quotient property of logarithms instead of the product property for this equation.

Explanation: The correct answer is C. For logarithms to be defined, we need x - 1 > 0 and x + 1 > 0, which means x > 1. While x = 3 satisfies this domain restriction, x = -3 does not (it would make both arguments negative). The student's algebra was correct, but they didn't check domain restrictions. Choice A is wrong because the product property was applied correctly. Choice B is wrong because (x-1)(x+1) = x² - 1 is correct. Choice D is wrong because the product property was the appropriate choice here.

Question 4

A student solving the equation $\sqrt{x + 5} = x - 1$ writes: "Squaring both sides gives $x + 5 = x^2 - 2x + 1$ , so $x^2 - 3x - 4 = 0$ . Factoring: $(x - 4)(x + 1) = 0$ , so $x = 4$ or $x = -1$ . Therefore, the solutions are $x = 4$ and $x = -1$ ." What error did the student make?

The student made an algebraic error when rearranging the equation after squaring both sides.
The student incorrectly factored the quadratic expression and found the wrong potential solutions.
The student failed to check whether both solutions satisfy the original equation's domain restrictions. (correct answer)
The student should have isolated the square root term on one side before squaring both sides.

Explanation: The correct answer is C. When solving radical equations, squaring both sides can introduce extraneous solutions. Checking: for x = 4,

\sqrt{9} = 3

and

4 - 1 = 3

, so x = 4 works. For x = -1,

\sqrt{4} = 2

but

-1 - 1 = -2

, so x = -1 is extraneous. The student's algebra and factoring were correct (eliminating A and B), and the square root was already isolated (eliminating D).

Question 5

Two students are debating about correlation and causation. Student A says: "The data shows a strong positive correlation between ice cream sales and drowning incidents. This proves that eating ice cream causes people to drown."

What is the most significant flaw in Student A's reasoning?

Student A misinterpreted the correlation as negative when it should be positive for this relationship.
Student A concluded that correlation implies causation without considering potential confounding variables like season. (correct answer)
Student A failed to calculate the correlation coefficient before making any claims about the strength of correlation.
Student A should have collected more data points before drawing any conclusions about the relationship.

Explanation: The correct answer is B. This is a classic example of confusing correlation with causation. The correlation likely exists because both ice cream sales and drowning incidents increase during summer months (confounding variable), not because ice cream consumption causes drowning. Choice A is wrong because the student correctly identified a positive correlation. Choice C is wrong because the flaw isn't about calculating the correlation coefficient. Choice D is wrong because the sample size isn't the main issue with the reasoning.

Question 6

A student analyzing the function $g(x) = \frac{x^2 - 9}{x - 3}$ writes: "This function is undefined at $x = 3$ because the denominator equals zero. Therefore, $x = 3$ is a vertical asymptote."

What is wrong with the student's conclusion about the vertical asymptote?

The student correctly identified that the function is undefined at $x = 3$ , but this creates a removable discontinuity, not a vertical asymptote. (correct answer)
The student made an error in finding where the denominator equals zero; it actually occurs at $x = -3$ .
The student should have found the horizontal asymptote first before analyzing vertical asymptotes for rational functions.
The student forgot to check whether the numerator and denominator share any common factors before concluding about asymptotes.

Explanation: The correct answer is A. Since

x^2 - 9 = (x-3)(x+3)

, the function simplifies to

g(x) = x + 3

for

x ≠ 3

. The factor (x-3) cancels, creating a hole (removable discontinuity) at x = 3, not a vertical asymptote. Choice B is wrong because x - 3 = 0 when x = 3. Choice C is wrong because the order of finding asymptotes doesn't matter. Choice D is partially correct about checking common factors but doesn't fully explain the error.

Question 7

A student analyzing a probability problem writes: "In a game where I flip a fair coin 3 times, the probability of getting exactly 2 heads is $\frac{3}{8}$ . Since there are 3 ways to arrange 2 heads among 3 flips (HHT, HTH, THH), and each outcome has probability $\frac{1}{8}$ , the total probability is $3 \times \frac{1}{8} = \frac{3}{8}$ ."

Evaluate the correctness of the student's reasoning and calculation.

The student correctly identified all possible arrangements but incorrectly calculated the probability of each individual outcome.
The student incorrectly counted the number of ways to arrange 2 heads among 3 flips; there should be 4 ways.
The student's reasoning and calculation are both correct and properly demonstrate the use of the multiplication principle. (correct answer)
The student should have used the binomial probability formula instead of counting arrangements manually.

Explanation: The correct answer is C. The student's work is completely correct. There are indeed 3 ways to get exactly 2 heads in 3 flips (HHT, HTH, THH), each with probability (1/2)³ = 1/8, giving a total probability of 3/8. Choice A is wrong because 1/8 is the correct probability for each outcome. Choice B is wrong because there are exactly 3 arrangements, not 4. Choice D is wrong because while the binomial formula could be used, the counting method is perfectly valid and demonstrates good understanding.

Question 8

A student is solving an inequality: " $\frac{x-2}{x+1} > 0$ . Since we want this fraction to be positive, both the numerator and denominator must have the same sign. The numerator $x-2 > 0$ when $x > 2$ , and the denominator $x+1 > 0$ when $x > -1$ . Therefore, the solution is $x > 2$ ."

What critical step did the student miss in their reasoning?

The student failed to consider the case where both the numerator and denominator are negative simultaneously. (correct answer)
The student incorrectly solved the individual inequalities for when the numerator and denominator are positive.
The student should have cross-multiplied both sides of the inequality before solving for the individual cases.
The student neglected to check that $x = -1$ should be included in the final solution set.

Explanation: The correct answer is A. The student only considered when both parts are positive (x > 2) but missed that the fraction is also positive when both numerator and denominator are negative. This occurs when x - 2 < 0 AND x + 1 < 0, which gives x < 2 AND x < -1, so x < -1. The complete solution is x ∈ (-∞, -1) ∪ (2, ∞). Choice B is wrong because the individual inequalities were solved correctly. Choice C is wrong because cross-multiplying requires knowing the sign of (x+1). Choice D is wrong because x = -1 makes the denominator zero.

Question 9

A student analyzing the sequence $a_n = \frac{(-1)^n}{n}$ states: "Since $\lim_{n \to \infty} \frac{1}{n} = 0$ , and the $(-1)^n$ term just alternates the sign, the limit of this sequence is 0." What aspect of their reasoning needs clarification or correction?

The student incorrectly calculated the limit of $\frac{1}{n}$ , which should be 1, not 0.
The student failed to recognize that alternating sequences cannot have limits by definition.
The student's reasoning is sound, but they should have used the formal definition of limits to be more rigorous.
The student correctly identified the limit but should have explicitly addressed how the alternating sign affects convergence. (correct answer)

Explanation: The correct answer is D. The student's conclusion is correct (the limit is 0), but they should have been more explicit about why the alternating sign doesn't prevent convergence. Since |(-1)ⁿ/n| = 1/n → 0, the sequence converges to 0 despite the alternating signs. Choice A is wrong because lim(1/n) = 0 is correct. Choice B is wrong because alternating sequences can converge if their absolute values approach 0. Choice C is wrong because the reasoning, while informal, captures the essential idea correctly.

Question 10

A student solving a system of equations writes: "From the system $\begin{cases} 2x + 3y = 7 \\ 4x + 6y = 12 \end{cases}$ , I'll multiply the first equation by 2 to get $4x + 6y = 14$ . Now I have $4x + 6y = 14$ and $4x + 6y = 12$ . Since $14 ≠ 12$ , these equations are inconsistent and the system has no solution."

Evaluate the student's reasoning about this system of equations.

The student made an arithmetic error when multiplying the first equation by 2; the result should be $4x + 6y = 10$ .
The student incorrectly concluded the system has no solution when it actually has infinitely many solutions.
The student should have used substitution instead of elimination to solve this particular system of equations.
The student correctly identified that the system is inconsistent and properly concluded it has no solution. (correct answer)

Explanation: When evaluating systems of equations, you need to recognize the three possible outcomes: one solution (consistent and independent), infinitely many solutions (consistent and dependent), or no solution (inconsistent). Let's check the student's work step by step. Starting with the system:

\begin{cases} 2x + 3y = 7 \\ 4x + 6y = 12 \end{cases}

The student multiplied the first equation by 2:

2(2x + 3y) = 2(7)

, which gives

4x + 6y = 14

. This arithmetic is correct. Now the student has two equations with identical left sides but different right sides:

4x + 6y = 14

and

4x + 6y = 12

. Since the same expression (

4x + 6y

) cannot equal both 14 and 12 simultaneously, the system is inconsistent and has no solution. The student's reasoning is sound. Looking at the wrong answers: Choice A claims an arithmetic error, but

2 \times 7 = 14

, not 10. Choice B suggests the system has infinitely many solutions, but that would require the equations to be equivalent (same coefficients AND same constants), which they're not. Choice C criticizes the method choice, but elimination is perfectly appropriate here and actually reveals the inconsistency more clearly than substitution would. Remember this pattern: when you manipulate a system and get equations with identical variable terms but different constants, you've found an inconsistent system with no solution. This is a key diagnostic tool in systems of equations.

Question 11

A student is analyzing the limit $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ and writes: "Since direct substitution gives $\frac{0}{0}$ , this limit does not exist."

What misconception does this student's reasoning reveal?

The student incorrectly calculated the result of direct substitution, which should give $\frac{8}{0}$ , not $\frac{0}{0}$ .
The student confused indeterminate forms with undefined expressions and concluded the limit doesn't exist prematurely. (correct answer)
The student should have approached the limit from both sides before concluding that it doesn't exist.
The student incorrectly assumed that limits can only be evaluated using direct substitution methods.

Explanation: The correct answer is B. The form 0/0 is indeterminate, meaning more work is needed to find the limit, not that it doesn't exist. Using factoring: x³ - 8 = (x-2)(x² + 2x + 4), so the limit becomes lim[x→2](x² + 2x + 4) = 12. Choice A is wrong because substituting x = 2 gives (8-8)/(2-2) = 0/0. Choice C is wrong because one-sided limits aren't the issue here. Choice D is wrong because the student didn't assume direct substitution was the only method; they gave up when it failed.

Question 12

A student proves: "For any triangle with sides $a$ , $b$ , and $c$ , if $a^2 + b^2 = c^2$ , then the triangle is a right triangle." They then conclude: "This means that any triangle satisfying $a^2 + b^2 = c^2$ is the only way to have a right triangle." What logical error has occurred?

The student incorrectly proved the original statement; the Pythagorean theorem works in the opposite direction.
The student incorrectly assumed that geometric theorems can be applied in both directions automatically.
The student should have considered obtuse and acute triangles before making claims about right triangles.
The student confused a sufficient condition with a necessary and sufficient condition without proving both directions. (correct answer)

Explanation: This question tests your understanding of logical conditionals and the difference between one-way implications and biconditional statements. When working with mathematical theorems, you need to carefully distinguish between proving one direction versus proving equivalence. The student correctly proved that if

a^2 + b^2 = c^2

, then the triangle is a right triangle. This is indeed true - it's the converse of the Pythagorean theorem. However, the student then claimed this means

a^2 + b^2 = c^2

is "the only way" to have a right triangle. This conclusion requires proving the opposite direction: that if a triangle is a right triangle, then

a^2 + b^2 = c^2

. The student only proved one direction (sufficient condition) but concluded they had proven both directions (necessary and sufficient condition). Answer D correctly identifies this error. Answer A is wrong because the student actually did prove their stated direction correctly - the converse of the Pythagorean theorem is valid. Answer B misses the specific logical error; the issue isn't about geometric theorems generally, but about the logical structure of implications. Answer C is irrelevant because the student's error has nothing to do with considering other types of triangles - it's purely about logical reasoning. Study tip: When you see "if and only if" claims or conclusions about something being "the only way," check whether both directions of the logical implication have been proven. Many math errors involve assuming that proving "if P then Q" automatically means "if Q then P."

Question 13

A student claims: "The function $f(x) = \frac{1}{x}$ has domain $(-\infty, 0) \cup (0, \infty)$ and range $(-\infty, 0) \cup (0, \infty)$ . Since the domain and range are the same, this function must be its own inverse." What error is present in this reasoning?

The student incorrectly determined the domain of the function, which should include $x = 0$ .
The student incorrectly determined the range of the function, which should be $[0, \infty)$ .
The student incorrectly concluded that having the same domain and range is sufficient for a function to be its own inverse. (correct answer)
The student should have checked whether the function passes the horizontal line test before making claims about inverses.

Explanation: The correct answer is C. While f(x) = 1/x does have the same domain and range, having the same domain and range is not sufficient to conclude a function is its own inverse. For example, f(x) = x² has domain ℝ and range [0,∞), but it's not even one-to-one. The student's reasoning is logically flawed even though their conclusion happens to be correct in this case. Choice A is wrong because x = 0 makes the function undefined. Choice B is wrong because the range correctly excludes 0. Choice D misses the main logical error in the reasoning.

Question 14

A student proves that if $n$ is even, then $n^2$ is even. They then conclude: "Since we've shown that even numbers have even squares, this means that if $n^2$ is even, then $n$ must be even." What logical error has the student made?

The student incorrectly assumed that proving the converse of a statement automatically proves the original statement.
The student confused the converse with the contrapositive and incorrectly assumed they are logically equivalent.
The student incorrectly assumed that proving a conditional statement automatically proves its converse without separate justification. (correct answer)
The student made an error in the original proof and should have proven the contrapositive statement instead.

Explanation: The correct answer is C. The student proved "if n is even, then n² is even" but then incorrectly concluded that "if n² is even, then n is even" without proving it separately. While this converse happens to be true, it requires its own proof. Choice A is backwards - they proved the original and assumed the converse. Choice B incorrectly describes what the student confused. Choice D is wrong because the original proof was fine; the contrapositive wasn't needed.

Question 15

A student is analyzing the rational function $h(x) = \frac{x^2-5x+6}{x^2-4}$ and makes the following claims: "The function has vertical asymptotes at $x = 2$ and $x = -2$ because these values make the denominator zero. The function also has a horizontal asymptote at $y = 1$ because both the numerator and denominator are degree 2, so the horizontal asymptote is the ratio of leading coefficients: $\frac{1}{1} = 1$ ." To check the first claim, the student factors: " $x^2-5x+6 = (x-2)(x-3)$ and $x^2-4 = (x-2)(x+2)$ , so $h(x) = \frac{(x-2)(x-3)}{(x-2)(x+2)}$ ." What error in reasoning should be identified?

The horizontal asymptote analysis is incorrect because the function has a slant asymptote instead
The student should have checked end behavior using limits rather than the ratio of leading coefficients
The factoring of the numerator $x^2-5x+6$ contains an algebraic error
The student failed to recognize that $x = 2$ creates a removable discontinuity, not a vertical asymptote (correct answer)

Explanation: When analyzing rational functions for vertical asymptotes and discontinuities, you need to examine what happens when both the numerator and denominator share common factors. The student correctly factored both expressions:

x^2-5x+6 = (x-2)(x-3)

and

x^2-4 = (x-2)(x+2)

. This gives

h(x) = \frac{(x-2)(x-3)}{(x-2)(x+2)}

. The critical insight is that both numerator and denominator contain the factor

(x-2)

, which can be canceled out for

x \neq 2

, leaving

h(x) = \frac{x-3}{x+2}

for all values except

x = 2

. At

x = 2

, the original function is undefined, but since the common factor cancels, this creates a hole (removable discontinuity) rather than a vertical asymptote. The function approaches

\frac{2-3}{2+2} = -\frac{1}{4}

x

approaches 2. Only

x = -2

creates a true vertical asymptote because it makes the simplified denominator zero without being canceled by the numerator. The horizontal asymptote analysis at

y = 1

is actually correct—both the simplified numerator and denominator have degree 1 with leading coefficients of 1. Looking at the wrong answers: (A) is incorrect because there is indeed a horizontal asymptote, not a slant asymptote. (B) is wrong because the ratio method is valid for polynomials of equal degree. (C) is incorrect since the factoring

x^2-5x+6 = (x-2)(x-3)

is algebraically correct. Study tip: Always factor completely and cancel common factors before identifying vertical asymptotes. Canceled factors indicate holes, not asymptotes.

Question 16

Two students are debating whether the equation $\sqrt{x-3} = x-5$ has any solutions. Student A argues: "Squaring both sides gives $x-3 = (x-5)^2 = x^2-10x+25$ , which simplifies to $x^2-11x+28=0$ . Using the quadratic formula, I get $x = 4$ or $x = 7$ . Therefore, the equation has two solutions." Student B responds: "You need to check your solutions in the original equation because squaring can introduce extraneous solutions." When Student B checks, which of the following represents the most complete and accurate analysis?

Both $x = 4$ and $x = 7$ satisfy the original equation, confirming two valid solutions
Only $x = 7$ satisfies the original equation; $x = 4$ is extraneous because $\sqrt{4-3} \neq 4-5$ (correct answer)
Only $x = 4$ satisfies the original equation; $x = 7$ fails because the domain requires $x \geq 3$
Neither solution is valid because both violate the domain restriction that $x-5 \geq 0$ for the equation to be meaningful

Explanation: Student B's complete analysis should verify both solutions in the original equation

\sqrt{x-3} = x-5

. For

x = 4

\sqrt{4-3} = \sqrt{1} = 1

, but

x-5 = 4-5 = -1

. Since

1 \neq -1

, this is extraneous. For

x = 7

\sqrt{7-3} = \sqrt{4} = 2

, and

x-5 = 7-5 = 2

. Since

2 = 2

, this is valid. Choice A is wrong because

x = 4

doesn't satisfy the original equation. Choice C is wrong because

x = 7

does satisfy the equation and both values are in the domain

x \geq 3

. Choice D is wrong because we don't need

x-5 \geq 0

for the equation to be meaningful—we just need

x \geq 3

for the square root to be defined.

Question 17

A student is asked to determine whether the function $f(x) = \frac{x^2-4}{x-2}$ is continuous at $x = 2$ . The student provides this analysis: "First, I'll find $f(2) = \frac{2^2-4}{2-2} = \frac{0}{0}$ , which is undefined. Since $f(2)$ doesn't exist, the function is not continuous at $x = 2$ . However, I can simplify the function by factoring: $f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$ . Therefore, $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x+2) = 4$ ." What additional step would make the student's analysis of continuity complete?

Verify that the original function and simplified function have the same domain restrictions throughout
Conclude that since $\lim_{x \to 2} f(x) = 4$ exists but $f(2)$ is undefined, the function is discontinuous at $x = 2$ (correct answer)
Check whether the function can be made continuous by defining $f(2) = 4$ to create a removable discontinuity
Determine if the discontinuity affects the overall continuity of the function on its natural domain

Explanation: The student correctly found that

f(2)

is undefined and that

\lim_{x \to 2} f(x) = 4

. For a function to be continuous at a point, three conditions must be met: (1)

f(a)

exists, (2)

\lim_{x \to a} f(x)

exists, and (3)

\lim_{x \to a} f(x) = f(a)

. Since condition (1) fails, the student should explicitly conclude that the function is discontinuous at

x = 2

. The student did most of the work but never stated the final conclusion about continuity. Choice A is unnecessarily complex for this analysis. Choice C goes beyond what's asked (the question asks about continuity of the given function, not a modified version). Choice D is irrelevant to determining continuity at the specific point

x = 2

Question 18

In analyzing the inequality $|2x-6| < 4$ , a student writes: "The absolute value inequality $|2x-6| < 4$ means that $2x-6$ is within 4 units of zero, so $-4 < 2x-6 < 4$ . Adding 6 to all parts: $2 < 2x < 10$ . Dividing by 2: $1 < x < 5$ . To verify, I'll test $x = 0$ : $|2(0)-6| = 6$ , and since $6 > 4$ , this confirms $x = 0$ is not in the solution set, which matches my answer since $0 < 1$ ." What error should be identified in this reasoning?

The student incorrectly solved $-4 < 2x-6 < 4$ by adding 6 instead of subtracting 6
The verification using $x = 0$ is insufficient because it only tests one point outside the solution interval (correct answer)
The student should have considered both $2x-6 < 4$ and $2x-6 > -4$ as separate inequalities rather than compound
The interpretation of absolute value as "within 4 units of zero" leads to an incorrect compound inequality setup

Explanation: The student's algebraic work is actually correct:

|2x-6| < 4

does give

1 < x < 5

. However, the verification strategy is flawed. Testing only one point (

x = 0

) outside the solution set is insufficient to verify the solution. A complete verification should test points inside the solution interval (like

x = 3

|2(3)-6| = 0 < 4

✓) and points outside both ends (like

x = 0

and

x = 6

). The student's reasoning would be stronger with more comprehensive testing. Choice A is wrong because adding 6 was correct. Choice C is wrong because the compound inequality approach is valid and equivalent. Choice D is wrong because the interpretation and setup are actually correct.

Question 19

A student attempts to solve the system of equations $\begin{cases} 2x + 3y = 7 \\ 4x + 6y = 15 \end{cases}$ and provides the following work: "I'll use elimination. Multiplying the first equation by -2: $-4x - 6y = -14$ . Adding this to the second equation: $(-4x - 6y) + (4x + 6y) = -14 + 15$ , which gives $0 = 1$ . Since this is a contradiction, the system has no solution." A classmate objects: "You made an error in your arithmetic. Let me check your multiplication: $-2(2x + 3y) = -2(7)$ gives $-4x - 6y = -14$ . That's correct. And $-14 + 15 = 1$ , so $0 = 1$ is right." What should the classmate conclude?

The original student's method was wrong; substitution should be used instead of elimination for this system
Both students made the same computational error in determining that $0 = 1$ represents a contradiction
The classmate should recognize that the original student's work and conclusion are actually correct (correct answer)
The system actually has infinitely many solutions because the equations are multiples of each other

Explanation: The classmate's verification confirms that the original student's work is correct. The first equation multiplied by -2 gives

-4x - 6y = -14

. Adding this to

4x + 6y = 15

yields

0 = 1

, which is indeed a contradiction, meaning the system has no solution. The classmate verified each step and found no errors, so should conclude the original work is correct. Choice A is wrong because elimination is perfectly valid here. Choice B is wrong because

0 = 1

is correctly identified as a contradiction. Choice D is wrong because if the equations were multiples of each other, we would get

0 = 0

(identity), not

0 = 1

(contradiction). The second equation is not a multiple of the first since

\frac{15}{7} \neq 2

Question 20

Two students are debating the solution to $3^{x+1} = 27^{x-2}$ . Student A argues: "I can rewrite this as $3^{x+1} = (3^3)^{x-2} = 3^{3(x-2)} = 3^{3x-6}$ . Since the bases are equal, the exponents must be equal: $x+1 = 3x-6$ . Solving: $1+6 = 3x-x$ , so $7 = 2x$ and $x = 3.5$ ." Student B responds: "Your algebra is right, but you should verify the answer. Let me check: $3^{3.5+1} = 3^{4.5}$ and $27^{3.5-2} = 27^{1.5}$ . Now, $27^{1.5} = (3^3)^{1.5} = 3^{4.5}$ . So both sides equal $3^{4.5}$ , confirming the solution." Which statement best describes the mathematical reasoning demonstrated?

Student A made an error in the exponent arithmetic when expanding $3^{3(x-2)}$
Student B's verification process contains a computational error in evaluating $27^{1.5}$
Both students used valid mathematical reasoning, with Student B providing appropriate verification (correct answer)
Student A should have used logarithms instead of the equal-base method for this exponential equation

Explanation: Both students demonstrated sound mathematical reasoning. Student A correctly rewrote

27^{x-2}

(3^3)^{x-2} = 3^{3(x-2)} = 3^{3x-6}

, then properly applied the equal-base property to get

x+1 = 3x-6

, and solved correctly to get

x = 3.5

. Student B appropriately verified this solution by substituting back into the original equation and confirming both sides equal

3^{4.5}

. The verification shows good mathematical practice. Choice A is wrong because

3(x-2) = 3x-6

is correct. Choice B is wrong because

27^{1.5} = (3^3)^{1.5} = 3^{4.5}

is correct. Choice D is wrong because the equal-base method is perfectly valid and more efficient than logarithms for this problem.