id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-012001 | Optimization: Two Variables — Concavity | 7 | Give a fully justified solution: Let $x,y>0$ satisfy $x+y=324$.
(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 $y=324-x$ with $x\\in(0,324)$. Then $P(x)=xy=x(324... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{26244}$.\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 $S^2/4=26244$.",
"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 $x=y=162.0$. (Here the result is $\boxed{26244}$.) |
math-012002 | Algebra: Rational Equations — Clearing Denominators | 7 | Compute the requested quantity: 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:
$$\frac{x-(10)}{x-(-1)}=\frac{13}{2}.$$
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 $x\\neq -1$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-3}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{2}$), giving the unique solution $x=-3$. 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 $x=-1$). (Here the result is $\boxed{x=-3}$.) |
math-012003 | Algebra: Rational Equations — Linear-Fractional Forms | 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:
$$\frac{x-(12)}{x-(-6)}=4.$$
Your final response must include (i) the solution set and (ii) a brief check for ex... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -6$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=4$), giving the unique solution $x=-12$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_analysis": "Generality note: Clearing denomi... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). (Here the result is $\boxed{x=-12}$.) |
math-012004 | Inequalities: AM–GM — Equality Conditions | 7 | Start by stating any domain restrictions: Let $x,y>0$ satisfy $x+y=583$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus ... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=583$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{339889}$.\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 $S^2/4=\\frac{339889}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=291.5$. (Here the result is $\boxed{\frac{339889}$.) |
math-012005 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Task: 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:
$$\frac{x-(2)}{x-(9)}=\frac{12}{19}.$$
Your final response must include (i) the solution set and (ii) a b... | [
{
"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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{12}{19}$), giving the unique solution $x=-10$. 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'.",
... | 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-012006 | Algebra: Extremal Values — Global Bounds | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=809$.
(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 $y=809-x$ with $x\\in(0,809)$. Then $P(x)=xy=x(809... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{654481}$.\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 $S^2/4=\\frac{654481}{4}$.",
"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 $x=y=404.5$. |
math-012007 | Inequalities: AM–GM — Equality Conditions | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=464$.
(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 $y=464-x$ with $x\\in(0,464)$. Then $P(x)=xy=x(464... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{53824}$.\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 $S^2/4=53824$.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=232.0$. (Here the result is $\boxed{53824}$.) |
math-012008 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Exercise: 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:
$$\frac{x-(-14)}{x-(15)}=\frac{-15}{14}.$$
Your final response must include (i) the solution s... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 15$ so that $x-... | {
"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 $k=\\frac{-15}{14}$), giving the unique solution $x=1$. 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'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=15$). |
math-012009 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Track quantifiers carefully: 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:
$$\frac{x-(10)}{x-(-15)}=\frac{-22}{3}.$$
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: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-22}{3}$), giving the unique solution $x=-12$. The domain check $x\\neq -15$ is satisfied here.",
"robustness_analysis": "Robustness n... | [
{
"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 $\boxed{x=-12}$.) |
math-012010 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Give a fully justified solution: 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:
$$\frac{x-(-4)}{x-(4)}=\frac{13}{5}.$$
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 $(x-a) - \\frac{13}{5}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{5}$), giving the unique solution $x=9$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_analysis": "Robustness note: Cl... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=4$). |
math-012011 | Algebra: Equations — Conditions for Valid Multiplication | 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:
$$\frac{x-(13)}{x-(14)}=\frac{21}{22}.$$
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 $x\\neq 14$ so that $x-... | {
"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 $k=\\frac{21}{22}$), giving the unique solution $x=-8$. The domain check $x\\neq 14$ 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'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=14$). |
math-012012 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Try to avoid pattern-matching; explain why: 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:
$$\frac{x-(-8)}{x-(-2)}=\frac{3}{5}.$$
Your final response m... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{3}{5}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "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 $k=\\frac{3}{5}$), giving the unique solution $x=-17$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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 $x=-2$). |
math-012013 | Algebra: Rational Equations — Clearing Denominators | 7 | Solve with verification: 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:
$$\frac{x-(10)}{x-(4)}=-5.$$
Your final response must include (i) the solution set ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 4$ so that $x-(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-5$), giving the unique solution $x=5$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_analysis": "Generality no... | [
{
"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 $x=4$). (Here the result is $\boxed{x=5}$.) |
math-012014 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Where appropriate, name the theorem you use: 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:
$$\frac{x-(-9)}{x-(2)}=\frac{24}{13}.$$
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 $(x-a) - \\frac{24}{13}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{24}{13}$), giving the unique solution $x=15$. 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'.",
... | 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-012015 | Optimization: Two Variables — Concavity | 7 | Track quantifiers carefully: Let $x,y>0$ satisfy $x+y=870$.
(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 $x+y=870$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{189225}$.\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 $S^2/4=189225$.",
"robustness_analysis": "Sensitivity analysis: AM... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=435.0$. (Here the result is $\boxed{189225}$.) |
math-012016 | Algebra: Rational Equations — Domain Restrictions | 7 | Derive the result step-by-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:
$$\frac{x-(14)}{x-(-14)}=\frac{-17}{11}.$$
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 $x\\neq -14$ so that $x... | {
"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 $k=\\frac{-17}{11}$), giving the unique solution $x=-3$. The domain check $x\\neq -14$ is satisfied here.",
"robustness... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-14$). (Here the result is $\boxed{x=-3}$.) |
math-012017 | Algebra: Rational Equations — Clearing Denominators | 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:
$$\frac{x-(-15)}{x-(13)}=\frac{17}{3}.$$
Your final response must include (i) the... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 13$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{17}{3}$), giving the unique solution $x=19$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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-012018 | Algebra: Rational Equations — Clearing Denominators | 7 | Work carefully and justify each inference: 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:
$$\frac{x-(-11)}{x-(8)}=\frac{... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{3}{22}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-14}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{22}$), giving the unique solution $x=-14$. The domain check $x\\neq 8$ is satisfied here.",
"robustness_a... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=8$). (Here the result is $\boxed{x=-14}$.) |
math-012019 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Write the solution set clearly: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(7)}=\frac{10}{3}.$$
Your final response must include (i) the... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 7$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{10}{3}$), giving the unique solution $x=13$. The domain check $x\\neq 7$ is satisfied here.",
"robustness_analysis": "Robustness note: Cleari... | [
{
"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=13}$.) |
math-012020 | Inequalities: Product Given Sum | 7 | Determine the requested value: Let $x,y>0$ satisfy $x+y=517$.
(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=517$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{267289}$.\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{267289}{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=258.5$. (Here the result is $\boxed{\frac{267289}$.) |
math-012021 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Task: 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-(-9)}=-2.$$
Your final response must include (i) the solution set and (ii) a brie... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -9$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-2$), giving the unique solution $x=-4$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": "Generality note: Clearing d... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). |
math-012022 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Give reasoning, not just computation: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-3)}{x-(1)}=\frac{7}{5}.$$
Your final re... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{7}{5}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{5}$), giving the unique solution $x=11$. 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=11}$.) |
math-012023 | Inequalities: AM–GM — Equality Conditions | 7 | Indicate where a theorem is used: Let $x,y>0$ satisfy $x+y=478$.
(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=478$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{57121}$.\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=57121$.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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=239.0$. (Here the result is $\boxed{57121}$.) |
math-012024 | Algebra: Extremal Values — Global Bounds | 7 | Where appropriate, name the theorem you use: Let $x,y>0$ satisfy $x+y=40$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculu... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=40$ to get $\\sqrt{xy}\\le \\frac{4... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{400}$.\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=400$.",
"robustness_analysis": "Sensitivity analysis: AM–GM 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_... | 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=20.0$. |
math-012025 | Algebra: Rational Equations — Verification by Substitution | 7 | Solve and sanity-check: 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{18}{17}.$$
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 4$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{18}{17}$), giving the unique solution $x=-13$. The domain check $x\\neq 4$ 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'.",
... | 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$). (Here the result is $\boxed{x=-13}$.) |
math-012026 | Optimization: Two Variables — Concavity | 7 | Find the exact value: Let $x,y>0$ satisfy $x+y=724$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., conca... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=724-x$ with $x\\in(0,724)$. Then $P(x)=xy=x(724... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{131044}$.\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=131044$.",
"robustness_analysis": "Robustn... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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=362.0$. (Here the result is $\boxed{131044}$.) |
math-012027 | Optimization: Two Variables — Concavity | 7 | Challenge: Let $x,y>0$ satisfy $x+y=729$.
(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=729-x$ with $x\\in(0,729)$. Then $P(x)=xy=x(729... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{531441}$.\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{531441}{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=364.5$. |
math-012028 | Inequalities: AM–GM — Equality Conditions | 7 | Carefully track domains: Let $x,y>0$ satisfy $x+y=334$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "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=334$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{27889}$.\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=27889$.",
"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_... | 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=167.0$. (Here the result is $\boxed{27889}$.) |
math-012029 | Algebra: Extremal Values — Global Bounds | 7 | Proceed methodically: Let $x,y>0$ satisfy $x+y=598$.
(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=598$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{89401}$.\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=89401$.",
"robustness_analysis": "If the problem were perturbed: A... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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=299.0$. (Here the result is $\boxed{89401}$.) |
math-012030 | Inequalities: Product Given Sum | 7 | Determine the requested value: Let $x,y>0$ satisfy $x+y=267$.
(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=267$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{71289}$.\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{71289}{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_... | 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=133.5$. (Here the result is $\boxed{\frac{71289}$.) |
math-012031 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Keep the final answer in boxed form: 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-(-7)}{x-(-4)}=\frac{22}{19}.$$
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{22}{19}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{22}{19}$), giving the unique solution $x=15$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis": "Robustness note: Clea... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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=15}$.) |
math-012032 | Algebra: Extremal Values — Global Bounds | 7 | Find the exact value: Let $x,y>0$ satisfy $x+y=443$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., conca... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=443-x$ with $x\\in(0,443)$. Then $P(x)=xy=x(443... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{196249}$.\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{196249}{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=221.5$. (Here the result is $\boxed{\frac{196249}$.) |
math-012033 | Inequalities: AM–GM — Equality Conditions | 7 | Track quantifiers carefully: Let $x,y>0$ satisfy $x+y=205$.
(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=205$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{42025}$.\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{42025}{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_... | 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=102.5$. |
math-012034 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Solve and then verify: 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-(-7)}=4.$$
Your final response must include (i) the solution set and... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - 4(x-b)=0$ with the domain restriction $x\\neq -7$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=4$), giving the unique solution $x=-12$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). |
math-012035 | Algebra: Rational Equations — Clearing Denominators | 7 | Track quantifiers carefully: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(2)}=\frac{2}{13}.$$
Your final response must include (i) the... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{2}{13}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{13}$), giving the unique solution $x=-11$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_a... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=2$). (Here the result is $\boxed{x=-11}$.) |
math-012036 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Explain why your operations are valid: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(15)}{x-(-10)}=\frac{-6}{19}.$$
Your final response must include (i) the ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-6}{19}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-6}{19}$), giving the unique solution $x=9$. The domain check $x\\neq -10$ is satisfied here.",
"robustness_analysis": "Robustness note: Clear... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-10$). |
math-012037 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Compute the requested quantity: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-10)}{x-(2)}=\frac{-5}{7}.$$
Your final response must include (i)... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-5}{7}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-5}{7}$), giving the unique solution $x=-5$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=2$). |
math-012038 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Solve with verification: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(3)}=\frac{9}{4}.$$
Your fina... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 3$ so that $x-(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{4}$), giving the unique solution $x=11$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_anal... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). |
math-012039 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Indicate where a theorem is used: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(15)}{x-(8)}=\frac{5}{4}.$$
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{5}{4}(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=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{4}$), giving the unique solution $x=-20$. The domain check $x\\neq 8$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=8$). (Here the result is $\boxed{x=-20}$.) |
math-012040 | Inequalities: AM–GM — Equality Conditions | 7 | Track units/moduli carefully: Let $x,y>0$ satisfy $x+y=654$.
(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=654-x$ with $x\\in(0,654)$. Then $P(x)=xy=x(654... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{106929}$.\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=106929$.",
"robustness_analysis": "Robustness note: AM–GM general... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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=327.0$. |
math-012041 | Algebra: Rational Equations — Domain Restrictions | 7 | State any required conditions first: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(13)}{x-(14)}=\frac{4}{5}.... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 14$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{5}$), giving the unique solution $x=9$. The domain check $x\\neq 14$ 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 $x=14$). (Here the result is $\boxed{x=9}$.) |
math-012042 | Algebra: Rational Equations — Verification by Substitution | 7 | Question: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-2)}{x-(0)}=\frac{13}{15}.$$
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 $(x-a) - \\frac{13}{15}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{15}$), giving the unique solution $x=-15$. The domain check $x\\neq 0$ 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 $x=0$). |
math-012043 | Inequalities: AM–GM — Equality Conditions | 7 | Work carefully and justify each inference: Let $x,y>0$ satisfy $x+y=635$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=635-x$ with $x\\in(0,635)$. Then $P(x)=xy=x(635... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{403225}$.\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 $S^2/4=\\frac{403225}{4}$.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=317.5$. (Here the result is $\boxed{\frac{403225}$.) |
math-012044 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Explain each transformation: 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:
$$\frac{x-(-15)}{x-(0)}=\frac{-1}{4}.$$
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 $(x-a) - \\frac{-1}{4}(x-b)=0$ with the domain restriction $x\\neq... | {
"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 $k=\\frac{-1}{4}$), giving the unique solution $x=-12$. The domain check $x\\neq 0$ is satisfied here.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=0$). |
math-012045 | Algebra: Extremal Values — Global Bounds | 7 | Track quantifiers carefully: Let $x,y>0$ satisfy $x+y=688$.
(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 $y=688-x$ with $x\\in(0,688)$. Then $P(x)=xy=x(688... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{118336}$.\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 $S^2/4=118336$.",
"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 $x=y=344.0$. (Here the result is $\boxed{118336}$.) |
math-012046 | Inequalities: Product Given Sum | 7 | Work this out carefully: Let $x,y>0$ satisfy $x+y=344$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=344-x$ with $x\\in(0,344)$. Then $P(x)=xy=x(344... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{29584}$.\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 $S^2/4=29584$.",
"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 $x=y=172.0$. |
math-012047 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Indicate where a theorem is used: 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:
$$\frac{x-(12)}{x-(1)}=\frac{16}{5}.$$
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 $(x-a) - \\frac{16}{5}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-4}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{16}{5}$), giving the unique solution $x=-4$. 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'.",
... | 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 $\boxed{x=-4}$.) |
math-012048 | Algebra: Rational Equations — Clearing Denominators | 7 | Complete the analysis: 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:
$$\frac{x-(5)}{x-(-9)}=\frac{11}{4}.$$
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 $(x-a) - \\frac{11}{4}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{11}{4}$), giving the unique solution $x=-17$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis"... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). (Here the result is $\boxed{x=-17}$.) |
math-012049 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Checkpoint: 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:
$$\frac{x-(-7)}{x-(-15)}=\frac{11}{15}.$$
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 $(x-a) - \\frac{11}{15}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{11}{15}$), giving the unique solution $x=15$. The domain check $x\\neq -15$ 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'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-15$). (Here the result is $\boxed{x=15}$.) |
math-012050 | Algebra: Rational Equations — Clearing Denominators | 7 | Keep the final answer in boxed form: 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:
$$\frac{x-(-8)}{x-(6)}=3.$$
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 $(x-a) - 3(x-b)=0$ with the domain restriction $x\\neq 6$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=3$), giving the unique solution $x=13$. The domain check $x\\neq 6$ is satisfied here.",
"robustness_analysis": "Robustness note: Clearing den... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). |
math-012051 | Inequalities: AM–GM — Equality Conditions | 7 | Do not skip justification steps: Let $x,y>0$ satisfy $x+y=736$.
(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 $x+y=736$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{135424}$.\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 $S^2/4=135424$.",
"robustness_analysis": "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_... | 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 $x=y=368.0$. (Here the result is $\boxed{135424}$.) |
math-012052 | Algebra: Extremal Values — Global Bounds | 7 | Start by stating any domain restrictions: Let $x,y>0$ satisfy $x+y=535$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus ... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=535$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{286225}$.\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 $S^2/4=\\frac{286225}{4}$.",
"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_... | 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 $x=y=267.5$. (Here the result is $\boxed{\frac{286225}$.) |
math-012053 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Start by stating any domain restrictions: 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:
$$\frac{x-(-1)}{x-(-13)}=\frac{1}{3}.$$
Your fi... | [
{
"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 $(x-a) - \\frac{1}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{3}$), giving the unique solution $x=5$. The domain check $x\\neq -13$ is satisfied here.",
"robustness_analysis": "R... | [
{
"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-012054 | Inequalities: Product Given Sum | 7 | Challenge: Let $x,y>0$ satisfy $x+y=613$.
(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 $x+y=613$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{375769}$.\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 $S^2/4=\\frac{375769}{4}$.",
"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 $x=y=306.5$. (Here the result is $\boxed{\frac{375769}$.) |
math-012055 | Algebra: Extremal Values — Global Bounds | 7 | Solve and then verify: Let $x,y>0$ satisfy $x+y=249$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., conc... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=249-x$ with $x\\in(0,249)$. Then $P(x)=xy=x(249... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{62001}$.\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 $S^2/4=\\frac{62001}{4}$.",
"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_... | 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 $x=y=124.5$. (Here the result is $\boxed{\frac{62001}$.) |
math-012056 | Algebra: Extremal Values — Global Bounds | 7 | Derive the result step-by-step: Let $x,y>0$ satisfy $x+y=375$.
(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 $x+y=375$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{140625}$.\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 $S^2/4=\\frac{140625}{4}$.",
"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 $x=y=187.5$. |
math-012057 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Track quantifiers carefully: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(5)}=\frac{3}{5}.$$
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 $x\\neq 5$ so that $x-(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{5}$), giving the unique solution $x=-20$. 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'.",
... | 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 $\boxed{x=-20}$.) |
math-012058 | Algebra: Extremal Values — Global Bounds | 7 | State any required conditions first: Let $x,y>0$ satisfy $x+y=439$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theor... | [
{
"method_name": "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 $x+y=439$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{192721}$.\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 $S^2/4=\\frac{192721}{4}$.",
"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 $x=y=219.5$. (Here the result is $\boxed{\frac{192721}$.) |
math-012059 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Keep the final answer in boxed form: 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:
$$\frac{x-(-1)}{x-(3)}=\frac{15}{19}.$$
Your final response... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 3$ 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 $k=\\frac{15}{19}$), giving the unique solution $x=-16$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_analysis": "Generality note: Cle... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). |
math-012060 | Inequalities: AM–GM — Equality Conditions | 7 | Solve with verification: Let $x,y>0$ satisfy $x+y=661$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=661-x$ with $x\\in(0,661)$. Then $P(x)=xy=x(661... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{436921}$.\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 $S^2/4=\\frac{436921}{4}$.",
"robustness_a... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=330.5$. (Here the result is $\boxed{\frac{436921}$.) |
math-012061 | Inequalities: AM–GM — Equality Conditions | 7 | Solve (and briefly cross-validate): Let $x,y>0$ satisfy $x+y=220$.
(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 $x+y=220$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{12100}$.\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 $S^2/4=12100$.",
"robustness_analysis": "Generality note: AM–GM gen... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=110.0$. |
math-012062 | Algebra: Extremal Values — Global Bounds | 7 | Solve and then verify: Let $x,y>0$ satisfy $x+y=633$.
(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 $x+y=633$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{400689}$.\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 $S^2/4=\\frac{400689}{4}$.",
"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_... | 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 $x=y=316.5$. |
math-012063 | Algebra: Rational Equations — Verification by Substitution | 7 | Start by stating any domain restrictions: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-13)}{x-(7)}=0.$$
Your final response must include (i) the solution s... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 7$ 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 $k=0$), giving the unique solution $x=-13$. The domain check $x\\neq 7$ is satisfied here.",
"robustness_analysis": "G... | [
{
"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 $\boxed{x=-13}$.) |
math-012064 | Algebra: Extremal Values — Global Bounds | 7 | Indicate where a theorem is used: Let $x,y>0$ satisfy $x+y=412$.
(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 $x+y=412$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{42436}$.\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 $S^2/4=42436$.",
"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 $x=y=206.0$. (Here the result is $\boxed{42436}$.) |
math-012065 | Algebra: Rational Equations — Clearing Denominators | 7 | Track quantifiers carefully: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-12)}{x-(-2)}=-1.$$
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 $x\\neq -2$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-1$), giving the unique solution $x=-7$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis": "Generality note: Clearing d... | [
{
"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 $x=-2$). (Here the result is $\boxed{x=-7}$.) |
math-012066 | Inequalities: Product Given Sum | 7 | Carefully track domains: Let $x,y>0$ satisfy $x+y=97$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., con... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=97-x$ with $x\\in(0,97)$. Then $P(x)=xy=x(97-x)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{9409}$.\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 $S^2/4=\\frac{9409}{4}$.",
"robustness_analysis": "If the problem w... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=48.5$. (Here the result is $\boxed{\frac{9409}$.) |
math-012067 | Algebra: Extremal Values — Global Bounds | 7 | Give reasoning, not just computation: Let $x,y>0$ satisfy $x+y=739$.
(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 $y=739-x$ with $x\\in(0,739)$. Then $P(x)=xy=x(739... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{546121}$.\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 $S^2/4=\\frac{546121}{4}$.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=369.5$. |
math-012068 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Keep the final answer in boxed form: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(0)}{x-(5)}=\frac{2}{7}.$$
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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-2}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{7}$), giving the unique solution $x=-2$. The domain check $x\\neq 5$ 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 $x=5$). |
math-012069 | Inequalities: Product Given Sum | 7 | Do not skip justification steps: Let $x,y>0$ satisfy $x+y=630$.
(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 $y=630-x$ with $x\\in(0,630)$. Then $P(x)=xy=x(630... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{99225}$.\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 $S^2/4=99225$.",
"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_... | 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 $x=y=315.0$. |
math-012070 | Optimization: Two Variables — Concavity | 7 | Checkpoint: Let $x,y>0$ satisfy $x+y=766$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/secon... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=766-x$ with $x\\in(0,766)$. Then $P(x)=xy=x(766... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{146689}$.\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 $S^2/4=146689$.",
"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_... | 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 $x=y=383.0$. |
math-012071 | Inequalities: AM–GM — Equality Conditions | 7 | Give an answer and a quick verification: Let $x,y>0$ satisfy $x+y=828$.
(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 $x+y=828$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{171396}$.\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 $S^2/4=171396$.",
"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 $x=y=414.0$. (Here the result is $\boxed{171396}$.) |
math-012072 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Explain each transformation: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(14)}{x-(-7)}=\frac{1}{8}.$$
Your... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{1}{8}(x-b)=0$ with the domain restriction $x\\neq ... | {
"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 $k=\\frac{1}{8}$), giving the unique solution $x=17$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analysis": "Generality note: Cleari... | [
{
"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 $x=-7$). (Here the result is $\boxed{x=17}$.) |
math-012073 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Give reasoning, not just computation: 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:
$$\frac{x-(-4)}{x-(6)}=\frac{-3}{7}.$$
Your final response must in... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-3}{7}(x-b)=0$ 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 $k=\\frac{-3}{7}$), giving the unique solution $x=-1$. The domain check $x\\neq 6$ 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'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). (Here the result is $\boxed{x=-1}$.) |
math-012074 | Algebra: Rational Equations — Domain Restrictions | 7 | Proceed methodically: 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:
$$\frac{x-(3)}{x-(-5)}=\frac{1}{5}.$$
Your final r... | [
{
"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: $\\boxed{x=5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{5}$), giving the unique solution $x=5$. The domain check $x\\neq -5$ is satisfied here.",
"robustness_analysis": "If the problem were pertu... | [
{
"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 $\boxed{x=5}$.) |
math-012075 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Provide a rigorous solution: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(-13)}=-13.$$
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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-13$), giving the unique solution $x=-12$. 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 $x=-13$). (Here the result is $\boxed{x=-12}$.) |
math-012076 | Algebra: Rational Equations — Domain Restrictions | 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:
$$\frac{x-(6)}{x-(-7)}=\frac{-5}{8}.$$
Your final respons... | [
{
"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": "Both approaches agree after simplification. Final answer: $\\boxed{x=1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-5}{8}$), giving the unique solution $x=1$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analysis": "Sensitivity analysis: Cl... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). (Here the result is $\boxed{x=1}$.) |
math-012077 | Algebra: Extremal Values — Global Bounds | 7 | Give a fully justified solution: Let $x,y>0$ satisfy $x+y=258$.
(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 $y=258-x$ with $x\\in(0,258)$. Then $P(x)=xy=x(258... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{16641}$.\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 $S^2/4=16641$.",
"robustness_analysis": "... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=129.0$. (Here the result is $\boxed{16641}$.) |
math-012078 | Inequalities: Product Given Sum | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=363$.
(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 $y=363-x$ with $x\\in(0,363)$. Then $P(x)=xy=x(363... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{131769}$.\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 $S^2/4=\\frac{131769}{4}$.",
"rob... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=181.5$. |
math-012079 | Algebra: Rational Equations — Clearing Denominators | 7 | Indicate where a theorem is used: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(-5)}=\frac{19}{8}.$$
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 $(x-a) - \\frac{19}{8}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{8}$), giving the unique solution $x=-13$. The domain check $x\\neq -5$ is satisfied here.",
"robustness_analysis"... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-5$). (Here the result is $\boxed{x=-13}$.) |
math-012080 | Inequalities: Product Given Sum | 7 | Compute the requested quantity: Let $x,y>0$ satisfy $x+y=225$.
(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 $x+y=225$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{50625}$.\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 $S^2/4=\\frac{50625}{4}$.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=112.5$. |
math-012081 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Solve with 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:
$$\frac{x-(15)}{x-(8)}=\frac{34}{27}.$$
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 $(x-a) - \\frac{34}{27}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{34}{27}$), giving the unique solution $x=-19$. The domain check $x\\neq 8$ is satisfied here.",
"robustness_analysis": "Robustness note: Cle... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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 $x=8$). (Here the result is $\boxed{x=-19}$.) |
math-012082 | Inequalities: AM–GM — Equality Conditions | 7 | Be explicit about assumptions: Let $x,y>0$ satisfy $x+y=163$.
(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 $y=163-x$ with $x\\in(0,163)$. Then $P(x)=xy=x(163... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{26569}$.\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 $S^2/4=\\frac{26569}{4}$.",
"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_... | 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 $x=y=81.5$. |
math-012083 | Inequalities: Product Given Sum | 7 | Give an answer and a quick verification: Let $x,y>0$ satisfy $x+y=90$.
(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": "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 $y=90-x$ with $x\\in(0,90)$. Then $P(x)=xy=x(90-x)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2025}$.\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 $S^2/4=2025$.",
"robustness_analysis": "Se... | [
{
"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 $x=y=45.0$. (Here the result is $\boxed{2025}$.) |
math-012084 | Algebra: Rational Equations — Verification by Substitution | 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:
$$\frac{x-(7)}{x-(-4)}=2.$$
Your final response must include (i) the solution set and (ii) a brief check for ext... | [
{
"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 $x\\neq -4$ 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 $k=2$), giving the unique solution $x=-15$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis": "If the problem were pertur... | [
{
"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 $\boxed{x=-15}$.) |
math-012085 | Optimization: Two Variables — Concavity | 7 | Carefully track domains: Let $x,y>0$ satisfy $x+y=864$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., co... | [
{
"method_name": "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 $x+y=864$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{186624}$.\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 $S^2/4=186624$.",
"robustness_analysis": "Generality note: AM–GM g... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=432.0$. |
math-012086 | Inequalities: AM–GM — Equality Conditions | 7 | Challenge: Let $x,y>0$ satisfy $x+y=396$.
(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 $y=396-x$ with $x\\in(0,396)$. Then $P(x)=xy=x(396... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{39204}$.\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 $S^2/4=39204$.",
"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 $x=y=198.0$. |
math-012087 | Inequalities: Product Given Sum | 7 | Task: Let $x,y>0$ satisfy $x+y=612$.
(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 $y=612-x$ with $x\\in(0,612)$. Then $P(x)=xy=x(612... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{93636}$.\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 $S^2/4=93636$.",
"robustness_analysis": "If the problem were perturbed: A... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=306.0$. (Here the result is $\boxed{93636}$.) |
math-012088 | Inequalities: AM–GM — Equality Conditions | 7 | Proceed methodically: Let $x,y>0$ satisfy $x+y=656$.
(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 $x+y=656$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{107584}$.\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 $S^2/4=107584$.",
"robustness_analysis": "Generality note: AM–GM general... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=328.0$. |
math-012089 | Precalculus: Rational Expressions — Valid Cancellation | 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:
$$\frac{x-(3)}{x-(-4)}=2.$$
Your final response must include (i) the... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - 2(x-b)=0$ with the domain restriction $x\\neq -4$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=2$), giving the unique solution $x=-11$. 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'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-4$). (Here the result is $\boxed{x=-11}$.) |
math-012090 | Algebra: Rational Equations — Clearing Denominators | 7 | Write the solution set clearly: 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:
$$\frac{x-(4)}{x-(-8)}=\frac{-5}{7}.$$
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 $(x-a) - \\frac{-5}{7}(x-b)=0$ with the domain restriction $x\\neq... | {
"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 $k=\\frac{-5}{7}$), giving the unique solution $x=-1$. The domain check $x\\neq -8$ 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 $x=-8$). (Here the result is $\boxed{x=-1}$.) |
math-012091 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Task: 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:
$$\frac{x-(2)}{x-(-8)}=\frac{-2}{3}.$$
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 $x\\neq -8$ 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 $k=\\frac{-2}{3}$), giving the unique solution $x=-2$. The domain check $x\\neq -8$ 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'.",
... | 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-012092 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Keep the final answer in boxed form: 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:
$$\frac{x-(-3)}{x-(-4)}=\frac{6}{5}.$$
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 $(x-a) - \\frac{6}{5}(x-b)=0$ 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 $k=\\frac{6}{5}$), giving the unique solution $x=-9$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis": "Robustness note: Cleari... | [
{
"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 $\boxed{x=-9}$.) |
math-012093 | 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:
$$\frac{x-(-5)}{x-(-13)}=\frac{5}{7}.$$
Your final response must include (i) the solution set and (ii) ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -13$ so that $x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{7}$), giving the unique solution $x=15$. 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'.",
... | 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-012094 | Inequalities: AM–GM — Equality Conditions | 7 | Track units/moduli carefully: Let $x,y>0$ satisfy $x+y=485$.
(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 $x+y=485$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{235225}$.\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 $S^2/4=\\frac{235225}{4}$.",
"robustness_analysis": "If the... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=242.5$. (Here the result is $\boxed{\frac{235225}$.) |
math-012095 | Algebra: Rational Equations — Clearing Denominators | 7 | Compute the requested quantity: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-6)}{x-(-3)}=\frac{-1}{2}.$$
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 $(x-a) - \\frac{-1}{2}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-1}{2}$), giving the unique solution $x=-5$. The domain check $x\\neq -3$ is satisfied here.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-3$). (Here the result is $\boxed{x=-5}$.) |
math-012096 | Algebra: Extremal Values — Global Bounds | 7 | Show all reasoning: Let $x,y>0$ satisfy $x+y=66$.
(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., concavit... | [
{
"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 $y=66-x$ with $x\\in(0,66)$. Then $P(x)=xy=x(66-x)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1089}$.\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 $S^2/4=1089$.",
"robustness_analysis": "Generality note: AM–GM gener... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=33.0$. (Here the result is $\boxed{1089}$.) |
math-012097 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Provide a rigorous solution: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-15)}{x-(5)}=-4.$$
Your final response must include (i) the solu... | [
{
"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 $(x-a) - -4(x-b)=0$ with the domain restriction $x\\neq 5$.",
... | {
"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 $k=-4$), giving the unique solution $x=1$. The domain check $x\\neq 5$ 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 $x=5$). (Here the result is $\boxed{x=1}$.) |
math-012098 | 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:
$$\frac{x-(6)}{x-(-6)}=3.$$
Your final response must include (i... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -6$ 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 $k=3$), giving the unique solution $x=-12$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). (Here the result is $\boxed{x=-12}$.) |
math-012099 | Algebra: Rational Equations — Clearing Denominators | 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:
$$\frac{x-(12)}{x-(13)}=0.$$
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 $x\\neq 13$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=0$), giving the unique solution $x=12$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "Se... | [
{
"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 $x=13$). (Here the result is $\boxed{x=12}$.) |
math-012100 | Algebra: Extremal Values — Global Bounds | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=27$.
(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 $x+y=27$ to get $\\sqrt{xy}\\le \\frac{2... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{729}$.\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 $S^2/4=\\frac{729}{4}$.",
"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_... | 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 $x=y=13.5$. |
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