id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-012201 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Find the exact value: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(15)}{x-(-2)}=-16.$$
Your final response must include (i)... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -2$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-16$), giving the unique solution $x=-1$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis": "Sensitivity analysis: Clea... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-2$). (Here the result is $\boxed{x=-1}$.) |
math-012202 | Algebra: Rational Equations — Verification by Substitution | 7 | Start by stating any domain restrictions: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(8)}{x-(-2)}=\frac{13... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{13}{8}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{8}$), giving the unique solution $x=-18$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis": "Sensitivity analysis... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-2$). (Here the result is $\boxed{x=-18}$.) |
math-012203 | Inequalities: AM–GM — Equality Conditions | 7 | Solve and justify each step: Let $x,y>0$ satisfy $x+y=875$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g.... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=875-x$ with $x\\in(0,875)$. Then $P(x)=xy=x(875... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{765625}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{765625}{4}$.",
"rob... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=437.5$. |
math-012204 | Algebra: Extremal Values — Global Bounds | 7 | Solve and justify each step: Let $x,y>0$ satisfy $x+y=689$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g.... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=689-x$ with $x\\in(0,689)$. Then $P(x)=xy=x(689... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{474721}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{474721}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=344.5$. (Here the result is $\boxed{\frac{474721}$.) |
math-012205 | Algebra: Extremal Values — Global Bounds | 7 | Work carefully and justify each inference: Let $x,y>0$ satisfy $x+y=461$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=461-x$ with $x\\in(0,461)$. Then $P(x)=xy=x(461... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{212521}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{212521}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=230.5$. |
math-012206 | Optimization: Two Variables — Concavity | 7 | Be explicit about assumptions: Let $x,y>0$ satisfy $x+y=538$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=538$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{72361}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=72361$.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=269.0$. |
math-012207 | Optimization: Two Variables — Concavity | 7 | Give reasoning, not just computation: Let $x,y>0$ satisfy $x+y=743$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theo... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=743-x$ with $x\\in(0,743)$. Then $P(x)=xy=x(743... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{552049}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{552049}{4}$.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=371.5$. (Here the result is $\boxed{\frac{552049}$.) |
math-012208 | Algebra: Extremal Values — Global Bounds | 7 | Task: Let $x,y>0$ satisfy $x+y=68$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second deriv... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=68-x$ with $x\\in(0,68)$. Then $P(x)=xy=x(68-x)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1156}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=1156$.",
"robustness_analysis": "Ge... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=34.0$. (Here the result is $\boxed{1156}$.) |
math-012209 | Algebra: Extremal Values — Global Bounds | 7 | Show all reasoning: Let $x,y>0$ satisfy $x+y=352$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavi... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=352-x$ with $x\\in(0,352)$. Then $P(x)=xy=x(352... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{30976}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=30976$.",
"robustness_analysis": "If the problem were perturbed: A... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=176.0$. (Here the result is $\boxed{30976}$.) |
math-012210 | Algebra: Extremal Values — Global Bounds | 7 | Start by stating any domain restrictions: Let $x,y>0$ satisfy $x+y=715$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus ... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=715$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{511225}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{511225}{4}$.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=357.5$. (Here the result is $\boxed{\frac{511225}$.) |
math-012211 | Algebra: Extremal Values — Global Bounds | 7 | Prompt: Let $x,y>0$ satisfy $x+y=47$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second der... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=47-x$ with $x\\in(0,47)$. Then $P(x)=xy=x(47-x)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{2209}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{2209}{4}$.",
"robustness_analysis": "Generality note:... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=23.5$. (Here the result is $\boxed{\frac{2209}$.) |
math-012212 | Optimization: Two Variables — Concavity | 7 | Answer with a short justification: Let $x,y>0$ satisfy $x+y=818$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=818-x$ with $x\\in(0,818)$. Then $P(x)=xy=x(818... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{167281}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=167281$.",
"robustness_analysis": "Robustness note: AM–GM general... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=409.0$. (Here the result is $\boxed{167281}$.) |
math-012213 | Algebra: Extremal Values — Global Bounds | 7 | Determine the requested value: Let $x,y>0$ satisfy $x+y=229$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=229-x$ with $x\\in(0,229)$. Then $P(x)=xy=x(229... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{52441}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{52441}{4}$.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=114.5$. (Here the result is $\boxed{\frac{52441}$.) |
math-012214 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Work this out carefully: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(0)}{x-(12)}=\frac{19}{7}.$$
Your final response must include... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 12$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{7}$), giving the unique solution $x=19$. The domain check $x\\neq 12$ is satisfied here.",
"robustness_analysis": "Generality note: Clear... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=12$). (Here the result is $\boxed{x=19}$.) |
math-012215 | Optimization: Two Variables — Concavity | 7 | Give a theorem-based solution: Let $x,y>0$ satisfy $x+y=553$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=553-x$ with $x\\in(0,553)$. Then $P(x)=xy=x(553... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{305809}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{305809}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=276.5$. (Here the result is $\boxed{\frac{305809}$.) |
math-012216 | Algebra: Rational Equations — Verification by Substitution | 7 | Question: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(0)}{x-(-10)}=11.$$
Your final response must include (i) the solution set and (ii) a bri... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - 11(x-b)=0$ with the domain restriction $x\\neq -10$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=11$), giving the unique solution $x=-11$. The domain check $x\\neq -10$ is satisfied here.",
"robustness_analysis": "Robustn... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-10$). (Here the result is $\boxed{x=-11}$.) |
math-012217 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Prompt: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(8)}{x-(3)}=\frac{3}{8}.$$
Your final response must include (i) the solution set and (ii) ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{3}{8}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{8}$), giving the unique solution $x=11$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearin... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). (Here the result is $\boxed{x=11}$.) |
math-012218 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Give a theorem-based solution: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-11)}{x-(7)}=\frac{-2}{7}.$$
Your final response must include ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 7$ so that $x-(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-2}{7}$), giving the unique solution $x=-7$. The domain check $x\\neq 7$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=7$). |
math-012219 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Work this out carefully: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-15)}{x-(14)}=\frac{-26}{3}.$$
Your ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 14$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-26}{3}$), giving the unique solution $x=11$. The domain check $x\\neq 14$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=14$). (Here the result is $\boxed{x=11}$.) |
math-012220 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Question: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-3)}{x-(-6)}=\frac{2}{5}.$$
Your final response must include (i) the solution set and (ii) a brief ch... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -6$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{5}$), giving the unique solution $x=-1$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). |
math-012221 | Algebra: Extremal Values — Global Bounds | 7 | Solve and sanity-check: Let $x,y>0$ satisfy $x+y=512$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., con... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=512-x$ with $x\\in(0,512)$. Then $P(x)=xy=x(512... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{65536}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=65536$.",
"robustness_analysis": "Robustness note: AM–GM generaliz... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=256.0$. |
math-012222 | Inequalities: Product Given Sum | 7 | Solve with verification: Let $x,y>0$ satisfy $x+y=270$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=270-x$ with $x\\in(0,270)$. Then $P(x)=xy=x(270... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{18225}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=18225$.",
"robustness_analysis": "Robustness note: AM–GM gen... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=135.0$. (Here the result is $\boxed{18225}$.) |
math-012223 | Algebra: Rational Equations — Clearing Denominators | 7 | Find the exact value: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(3)}{x-(5)}=\frac{3}{5}.$$
Your final response must include (i) the solution set and (ii) ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{3}{5}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=0}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{5}$), giving the unique solution $x=0$. The domain check $x\\neq 5$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearing ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=5$). |
math-012224 | Optimization: Two Variables — Concavity | 7 | Work this out carefully: Let $x,y>0$ satisfy $x+y=834$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=834-x$ with $x\\in(0,834)$. Then $P(x)=xy=x(834... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{173889}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=173889$.",
"robustness_analysis": "Robustness note: AM–GM general... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=417.0$. |
math-012225 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Solve and sanity-check: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(15)}=\frac{14}{23}.$$
Your final response must include (i) the... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 15$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-8}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{14}{23}$), giving the unique solution $x=-8$. The domain check $x\\neq 15$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=15$). (Here the result is $\boxed{x=-8}$.) |
math-012226 | Algebra: Extremal Values — Global Bounds | 7 | Work this out carefully: Let $x,y>0$ satisfy $x+y=395$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=395-x$ with $x\\in(0,395)$. Then $P(x)=xy=x(395... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{156025}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{156025}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=197.5$. (Here the result is $\boxed{\frac{156025}$.) |
math-012227 | Inequalities: Product Given Sum | 7 | Give an answer and a quick verification: Let $x,y>0$ satisfy $x+y=488$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus t... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=488$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{59536}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=59536$.",
"robustness_analysis": "If the problem were perturbed: A... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=244.0$. (Here the result is $\boxed{59536}$.) |
math-012228 | Algebra: Extremal Values — Global Bounds | 7 | Work this out carefully: Let $x,y>0$ satisfy $x+y=843$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=843-x$ with $x\\in(0,843)$. Then $P(x)=xy=x(843... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{710649}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{710649}{4}$.",
"rob... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=421.5$. |
math-012229 | Algebra: Extremal Values — Global Bounds | 7 | Provide both a computational and a conceptual explanation: Let $x,y>0$ satisfy $x+y=454$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=454$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{51529}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=51529$.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=227.0$. (Here the result is $\boxed{51529}$.) |
math-012230 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Where appropriate, name the theorem you use: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-14)}{x-(-6)}=\fr... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{5}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{3}$), giving the unique solution $x=6$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_analysis": "Ro... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). (Here the result is $\boxed{x=6}$.) |
math-012231 | Optimization: Two Variables — Concavity | 7 | Explain what is being counted/optimized: Let $x,y>0$ satisfy $x+y=634$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus t... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=634-x$ with $x\\in(0,634)$. Then $P(x)=xy=x(634... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{100489}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=100489$.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=317.0$. |
math-012232 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Do not skip justification steps: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-11)}{x-(-10)}=\frac{11}{10}.$$
Your final response must include (i... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -10$ so that $x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=0}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{11}{10}$), giving the unique solution $x=0$. The domain check $x\\neq -10$ is satisfied here.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-10$). |
math-012233 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | State any required conditions first: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(7)}{x-(-11)}=\frac{1}{10}... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{1}{10}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{10}$), giving the unique solution $x=9$. The domain check $x\\neq -11$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-11$). (Here the result is $\boxed{x=9}$.) |
math-012234 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Determine the requested value: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(5)}{x-(-1)}=\frac{1}{3}.$$
Your final response must include (i) the solution set... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -1$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=8}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{3}$), giving the unique solution $x=8$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_analysis": "Ge... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). (Here the result is $\boxed{x=8}$.) |
math-012235 | Algebra: Rational Equations — Domain Restrictions | 7 | Warm-up: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-3)}{x-(6)}=\frac{16}{25}.$$
Your final response must include (i) the solution set a... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 6$ so that $x-(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{16}{25}$), giving the unique solution $x=-19$. The domain check $x\\neq 6$ is satisfied here.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). (Here the result is $\boxed{x=-19}$.) |
math-012236 | Algebra: Rational Equations — Verification by Substitution | 7 | Give a theorem-based solution: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(-11)}=\frac{13}{9}.$$
Your final respon... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -11$ so that $x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{9}$), giving the unique solution $x=-20$. The domain check $x\\neq -11$ is satisfied here.",
"robustness_analysis... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-11$). (Here the result is $\boxed{x=-20}$.) |
math-012237 | Algebra: Rational Equations — Clearing Denominators | 7 | Show all reasoning: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(-3)}=\frac{5}{2}.$$
Your final res... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -3$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{2}$), giving the unique solution $x=-9$. The domain check $x\\neq -3$ is satisfied here.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-3$). (Here the result is $\boxed{x=-9}$.) |
math-012238 | Algebra: Rational Equations — Clearing Denominators | 7 | Give reasoning, not just computation: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(10)}{x-(-9)}=\frac{-11}{8}.$$
Your final respon... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-11}{8}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-11}{8}$), giving the unique solution $x=-1$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": "Generality note... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). (Here the result is $\boxed{x=-1}$.) |
math-012239 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Solve with verification: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-14)}{x-(10)}=9.$$
Your final response must include (i) the solution set an... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - 9(x-b)=0$ with the domain restriction $x\\neq 10$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=9$), giving the unique solution $x=13$. The domain check $x\\neq 10$ is satisfied here.",
"robustness_analysis": "Generality note: Clearing denomina... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=10$). (Here the result is $\boxed{x=13}$.) |
math-012240 | Algebra: Rational Equations — Verification by Substitution | 7 | Write the solution set clearly: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-1)}{x-(0)}=\frac{10}{9}.$$
Your final response must include (i) the solution s... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 0$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{10}{9}$), giving the unique solution $x=9$. The domain check $x\\neq 0$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearing... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=0$). (Here the result is $\boxed{x=9}$.) |
math-012241 | Optimization: Two Variables — Concavity | 7 | Find the exact value: Let $x,y>0$ satisfy $x+y=314$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., conca... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=314-x$ with $x\\in(0,314)$. Then $P(x)=xy=x(314... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{24649}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=24649$.",
"robustness_analysis": "... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=157.0$. (Here the result is $\boxed{24649}$.) |
math-012242 | Algebra: Extremal Values — Global Bounds | 7 | State any required conditions first: Let $x,y>0$ satisfy $x+y=312$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theor... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=312-x$ with $x\\in(0,312)$. Then $P(x)=xy=x(312... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{24336}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=24336$.",
"robustness_analysis": "Sensitivity analysis: AM–G... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=156.0$. (Here the result is $\boxed{24336}$.) |
math-012243 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Derive the result step-by-step: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-2)}{x-(-1)}=\frac{6}{7}.$$
Your final response must include (i) the solution s... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{6}{7}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-8}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{6}{7}$), giving the unique solution $x=-8$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). |
math-012244 | Inequalities: Product Given Sum | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=84$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=84$ to get $\\sqrt{xy}\\le \\frac{8... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1764}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=1764$.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=42.0$. |
math-012245 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Prompt: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(7)}{x-(-9)}=\frac{11}{27}.$$
Your final response must include (i) the solution set and (ii) a brief che... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{11}{27}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{11}{27}$), giving the unique solution $x=18$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": "Robustness note: Clea... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). (Here the result is $\boxed{x=18}$.) |
math-012246 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Complete the analysis: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-4)}{x-(8)}=\frac{-1}{2}.$$
Your final response must in... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-1}{2}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=0}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-1}{2}$), giving the unique solution $x=0$. The domain check $x\\neq 8$ is satisfied here.",
"robustness_analy... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=8$). |
math-012247 | Optimization: Two Variables — Concavity | 7 | Compute the requested quantity: Let $x,y>0$ satisfy $x+y=565$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=565$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{319225}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{319225}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=282.5$. (Here the result is $\boxed{\frac{319225}$.) |
math-012248 | Optimization: Two Variables — Concavity | 7 | Where appropriate, name the theorem you use: Let $x,y>0$ satisfy $x+y=41$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculu... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=41$ to get $\\sqrt{xy}\\le \\frac{4... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{1681}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{1681}{4}$.",
"robustness_analysis": "If the problem w... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=20.5$. (Here the result is $\boxed{\frac{1681}$.) |
math-012249 | Optimization: Two Variables — Concavity | 7 | Start by stating any domain restrictions: Let $x,y>0$ satisfy $x+y=300$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus ... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=300-x$ with $x\\in(0,300)$. Then $P(x)=xy=x(300... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{22500}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=22500$.",
"robustness_analysis": "... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=150.0$. (Here the result is $\boxed{22500}$.) |
math-012250 | Optimization: Two Variables — Concavity | 7 | Task: Let $x,y>0$ satisfy $x+y=177$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second deri... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=177$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{31329}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{31329}{4}$.",
"robustness_ana... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=88.5$. (Here the result is $\boxed{\frac{31329}$.) |
math-012251 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Work carefully and justify each inference: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(3)}=-2.$$
Your final response must ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - -2(x-b)=0$ with the domain restriction $x\\neq 3$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-2$), giving the unique solution $x=4$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_analysis": "If t... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). (Here the result is $\boxed{x=4}$.) |
math-012252 | Inequalities: AM–GM — Equality Conditions | 7 | Challenge: Let $x,y>0$ satisfy $x+y=801$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=801-x$ with $x\\in(0,801)$. Then $P(x)=xy=x(801... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{641601}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{641601}{4}$.",
"rob... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=400.5$. (Here the result is $\boxed{\frac{641601}$.) |
math-012253 | Inequalities: Product Given Sum | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=389$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=389-x$ with $x\\in(0,389)$. Then $P(x)=xy=x(389... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{151321}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{151321}{4}$.",
"robustness_analysis": "If the probl... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=194.5$. (Here the result is $\boxed{\frac{151321}$.) |
math-012254 | Inequalities: Product Given Sum | 7 | Checkpoint: Let $x,y>0$ satisfy $x+y=524$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/secon... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=524-x$ with $x\\in(0,524)$. Then $P(x)=xy=x(524... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{68644}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=68644$.",
"robustness_analysis": "Sensitivity analysis: AM–G... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=262.0$. (Here the result is $\boxed{68644}$.) |
math-012255 | Algebra: Rational Equations — Domain Restrictions | 7 | Explain why your operations are valid: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(13)}=\frac{-2}{5... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-2}{5}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=8}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-2}{5}$), giving the unique solution $x=8$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "Sensitivity analysis: Cl... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). |
math-012256 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Use two approaches if possible: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(-14)}=\frac{8}{3}.$$
Your final response must include (i) th... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -14$ so that $x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{8}{3}$), giving the unique solution $x=-17$. The domain check $x\\neq -14$ is satisfied here.",
"robustness_... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-14$). |
math-012257 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Indicate where a theorem is used: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(-14)}=\frac{1}{5}.$$
Your final response mus... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{1}{5}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{5}$), giving the unique solution $x=11$. The domain check $x\\neq -14$ is satisfied here.",
"robustness_an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-14$). (Here the result is $\boxed{x=11}$.) |
math-012258 | Algebra: Extremal Values — Global Bounds | 7 | Show all reasoning: Let $x,y>0$ satisfy $x+y=372$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavi... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=372-x$ with $x\\in(0,372)$. Then $P(x)=xy=x(372... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{34596}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=34596$.",
"robustness_analysis": "Generality note: AM–GM generaliz... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=186.0$. (Here the result is $\boxed{34596}$.) |
math-012259 | Inequalities: Product Given Sum | 7 | Complete the analysis: Let $x,y>0$ satisfy $x+y=708$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., conc... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=708$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{125316}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=125316$.",
"robustness_analysis": "Sensitivity analysis: AM... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=354.0$. (Here the result is $\boxed{125316}$.) |
math-012260 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Warm-up: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(-11)}=\frac{7}{3}.$$
Your final response must include (i) the ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -11$ so that $x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{3}$), giving the unique solution $x=-20$. The domain check $x\\neq -11$ is satisfied here.",
"robustness_... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-11$). (Here the result is $\boxed{x=-20}$.) |
math-012261 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Track units/moduli carefully: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(13)}{x-(-13)}=\frac{-3}{10}.$$
Your final response must include (i)... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -13$ so that $x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-3}{10}$), giving the unique solution $x=7$. The domain check $x\\neq -13$ is satisfied here.",
"robustness_an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-13$). |
math-012262 | Inequalities: AM–GM — Equality Conditions | 7 | Carefully track domains: Let $x,y>0$ satisfy $x+y=567$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=567-x$ with $x\\in(0,567)$. Then $P(x)=xy=x(567... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{321489}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{321489}{4}$.",
"robustness_analysis": "Sensit... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=283.5$. (Here the result is $\boxed{\frac{321489}$.) |
math-012263 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Give an answer and a quick verification: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(-4)}=\frac{3}{2}.$$
Your final response must include... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{3}{2}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-14}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{2}$), giving the unique solution $x=-14$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-4$). (Here the result is $\boxed{x=-14}$.) |
math-012264 | Optimization: Two Variables — Concavity | 7 | Answer with a short justification: Let $x,y>0$ satisfy $x+y=848$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=848-x$ with $x\\in(0,848)$. Then $P(x)=xy=x(848... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{179776}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=179776$.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=424.0$. (Here the result is $\boxed{179776}$.) |
math-012265 | Algebra: Rational Equations — Verification by Substitution | 7 | Problem: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(4)}{x-(-6)}=\frac{7}{12}.$$
Your final response must include (i) the solution set and (ii) a brief che... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{7}{12}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{12}$), giving the unique solution $x=18$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). (Here the result is $\boxed{x=18}$.) |
math-012266 | Algebra: Rational Equations — Clearing Denominators | 7 | Give a theorem-based solution: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-1)}{x-(-9)}=\frac{21}{29}.$$
Your final respon... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -9$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{21}{29}$), giving the unique solution $x=20$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": "Generality note... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). (Here the result is $\boxed{x=20}$.) |
math-012267 | Algebra: Rational Equations — Clearing Denominators | 7 | Start by stating any domain restrictions: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(14)}{x-(2)}=\frac{29}{17}.$$
Your final response must include (i) the... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{29}{17}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{29}{17}$), giving the unique solution $x=-15$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=2$). (Here the result is $\boxed{x=-15}$.) |
math-012268 | Algebra: Rational Equations — Verification by Substitution | 7 | Start by stating any domain restrictions: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(7)}{x-(-1)}=\frac{17}{9}.$$
Your final resp... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{17}{9}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{17}{9}$), giving the unique solution $x=-10$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_analysis": "Robustness note: Cle... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). (Here the result is $\boxed{x=-10}$.) |
math-012269 | Algebra: Rational Equations — Domain Restrictions | 7 | Exercise: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(12)}{x-(6)}=\frac{1}{2}.$$
Your final response must... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 6$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{2}$), giving the unique solution $x=18$. The domain check $x\\neq 6$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearin... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). (Here the result is $\boxed{x=18}$.) |
math-012270 | Optimization: Two Variables — Concavity | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=896$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=896-x$ with $x\\in(0,896)$. Then $P(x)=xy=x(896... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{200704}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=200704$.",
"robustness_analysis": "Robustn... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=448.0$. |
math-012271 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Explain what is being counted/optimized: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(9)}{x-(2)}=\frac{9}{16}.$$
Your final response must incl... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{9}{16}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{16}$), giving the unique solution $x=18$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=2$). |
math-012272 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Answer with a short justification: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(10)}{x-(4)}=\frac{1}{2}.$$
Your final response must include (i) the solution... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{1}{2}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{2}$), giving the unique solution $x=16$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_analysis": "Robustness note: C... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=4$). |
math-012273 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Solve (and briefly cross-validate): Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(10)}{x-(9)}=\frac{9}{8}.$$
Your final response must inclu... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{9}{8}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{8}$), giving the unique solution $x=1$. The domain check $x\\neq 9$ is satisfied here.",
"robustness_analysis": "If ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=9$). (Here the result is $\boxed{x=1}$.) |
math-012274 | Algebra: Rational Equations — Domain Restrictions | 7 | Carefully track domains: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-11)}{x-(15)}=\frac{-15}{11}.$$
Your final response must include (i) the solution set ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-15}{11}(x-b)=0$ with the domain restriction $x\\n... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-15}{11}$), giving the unique solution $x=4$. The domain check $x\\neq 15$ is satisfied here.",
"robustness_analysis": "Generality note: Clear... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=15$). (Here the result is $\boxed{x=4}$.) |
math-012275 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Work this out carefully: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(14)}{x-(-5)}=-18.$$
Your final response must include (i) the solution se... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - -18(x-b)=0$ with the domain restriction $x\\neq -5$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-18$), giving the unique solution $x=-4$. The domain check $x\\neq -5$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearing ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-5$). (Here the result is $\boxed{x=-4}$.) |
math-012276 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Derive the result step-by-step: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(-4)}=\frac{7}{17}.$$
Your final response must include (i) ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{7}{17}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{17}$), giving the unique solution $x=13$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-4$). |
math-012277 | Algebra: Rational Equations — Clearing Denominators | 7 | Solve and then verify: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(2)}{x-(-8)}=\frac{-3}{7}.$$
Your final response must include (... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -8$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-3}{7}$), giving the unique solution $x=-1$. The domain check $x\\neq -8$ is satisfied here.",
"robustness_analysis": "Robustness note:... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-8$). (Here the result is $\boxed{x=-1}$.) |
math-012278 | Inequalities: AM–GM — Equality Conditions | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=374$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g.... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=374-x$ with $x\\in(0,374)$. Then $P(x)=xy=x(374... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{34969}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=34969$.",
"robustness_analysis": "... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=187.0$. (Here the result is $\boxed{34969}$.) |
math-012279 | Inequalities: AM–GM — Equality Conditions | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=318$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=318$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{25281}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=25281$.",
"robustness_analysis": "... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=159.0$. (Here the result is $\boxed{25281}$.) |
math-012280 | Algebra: Extremal Values — Global Bounds | 7 | Solve with verification: Let $x,y>0$ satisfy $x+y=452$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=452-x$ with $x\\in(0,452)$. Then $P(x)=xy=x(452... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{51076}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=51076$.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=226.0$. |
math-012281 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Give an answer and a quick verification: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-3)}{x-(1)}=\frac{13}{9}.$$
Your final respo... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{13}{9}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{9}$), giving the unique solution $x=10$. The domain check $x\\neq 1$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=1$). (Here the result is $\boxed{x=10}$.) |
math-012282 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Solve and justify each step: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-12)}{x-(-1)}=\frac{21}{10}.$$
Y... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -1$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{21}{10}$), giving the unique solution $x=9$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). |
math-012283 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Checkpoint: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(10)}{x-(9)}=\frac{17}{16}.$$
Your final response ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 9$ so that $x-(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{17}{16}$), giving the unique solution $x=-7$. The domain check $x\\neq 9$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=9$). (Here the result is $\boxed{x=-7}$.) |
math-012284 | Algebra: Rational Equations — Verification by Substitution | 7 | Track units/moduli carefully: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-6)}{x-(-2)}=0.$$
Your final response must include (i) the solution set and (ii) ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - 0(x-b)=0$ with the domain restriction $x\\neq -2$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=0$), giving the unique solution $x=-6$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearing de... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-2$). (Here the result is $\boxed{x=-6}$.) |
math-012285 | Inequalities: Product Given Sum | 7 | Explain each transformation: Let $x,y>0$ satisfy $x+y=77$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g.,... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=77-x$ with $x\\in(0,77)$. Then $P(x)=xy=x(77-x)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{5929}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{5929}{4}$.",
"robustn... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=38.5$. (Here the result is $\boxed{\frac{5929}$.) |
math-012286 | Optimization: Two Variables — Concavity | 7 | Checkpoint: Let $x,y>0$ satisfy $x+y=76$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=76-x$ with $x\\in(0,76)$. Then $P(x)=xy=x(76-x)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1444}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=1444$.",
"robustness_analysis": "Robustness note: AM–GM gener... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=38.0$. |
math-012287 | Optimization: Two Variables — Concavity | 7 | Track quantifiers carefully: Let $x,y>0$ satisfy $x+y=403$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g.... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=403$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{162409}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{162409}{4}$.",
"rob... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=201.5$. |
math-012288 | Algebra: Rational Equations — Domain Restrictions | 7 | Where appropriate, name the theorem you use: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-6)}{x-(6)}=\frac{11}{23}.$$
Your... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{11}{23}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{11}{23}$), giving the unique solution $x=-17$. The domain check $x\\neq 6$ is satisfied here.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). |
math-012289 | Algebra: Extremal Values — Global Bounds | 7 | Challenge: Let $x,y>0$ satisfy $x+y=716$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=716-x$ with $x\\in(0,716)$. Then $P(x)=xy=x(716... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{128164}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=128164$.",
"robustness_analysis": "Sensitivity analysis: AM–GM ge... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=358.0$. |
math-012290 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Warm-up: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(4)}{x-(-14)}=\frac{4}{13}.$$
Your final response must include (i) the solution set and (ii)... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -14$ so that $x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{13}$), giving the unique solution $x=12$. The domain check $x\\neq -14$ is satisfied here.",
"robustness_analysis": "If the problem were p... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-14$). (Here the result is $\boxed{x=12}$.) |
math-012291 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Solve and justify each step: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(13)}=\frac{14}{3}.$$
Your final response must include (i... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 13$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{14}{3}$), giving the unique solution $x=19$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "Sensitivity anal... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). (Here the result is $\boxed{x=19}$.) |
math-012292 | Inequalities: Product Given Sum | 7 | Try to avoid pattern-matching; explain why: Let $x,y>0$ satisfy $x+y=349$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculu... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=349-x$ with $x\\in(0,349)$. Then $P(x)=xy=x(349... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{121801}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{121801}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=174.5$. |
math-012293 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Problem: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(9)}{x-(-15)}=\frac{1}{9}.$$
Your final response must include (i) the solution set and (ii) ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -15$ so that $x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{9}$), giving the unique solution $x=12$. The domain check $x\\neq -15$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-15$). (Here the result is $\boxed{x=12}$.) |
math-012294 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Give a theorem-based solution: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(7)}=\frac{-1}{2}.$$
Yo... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-1}{2}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-1}{2}$), giving the unique solution $x=-1$. The domain check $x\\neq 7$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=7$). (Here the result is $\boxed{x=-1}$.) |
math-012295 | Inequalities: Product Given Sum | 7 | Exercise: Let $x,y>0$ satisfy $x+y=668$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second ... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=668$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{111556}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=111556$.",
"robustness_analysis": "Generality note: AM–GM general... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=334.0$. (Here the result is $\boxed{111556}$.) |
math-012296 | Inequalities: AM–GM — Equality Conditions | 7 | Complete the analysis: Let $x,y>0$ satisfy $x+y=332$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., conc... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=332-x$ with $x\\in(0,332)$. Then $P(x)=xy=x(332... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{27556}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=27556$.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=166.0$. (Here the result is $\boxed{27556}$.) |
math-012297 | Optimization: Two Variables — Concavity | 7 | Exercise: Let $x,y>0$ satisfy $x+y=261$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second ... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=261-x$ with $x\\in(0,261)$. Then $P(x)=xy=x(261... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{68121}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{68121}{4}$.",
"robustness_ana... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=130.5$. |
math-012298 | Inequalities: Product Given Sum | 7 | Do not skip justification steps: Let $x,y>0$ satisfy $x+y=449$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=449$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{201601}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{201601}{4}$.",
"robustness_analysis": "If the... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=224.5$. |
math-012299 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Start by stating any domain restrictions: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(10)}=\frac{-15}{2}.$$
Your final response must inc... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 10$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=8}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-15}{2}$), giving the unique solution $x=8$. The domain check $x\\neq 10$ is satisfied here.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=10$). |
math-012300 | Inequalities: AM–GM — Equality Conditions | 7 | Derive the result step-by-step: Let $x,y>0$ satisfy $x+y=719$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=719-x$ with $x\\in(0,719)$. Then $P(x)=xy=x(719... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{516961}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=\\frac{516961}{4}$.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=359.5$. (Here the result is $\boxed{\frac{516961}$.) |
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