id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-012101 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Give an answer and a quick verification: 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:
$$\frac{x-(3)}{x-(9)}=\frac{17}{23}.$$
Your final response must include (i) the 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 $(x-a) - \\frac{17}{23}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-14}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{17}{23}$), giving the unique solution $x=-14$. The domain check $x\\neq 9$ 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 $x=9$). |
math-012102 | Algebra: Extremal Values — Global Bounds | 7 | Task: Let $x,y>0$ satisfy $x+y=208$.
(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": "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 $y=208-x$ with $x\\in(0,208)$. Then $P(x)=xy=x(208... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{10816}$.\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 $S^2/4=10816$.",
"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_... | 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 $x=y=104.0$. |
math-012103 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Give an answer and a quick verification: 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:
$$\frac{x-(11)}{x-(-11)}=-21.$$
Your final response must includ... | [
{
"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 $(x-a) - -21(x-b)=0$ with the domain restriction $x\\neq -11$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-21$), giving the unique solution $x=-10$. The domain check $x\\neq -11$ is satisfied here.",
"robustness_analysis": "If the problem were per... | [
{
"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 $\boxed{x=-10}$.) |
math-012104 | Inequalities: Product Given Sum | 7 | Proceed methodically: Let $x,y>0$ satisfy $x+y=550$.
(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": "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 $x+y=550$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{75625}$.\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 $S^2/4=75625$.",
"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 $x=y=275.0$. |
math-012105 | Optimization: Two Variables — Concavity | 7 | Give a fully justified solution: Let $x,y>0$ satisfy $x+y=407$.
(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 $y=407-x$ with $x\\in(0,407)$. Then $P(x)=xy=x(407... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{165649}$.\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 $S^2/4=\\frac{165649}{4}$.",
"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 $x=y=203.5$. |
math-012106 | Algebra: Rational Equations — Clearing Denominators | 7 | Be explicit about assumptions: 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:
$$\frac{x-(-9)}{x-(15)}=13.$$
Your final response must include (i) the so... | [
{
"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: $\\boxed{x=17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=13$), giving the unique solution $x=17$. The domain check $x\\neq 15$ is satisfied here.",
"robustness_analysis": "Sensitivit... | [
{
"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 $\boxed{x=17}$.) |
math-012107 | Inequalities: AM–GM — Equality Conditions | 7 | Be explicit about assumptions: Let $x,y>0$ satisfy $x+y=530$.
(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 $x+y=530$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{70225}$.\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 $S^2/4=70225$.",
"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 $x=y=265.0$. |
math-012108 | Inequalities: AM–GM — Equality Conditions | 7 | Indicate where a theorem is used: Let $x,y>0$ satisfy $x+y=680$.
(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 $x+y=680$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{115600}$.\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 $S^2/4=115600$.",
"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_... | 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 $x=y=340.0$. |
math-012109 | Algebra: Extremal Values — Global Bounds | 7 | Write the solution set clearly: Let $x,y>0$ satisfy $x+y=574$.
(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 $x+y=574$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{82369}$.\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 $S^2/4=82369$.",
"robustness_analysis": "Robustnes... | [
{
"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 $x=y=287.0$. |
math-012110 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Find the exact value: 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:
$$\frac{x-(6)}{x-(-5)}=\frac{16}{5}.$$
Your final response must include (i) the so... | [
{
"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 $x\\neq -5$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{16}{5}$), giving the unique solution $x=-10$. The domain check $x\\neq -5$ 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 $x=-5$). (Here the result is $\boxed{x=-10}$.) |
math-012111 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Solve and then verify: 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:
$$\frac{x-(8)}{x-(6)}=\frac{4}{3}.$$
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 $(x-a) - \\frac{4}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=0}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{3}$), giving the unique solution $x=0$. The domain check $x\\neq 6$ is satisfied here.",
"robustness_analysis": "Sen... | [
{
"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-012112 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Warm-up: 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:
$$\frac{x-(7)}{x-(2)}=\frac{1}{6}.$$
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 $(x-a) - \\frac{1}{6}(x-b)=0$ 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 $k=\\frac{1}{6}$), giving the unique solution $x=8$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_analys... | [
{
"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-012113 | Algebra: Rational Equations — Verification by Substitution | 7 | Answer with a short justification: 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:
$$\frac{x-(-9)}{x-(-1)}=\frac{19}{11}.$$
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 $x\\neq -1$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{11}$), giving the unique solution $x=10$. The domain check $x\\neq -1$ 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 $x=-1$). (Here the result is $\boxed{x=10}$.) |
math-012114 | Algebra: Rational Equations — Domain Restrictions | 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:
$$\frac{x-(-9)}{x-(15)}=\frac{-1}{2}.$$
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 $x\\neq 15$ 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 $k=\\frac{-1}{2}$), giving the unique solution $x=-1$. The domain check $x\\neq 15$ 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 $x=15$). (Here the result is $\boxed{x=-1}$.) |
math-012115 | Inequalities: AM–GM — Equality Conditions | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=703$.
(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 $x+y=703$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{494209}$.\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 $S^2/4=\\frac{494209}{4}$.",
"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_... | 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 $x=y=351.5$. (Here the result is $\boxed{\frac{494209}$.) |
math-012116 | Algebra: Rational Equations — Verification by Substitution | 7 | Track units/moduli carefully: 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:
$$\frac{x-(8)}{x-(-3)}=\frac{18}{7}.$$
Your final response ... | [
{
"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 $(x-a) - \\frac{18}{7}(x-b)=0$ 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 $k=\\frac{18}{7}$), giving the unique solution $x=-10$. The domain check $x\\neq -3$ 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 $x=-3$). (Here the result is $\boxed{x=-10}$.) |
math-012117 | Optimization: Two Variables — Concavity | 7 | Solve (and briefly cross-validate): Let $x,y>0$ satisfy $x+y=871$.
(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 theore... | [
{
"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 $y=871-x$ with $x\\in(0,871)$. Then $P(x)=xy=x(871... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{758641}$.\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 $S^2/4=\\frac{758641}{4}$.",
"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_... | 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 $x=y=435.5$. (Here the result is $\boxed{\frac{758641}$.) |
math-012118 | Optimization: Two Variables — Concavity | 7 | Warm-up: Let $x,y>0$ satisfy $x+y=776$.
(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 d... | [
{
"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 $x+y=776$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{150544}$.\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 $S^2/4=150544$.",
"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_... | 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 $x=y=388.0$. (Here the result is $\boxed{150544}$.) |
math-012119 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Do not skip justification steps: 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:
$$\frac{x-(10)}{x-(11)}=\frac{29}{30}.$$
Your final response must inclu... | [
{
"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": "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 $k=\\frac{29}{30}$), giving the unique solution $x=-19$. The domain check $x\\neq 11$ is satisfied here.",
"robustness_analysis": "If the proble... | [
{
"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 $x=11$). (Here the result is $\boxed{x=-19}$.) |
math-012120 | Inequalities: AM–GM — Equality Conditions | 7 | Provide both a computational and a conceptual explanation: Let $x,y>0$ satisfy $x+y=406$.
(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": "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 $y=406-x$ with $x\\in(0,406)$. Then $P(x)=xy=x(406... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{41209}$.\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 $S^2/4=41209$.",
"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_... | 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 $x=y=203.0$. (Here the result is $\boxed{41209}$.) |
math-012121 | Precalculus: Rational Expressions — Valid Cancellation | 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:
$$\frac{x-(10)}{x-(2)}=\frac{23}{15}.$$
Your final response must include (i) the solution set and ... | [
{
"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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{23}{15}$), giving the unique solution $x=-13$. The domain check $x\\neq 2$ 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'.",
... | 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 $\boxed{x=-13}$.) |
math-012122 | Algebra: Equations — Conditions for Valid Multiplication | 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:
$$\frac{x-(5)}{x-(-4)}=\frac{-2}{7}.$$
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 $(x-a) - \\frac{-2}{7}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=3}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-2}{7}$), giving the unique solution $x=3$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_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 $x=-4$). (Here the result is $\boxed{x=3}$.) |
math-012123 | Algebra: Rational Equations — Verification by Substitution | 7 | Solve and justify each 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:
$$\frac{x-(8)}{x-(15)}=\frac{5}{12}.$$
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 $(x-a) - \\frac{5}{12}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=3}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{12}$), giving the unique solution $x=3$. The domain check $x\\neq 15$ is satisfied here.",
"robustness_analysis": "Generality 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'.",
... | 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 $x=15$). |
math-012124 | Inequalities: AM–GM — Equality Conditions | 7 | Explain what is being counted/optimized: Let $x,y>0$ satisfy $x+y=621$.
(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 $x+y=621$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{385641}$.\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 $S^2/4=\\frac{385641}{4}$.",
"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_... | 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 $x=y=310.5$. |
math-012125 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Explain why your operations are valid: 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:
$$\frac{x-(14)}{x-(5)}=\frac{32}{23}.$$
Your final response must incl... | [
{
"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 $x\\neq 5$ so that $x-(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{32}{23}$), giving the unique solution $x=-18$. The domain check $x\\neq 5$ 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'.",
... | 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 $x=5$). (Here the result is $\boxed{x=-18}$.) |
math-012126 | Inequalities: AM–GM — Equality Conditions | 7 | Give a theorem-based solution: Let $x,y>0$ satisfy $x+y=585$.
(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 $y=585-x$ with $x\\in(0,585)$. Then $P(x)=xy=x(585... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{342225}$.\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 $S^2/4=\\frac{342225}{4}$.",
"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_... | 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 $x=y=292.5$. (Here the result is $\boxed{\frac{342225}$.) |
math-012127 | Algebra: Equations — Conditions for Valid Multiplication | 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:
$$\frac{x-(13)}{x-(-13)}=\frac{-9}{17}.$$
Your final respo... | [
{
"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": "Both approaches agree after simplification. Final answer: $\\boxed{x=4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-9}{17}$), giving the unique solution $x=4$. The domain check $x\\neq -13$ is satisfied here.",
"robustness_analysis": "Robustness 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 $x=-13$). |
math-012128 | Algebra: Rational Equations — Clearing Denominators | 7 | Write the solution set clearly: 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:
$$\frac{x-(4)}{x-(-10)}=8.$$
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": "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 $k=8$), giving the unique solution $x=-12$. The domain check $x\\neq -10$ is satisfied here.",
"robustness_analysis": "Sensitiv... | [
{
"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 $x=-10$). |
math-012129 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Answer using clear logical 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:
$$\frac{x-(-8)}{x-(-12)}=0.$$
Your final response must include (i) the soluti... | [
{
"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 $x\\neq -12$.",
... | {
"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 $k=0$), giving the unique solution $x=-8$. The domain check $x\\neq -12$ is satisfied here.",
"robustness_analysis": "If the problem were perturbed: 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'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-12$). |
math-012130 | Inequalities: AM–GM — Equality Conditions | 7 | Use two approaches if possible: Let $x,y>0$ satisfy $x+y=88$.
(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 $y=88-x$ with $x\\in(0,88)$. Then $P(x)=xy=x(88-x)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1936}$.\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 $S^2/4=1936$.",
"robustness_analysis": "Sensitivity analysis: AM–GM ... | [
{
"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 $x=y=44.0$. |
math-012131 | Inequalities: Product Given Sum | 7 | Find the exact value: Let $x,y>0$ satisfy $x+y=72$.
(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., concav... | [
{
"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 $y=72-x$ with $x\\in(0,72)$. Then $P(x)=xy=x(72-x)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1296}$.\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 $S^2/4=1296$.",
"robustness_analysis": "If the problem were perturbe... | [
{
"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 $x=y=36.0$. |
math-012132 | Algebra: Rational Equations — Verification by Substitution | 7 | Provide a rigorous 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:
$$\frac{x-(-9)}{x-(4)}=2.$$
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 $(x-a) - 2(x-b)=0$ with the domain restriction $x\\neq 4$.",
... | {
"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 $k=2$), giving the unique solution $x=17$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_analysis": "Sen... | [
{
"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 $x=4$). (Here the result is $\boxed{x=17}$.) |
math-012133 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Do not skip justification steps: 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:
$$\frac{x-(6)}{x-(-5)}=\frac{7}{18}.$$
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 $x\\neq -5$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{18}$), giving the unique solution $x=13$. The domain check $x\\neq -5$ 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 $x=-5$). |
math-012134 | Inequalities: Product Given Sum | 7 | Explain why your operations are valid: Let $x,y>0$ satisfy $x+y=203$.
(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 the... | [
{
"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 $x+y=203$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{41209}$.\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 $S^2/4=\\frac{41209}{4}$.",
"robustness_analysis": "Sensitiv... | [
{
"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 $x=y=101.5$. (Here the result is $\boxed{\frac{41209}$.) |
math-012135 | Algebra: Rational Equations — Domain Restrictions | 7 | Solve with 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:
$$\frac{x-(-8)}{x-(-7)}=\frac{25}{24}.$$
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 $(x-a) - \\frac{25}{24}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{25}{24}$), giving the unique solution $x=17$. 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 $x=-7$). |
math-012136 | Algebra: Rational Equations — Verification by Substitution | 7 | Do not skip justification steps: 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:
$$\frac{x-(-1)}{x-(-8)}=\frac{9}{16}.$$
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 $(x-a) - \\frac{9}{16}(x-b)=0$ 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 $k=\\frac{9}{16}$), giving the unique solution $x=8$. The domain check $x\\neq -8$ is satisfied here.",
"robustness_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'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-8$). |
math-012137 | Optimization: Two Variables — Concavity | 7 | Derive the result step-by-step: Let $x,y>0$ satisfy $x+y=883$.
(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 $y=883-x$ with $x\\in(0,883)$. Then $P(x)=xy=x(883... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{779689}$.\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 $S^2/4=\\frac{779689}{4}$.",
"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_... | 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 $x=y=441.5$. |
math-012138 | Optimization: Two Variables — Concavity | 7 | Track units/moduli carefully: Let $x,y>0$ satisfy $x+y=526$.
(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 $x+y=526$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{69169}$.\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 $S^2/4=69169$.",
"robustness_analysis": "Generalit... | [
{
"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 $x=y=263.0$. |
math-012139 | Inequalities: AM–GM — Equality Conditions | 7 | Give a theorem-based solution: Let $x,y>0$ satisfy $x+y=303$.
(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 $x+y=303$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{91809}$.\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 $S^2/4=\\frac{91809}{4}$.",
"robustness_analysis": "If the problem... | [
{
"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 $x=y=151.5$. |
math-012140 | Inequalities: Product Given Sum | 7 | Explain each transformation: Let $x,y>0$ satisfy $x+y=287$.
(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 $x+y=287$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{82369}$.\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 $S^2/4=\\frac{82369}{4}$.",
"robustness_analysis": "Sensitivity an... | [
{
"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 $x=y=143.5$. (Here the result is $\boxed{\frac{82369}$.) |
math-012141 | Precalculus: Rational Expressions — Valid Cancellation | 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:
$$\frac{x-(3)}{x-(4)}=\frac{15}{16}.$$
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 $(x-a) - \\frac{15}{16}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{15}{16}$), giving the unique solution $x=-12$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_analysis": "Generality 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'.",
... | 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 $x=4$). (Here the result is $\boxed{x=-12}$.) |
math-012142 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Complete the analysis: 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:
$$\frac{x-(-4)}{x-(-14)}=\frac{2}{3}.$$
Your final response must include (i) the solution set and (... | [
{
"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 $(x-a) - \\frac{2}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{3}$), giving the unique solution $x=16$. 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'.",
... | 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 $\boxed{x=16}$.) |
math-012143 | Inequalities: AM–GM — Equality Conditions | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=321$.
(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 $x+y=321$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{103041}$.\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 $S^2/4=\\frac{103041}{4}$.",
"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 $x=y=160.5$. (Here the result is $\boxed{\frac{103041}$.) |
math-012144 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Track units/moduli carefully: 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:
$$\frac{x-(1)}{x-(2)}=\frac{5}{6}.$$
Your final response must include (i) the sol... | [
{
"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 $(x-a) - \\frac{5}{6}(x-b)=0$ with the domain restriction $x\\neq ... | {
"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 $k=\\frac{5}{6}$), giving the unique solution $x=-4$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_analysis": "If the problem were pert... | [
{
"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 $x=2$). (Here the result is $\boxed{x=-4}$.) |
math-012145 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Use two approaches if possible: 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:
$$\frac{x-(-11)}{x-(0)}=\frac{16}{5}.$$
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 $x\\neq 0$ so that $x-(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{16}{5}$), giving the unique solution $x=5$. The domain check $x\\neq 0$ 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'.",
... | 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 $x=0$). (Here the result is $\boxed{x=5}$.) |
math-012146 | 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:
$$\frac{x-(-8)}{x-(7)}=\frac{8}{23}.$$
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 $x\\neq 7$ so that $x-(... | {
"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 $k=\\frac{8}{23}$), giving the unique solution $x=-16$. The domain check $x\\neq 7$ 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'.",
... | 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 $x=7$). (Here the result is $\boxed{x=-16}$.) |
math-012147 | Algebra: Rational Equations — Domain Restrictions | 7 | Answer with a short justification: 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-(11)}{x-(-9)}=\frac{-13}{7}.... | [
{
"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}{7}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-2}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-13}{7}$), giving the unique solution $x=-2$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": "Generality 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'.",
... | 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$). |
math-012148 | Optimization: Two Variables — Concavity | 7 | Answer using clear logical steps: Let $x,y>0$ satisfy $x+y=734$.
(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=734$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{134689}$.\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=134689$.",
"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_... | 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=367.0$. |
math-012149 | Algebra: Rational Equations — Domain Restrictions | 7 | Checkpoint: 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-(1)}{x-(15)}=\frac{4}{11}.$$
Your final response must include (i) the solution set... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{4}{11}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{11}$), giving the unique solution $x=-7$. The domain check $x\\neq 15$ is satisfied here.",
"robustness_analysis": "Generality note: Clear... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=15$). |
math-012150 | Algebra: Rational Equations — Domain Restrictions | 7 | Question: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-1)}{x-(-6)}=0.$$
Your final response must include (i) the solution set and (ii) a brief check for ex... | [
{
"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 -6$.",
... | {
"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=0$), giving the unique solution $x=-1$. The domain check $x\\neq -6$ 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'.",
... | 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-012151 | Inequalities: AM–GM — Equality Conditions | 7 | Indicate where a theorem is used: Let $x,y>0$ satisfy $x+y=768$.
(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=768-x$ with $x\\in(0,768)$. Then $P(x)=xy=x(768... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{147456}$.\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=147456$.",
"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_... | 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=384.0$. (Here the result is $\boxed{147456}$.) |
math-012152 | Optimization: Two Variables — Concavity | 7 | Solve and then verify: Let $x,y>0$ satisfy $x+y=243$.
(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=243$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{59049}$.\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{59049}{4}$.",
"robus... | [
{
"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=121.5$. |
math-012153 | Optimization: Two Variables — Concavity | 7 | Exercise: Let $x,y>0$ satisfy $x+y=147$.
(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=147-x$ with $x\\in(0,147)$. Then $P(x)=xy=x(147... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{21609}$.\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{21609}{4}$.",
"robustness_analysis": "If the problem... | [
{
"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=73.5$. (Here the result is $\boxed{\frac{21609}$.) |
math-012154 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Solve (and briefly cross-validate): 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-(-5)}=\frac{1}{5}.$$
Your final response mu... | [
{
"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 -5$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{5}$), giving the unique solution $x=10$. The domain check $x\\neq -5$ 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=-5$). (Here the result is $\boxed{x=10}$.) |
math-012155 | Algebra: Extremal Values — Global Bounds | 7 | Answer using clear logical steps: Let $x,y>0$ satisfy $x+y=280$.
(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=280-x$ with $x\\in(0,280)$. Then $P(x)=xy=x(280... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{19600}$.\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=19600$.",
"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=140.0$. |
math-012156 | Algebra: Extremal Values — Global Bounds | 7 | Track units/moduli carefully: Let $x,y>0$ satisfy $x+y=515$.
(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=515-x$ with $x\\in(0,515)$. Then $P(x)=xy=x(515... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{265225}$.\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{265225}{4}$.",
"robustness_analysis": "Genera... | [
{
"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=257.5$. (Here the result is $\boxed{\frac{265225}$.) |
math-012157 | Inequalities: AM–GM — Equality Conditions | 7 | Do not skip justification steps: Let $x,y>0$ satisfy $x+y=194$.
(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=194$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9409}$.\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=9409$.",
"robustness_analysis": "Ro... | [
{
"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=97.0$. |
math-012158 | Inequalities: Product Given Sum | 7 | Use two approaches if possible: Let $x,y>0$ satisfy $x+y=639$.
(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=639$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{408321}$.\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{408321}{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_... | 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=319.5$. (Here the result is $\boxed{\frac{408321}$.) |
math-012159 | Algebra: Rational Equations — Clearing Denominators | 7 | Carefully track domains: 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-(7)}=-13.$$
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) - -13(x-b)=0$ with the domain restriction $x\\neq 7$.",
... | {
"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=-13$), giving the unique solution $x=6$. The domain check $x\\neq 7$ is satisfied here.",
"robustness_analysis": "If the problem were perturbed... | [
{
"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=6}$.) |
math-012160 | Inequalities: Product Given Sum | 7 | Indicate where a theorem is used: Let $x,y>0$ satisfy $x+y=861$.
(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=861-x$ with $x\\in(0,861)$. Then $P(x)=xy=x(861... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{741321}$.\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{741321}{4}$.",
"robustness_analysis": "Robustness 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_... | 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=430.5$. |
math-012161 | Optimization: Two Variables — Concavity | 7 | Checkpoint: Let $x,y>0$ satisfy $x+y=289$.
(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": "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=289$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{83521}$.\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{83521}{4}$.",
"robustness_analysis": "If the p... | [
{
"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=144.5$. (Here the result is $\boxed{\frac{83521}$.) |
math-012162 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Exercise: 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-(12)}=\frac{9}{11}.$$
Your final response must include (i) the solution set and (ii) a brief c... | [
{
"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}{11}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{11}$), giving the unique solution $x=1$. The domain check $x\\neq 12$ 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=12$). (Here the result is $\boxed{x=1}$.) |
math-012163 | Inequalities: Product Given Sum | 7 | Write the solution set clearly: Let $x,y>0$ satisfy $x+y=554$.
(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=554-x$ with $x\\in(0,554)$. Then $P(x)=xy=x(554... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{76729}$.\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=76729$.",
"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=277.0$. (Here the result is $\boxed{76729}$.) |
math-012164 | Algebra: Rational Equations — Clearing Denominators | 7 | Challenge: 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-(7)}=\frac{11}{24}.$$
Your final response must include (i) t... | [
{
"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}{24}(x-b)=0$ with the domain restriction $x\\ne... | {
"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{11}{24}$), giving the unique solution $x=-17$. The domain check $x\\neq 7$ 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'.",
... | 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$). |
math-012165 | Inequalities: Product Given Sum | 7 | Explain what is being counted/optimized: Let $x,y>0$ satisfy $x+y=29$.
(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 th... | [
{
"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=29$ to get $\\sqrt{xy}\\le \\frac{2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{841}$.\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{841}{4}$.",
"robustness_analysi... | [
{
"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=14.5$. (Here the result is $\boxed{\frac{841}$.) |
math-012166 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Solve with 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-(-15)}=\frac{2}{5}.$$
Your final response must include... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -15$ so that $x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{5}$), giving the unique solution $x=15$. The domain check $x\\neq -15$ is satisfied here.",
"robustness_analysis": "Robustness note:... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-15$). (Here the result is $\boxed{x=15}$.) |
math-012167 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Solve and then verify: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-11)}{x-(-1)}=\frac{9}{19}.$$
Your final response must include (i) the soluti... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -1$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{19}$), giving the unique solution $x=-20$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_analysis"... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). |
math-012168 | Algebra: Rational Equations — Clearing Denominators | 7 | Complete the analysis: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(-2)}=\frac{19}{16}.$$
Your final response must 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{19}{16}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{16}$), giving the unique solution $x=-18$. The domain check $x\\neq -2$ 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=-2$). |
math-012169 | Inequalities: Product Given Sum | 7 | Do not skip justification steps: Let $x,y>0$ satisfy $x+y=31$.
(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=31-x$ with $x\\in(0,31)$. Then $P(x)=xy=x(31-x)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{961}$.\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{961}{4}$.",
"robustness_analysi... | [
{
"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=15.5$. |
math-012170 | Inequalities: Product Given Sum | 7 | Complete the analysis: Let $x,y>0$ satisfy $x+y=109$.
(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=109$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{11881}$.\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{11881}{4}$.",
"robustness_analysis": "Generali... | [
{
"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=54.5$. (Here the result is $\boxed{\frac{11881}$.) |
math-012171 | Inequalities: Product Given Sum | 7 | Warm-up: Let $x,y>0$ satisfy $x+y=460$.
(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 d... | [
{
"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=460-x$ with $x\\in(0,460)$. Then $P(x)=xy=x(460... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{52900}$.\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=52900$.",
"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=230.0$. (Here the result is $\boxed{52900}$.) |
math-012172 | Inequalities: AM–GM — Equality Conditions | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=603$.
(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=603-x$ with $x\\in(0,603)$. Then $P(x)=xy=x(603... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{363609}$.\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{363609}{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=301.5$. |
math-012173 | Algebra: Rational Equations — Clearing Denominators | 7 | Answer with a short justification: 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-(-10)}{x-(13)}=\frac{-21}{2}.$$
Your final response must include (... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-21}{2}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-21}{2}$), giving the unique solution $x=11$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_a... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). |
math-012174 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Provide a rigorous solution: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(13)}{x-(-13)}=\frac{-21}{5}.$$
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 -13$ 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{-21}{5}$), giving the unique solution $x=-8$. The domain check $x\\neq -13$ 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=-13$). |
math-012175 | Algebra: Extremal Values — Global Bounds | 7 | Do not skip justification steps: Let $x,y>0$ satisfy $x+y=183$.
(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=183-x$ with $x\\in(0,183)$. Then $P(x)=xy=x(183... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{33489}$.\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{33489}{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_... | 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=91.5$. (Here the result is $\boxed{\frac{33489}$.) |
math-012176 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Proceed methodically: 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-(12)}=\frac{-6}{7}.$$
Your final response must include (i) the solution set and (... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 12$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-2}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-6}{7}$), giving the unique solution $x=-2$. The domain check $x\\neq 12$ is satisfied here.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=12$). (Here the result is $\boxed{x=-2}$.) |
math-012177 | Algebra: Rational Equations — Verification by Substitution | 7 | Write the solution set clearly: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(11)}{x-(-1)}=\frac{-5}{7}.$$
Your final response must include (i) th... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-5}{7}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-5}{7}$), giving the unique solution $x=6$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_anal... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). (Here the result is $\boxed{x=6}$.) |
math-012178 | Algebra: Rational Equations — Verification by Substitution | 7 | Complete the analysis: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(2)}{x-(-14)}=\frac{1}{5}.$$
Your final response must include (i) the solut... | [
{
"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=6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{5}$), giving the unique solution $x=6$. The domain check $x\\neq -14$ is satisfied here.",
"robustness_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'.",
... | 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=6}$.) |
math-012179 | Algebra: Extremal Values — Global Bounds | 7 | Question: Let $x,y>0$ satisfy $x+y=322$.
(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=322$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{25921}$.\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=25921$.",
"robustness_analysis": "Sensitivity analysis: AM–GM gene... | [
{
"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=161.0$. |
math-012180 | Algebra: Rational Equations — Verification by Substitution | 7 | Complete the analysis: 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-(13)}=\frac{-8}{13}.$$
Your final response must include (i) the sol... | [
{
"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=0}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-8}{13}$), giving the unique solution $x=0$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "If the problem we... | [
{
"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-012181 | Inequalities: Product Given Sum | 7 | Complete the analysis: Let $x,y>0$ satisfy $x+y=367$.
(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=367$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{134689}$.\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{134689}{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=183.5$. (Here the result is $\boxed{\frac{134689}$.) |
math-012182 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Determine the requested 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-(-7)}{x-(3)}=\frac{-3}{2}.$$
Your final response... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-3}{2}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-3}{2}$), giving the unique solution $x=-1$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). (Here the result is $\boxed{x=-1}$.) |
math-012183 | Algebra: Rational Equations — Clearing Denominators | 7 | Prompt: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(-4)}=\frac{4}{3}.$$
Your final response must include (i) the solution set and (ii) a... | [
{
"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{4}{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=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{3}$), giving the unique solution $x=-1$. 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-012184 | Optimization: Two Variables — Concavity | 7 | Show all reasoning: Let $x,y>0$ satisfy $x+y=699$.
(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=699-x$ with $x\\in(0,699)$. Then $P(x)=xy=x(699... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{488601}$.\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{488601}{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=349.5$. (Here the result is $\boxed{\frac{488601}$.) |
math-012185 | Algebra: Rational Equations — Domain Restrictions | 7 | Complete the analysis: 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-(11)}{x-(5)}=\frac{9}{7}.$$
Your final response must include (i) the soluti... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 5$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{7}$), giving the unique solution $x=-16$. The domain check $x\\neq 5$ is satisfied here.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=5$). |
math-012186 | Algebra: Rational Equations — Verification by Substitution | 7 | Prompt: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(12)}{x-(6)}=\frac{31}{25}.$$
Your final response must include (i) the solutio... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 6$ so that $x-(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{31}{25}$), giving the unique solution $x=-19$. The domain check $x\\neq 6$ is satisfied here.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). (Here the result is $\boxed{x=-19}$.) |
math-012187 | Algebra: Rational Equations — Verification by Substitution | 7 | Find the exact value: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(11)}{x-(2)}=\frac{31}{22}.$$
Your final response must include (i) the s... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 2$ so that $x-(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{31}{22}$), giving the unique solution $x=-20$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=2$). (Here the result is $\boxed{x=-20}$.) |
math-012188 | Algebra: Rational Equations — Clearing Denominators | 7 | Answer using clear logical steps: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-3)}{x-(-2)}=\frac{3}{4}.$$
Your final response must include (i... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{3}{4}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{4}$), giving the unique solution $x=-6$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-2$). (Here the result is $\boxed{x=-6}$.) |
math-012189 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Explain what is being counted/optimized: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(2)}{x-(4)}=\frac{9}{8}.$$
Your final response must include ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{9}{8}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{8}$), giving the unique solution $x=20$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_anal... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=4$). |
math-012190 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Work carefully and justify each inference: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(8)}{x-(-9)}=\frac{9}{26}.$$
Your final response must i... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -9$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{26}$), giving the unique solution $x=17$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": "Robustness note:... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). |
math-012191 | Algebra: Rational Equations — Domain Restrictions | 7 | Explain why your operations are valid: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(13)}{x-(-1)}=\frac{-2}{5}.$$
Your final response must ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-2}{5}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-2}{5}$), giving the unique solution $x=9$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). (Here the result is $\boxed{x=9}$.) |
math-012192 | Algebra: Rational Equations — Verification by Substitution | 7 | Warm-up: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(-12)}=\frac{13}{16}.$$
Your final response must include (i) the solution set and (i... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{13}{16}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{16}$), giving the unique solution $x=4$. The domain check $x\\neq -12$ is satisfied here.",
"robustness_analysis": "Robustness note: Clear... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-12$). (Here the result is $\boxed{x=4}$.) |
math-012193 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Determine the requested value: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(13)}{x-(-12)}=\frac{-2}{3}.$$
Your final response must include (i) th... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -12$ so that $x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=3}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-2}{3}$), giving the unique solution $x=3$. The domain check $x\\neq -12$ is satisfied here.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-12$). (Here the result is $\boxed{x=3}$.) |
math-012194 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Do not skip justification steps: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(13)}=\frac{5}{23}.$$
Your final response must include (i) the solution... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 13$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{23}$), giving the unique solution $x=-10$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis"... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). |
math-012195 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Where appropriate, name the theorem you use: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-13)}{x-(-7)}=\frac{7}{4}.$$
Your final response must include (i) ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -7$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{4}$), giving the unique solution $x=1$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analy... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). |
math-012196 | Algebra: Rational Equations — Domain Restrictions | 7 | Write the solution set clearly: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(15)}{x-(12)}=\frac{8}{7}.$$
Your final response must include (i) the solution s... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{8}{7}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{8}{7}$), giving the unique solution $x=-9$. The domain check $x\\neq 12$ is satisfied here.",
"robustness_analysis": "Sensitivity analysis: C... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=12$). (Here the result is $\boxed{x=-9}$.) |
math-012197 | Algebra: Rational Equations — Domain Restrictions | 7 | Challenge: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(11)}{x-(-13)}=-3.$$
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) - -3(x-b)=0$ with the domain restriction $x\\neq -13$.",
... | {
"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=-3$), giving the unique solution $x=-7$. The domain check $x\\neq -13$ is satisfied here.",
"robustness_analysis": "Generalit... | [
{
"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-012198 | Algebra: Extremal Values — Global Bounds | 7 | State any required conditions first: Let $x,y>0$ satisfy $x+y=56$.
(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 theore... | [
{
"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=56$ to get $\\sqrt{xy}\\le \\frac{5... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{784}$.\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=784$.",
"robustness_analysis": "Robu... | [
{
"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=28.0$. (Here the result is $\boxed{784}$.) |
math-012199 | Algebra: Rational Equations — Verification by Substitution | 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-(-11)}{x-(-5)}=\frac{14}{11}.$$
Your final response must includ... | [
{
"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 $(x-a) - \\frac{14}{11}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{14}{11}$), giving the unique solution $x=17$. The domain check $x\\neq -5$ 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'.",
... | 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-012200 | Optimization: Two Variables — Concavity | 7 | Give reasoning, not just computation: Let $x,y>0$ satisfy $x+y=711$.
(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 $y=711-x$ with $x\\in(0,711)$. Then $P(x)=xy=x(711... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{505521}$.\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 $S^2/4=\\frac{505521}{4}$.",
"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_... | 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 $x=y=355.5$. |
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