id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-012301 | Algebra: Extremal Values — Global Bounds | 7 | Give a fully justified solution: Let $x,y>0$ satisfy $x+y=420$.
(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 $y=420-x$ with $x\\in(0,420)$. Then $P(x)=xy=x(420... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{44100}$.\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 $S^2/4=44100$.",
"robustness_analysis": "Robustness note: AM–GM gen... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=210.0$. (Here the result is $\boxed{44100}$.) |
math-012302 | Algebra: Rational Equations — Verification by Substitution | 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:
$$\frac{x-(-4)}{x-(14)}=\frac{13}{31}.$$
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 $x\\neq 14$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{31}$), giving the unique solution $x=-17$. The domain check $x\\neq 14$ is satisfied here.",
"robustness... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=14$). (Here the result is $\boxed{x=-17}$.) |
math-012303 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Answer using clear logical steps: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(11)}=\frac{2}{5}.$$
Your final response must includ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 11$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{5}$), giving the unique solution $x=-19$. The domain check $x\\neq 11$ is satisfied here.",
"robustness_analysis": "Generality note: Clea... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=11$). |
math-012304 | Inequalities: Product Given Sum | 7 | Answer using clear logical steps: Let $x,y>0$ satisfy $x+y=86$.
(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 $y=86-x$ with $x\\in(0,86)$. Then $P(x)=xy=x(86-x)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1849}$.\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 $S^2/4=1849$.",
"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_... | 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 $x=y=43.0$. (Here the result is $\boxed{1849}$.) |
math-012305 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Start by stating any domain restrictions: 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:
$$\frac{x-(-2)}{x-(-6)}=\frac{21}{25}.$$
Your final response must ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{21}{25}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{21}{25}$), giving the unique solution $x=19$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). (Here the result is $\boxed{x=19}$.) |
math-012306 | Algebra: Rational Equations — Verification by Substitution | 7 | Checkpoint: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(12)}{x-(6)}=\frac{4}{3}.$$
Your final response must include (i) the solution set ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{4}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{3}$), giving the unique solution $x=-12$. The domain check $x\\neq 6$ 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'.",
... | 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-012307 | Optimization: Two Variables — Concavity | 7 | Provide both a computational and a conceptual explanation: Let $x,y>0$ satisfy $x+y=305$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=305$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{93025}$.\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 $S^2/4=\\frac{93025}{4}$.",
"robustness_analysis": "Robustne... | [
{
"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 $x=y=152.5$. (Here the result is $\boxed{\frac{93025}$.) |
math-012308 | Algebra: Rational Equations — Domain Restrictions | 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:
$$\frac{x-(-7)}{x-(3)}=\frac{3}{8}.$$
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 $x\\neq 3$ so that $x-(... | {
"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 $k=\\frac{3}{8}$), giving the unique solution $x=-13$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). (Here the result is $\boxed{x=-13}$.) |
math-012309 | Algebra: Rational Equations — Clearing Denominators | 7 | Compute the requested quantity: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-11)}{x-(-4)}=\frac{9}{16}.$$
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 $(x-a) - \\frac{9}{16}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{16}$), giving the unique solution $x=-20$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-4$). |
math-012310 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Provide both a computational and a conceptual explanation: 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:
$$\frac{x-(-7)}{x-(3)}=2.$$
Your final response m... | [
{
"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: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=2$), giving the unique solution $x=13$. The domain check $x\\neq 3$ is satisfied here.",
"robustness_analysis": "Sen... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=3$). (Here the result is $\boxed{x=13}$.) |
math-012311 | Inequalities: AM–GM — Equality Conditions | 7 | Solve and then verify: Let $x,y>0$ satisfy $x+y=860$.
(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 $y=860-x$ with $x\\in(0,860)$. Then $P(x)=xy=x(860... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{184900}$.\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 $S^2/4=184900$.",
"robustness_analysis": "If the problem were perturbed:... | [
{
"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 $x=y=430.0$. (Here the result is $\boxed{184900}$.) |
math-012312 | Algebra: Rational Equations — Clearing Denominators | 7 | State any required conditions first: 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:
$$\frac{x-(-9)}{x-(9)}=\frac{-1}{5}.$$
Your final response must include (i) the solut... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-1}{5}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-1}{5}$), giving the unique solution $x=-6$. 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'.",
... | 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-012313 | Inequalities: AM–GM — Equality Conditions | 7 | Answer with a short justification: Let $x,y>0$ satisfy $x+y=309$.
(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 $x+y=309$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{95481}$.\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 $S^2/4=\\frac{95481}{4}$.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=154.5$. (Here the result is $\boxed{\frac{95481}$.) |
math-012314 | Inequalities: Product Given Sum | 7 | Keep the final answer in boxed form: Let $x,y>0$ satisfy $x+y=127$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theor... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=127-x$ with $x\\in(0,127)$. Then $P(x)=xy=x(127... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{16129}$.\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 $S^2/4=\\frac{16129}{4}$.",
"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 $x=y=63.5$. (Here the result is $\boxed{\frac{16129}$.) |
math-012315 | Inequalities: AM–GM — Equality Conditions | 7 | Solve and include a self-check: Let $x,y>0$ satisfy $x+y=804$.
(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 $y=804-x$ with $x\\in(0,804)$. Then $P(x)=xy=x(804... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{161604}$.\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 $S^2/4=161604$.",
"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 $x=y=402.0$. (Here the result is $\boxed{161604}$.) |
math-012316 | Optimization: Two Variables — Concavity | 7 | Write the solution set clearly: Let $x,y>0$ satisfy $x+y=209$.
(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 $x+y=209$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{43681}$.\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 $S^2/4=\\frac{43681}{4}$.",
"robustness_ana... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=104.5$. (Here the result is $\boxed{\frac{43681}$.) |
math-012317 | Algebra: Extremal Values — Global Bounds | 7 | Indicate where a theorem is used: Let $x,y>0$ satisfy $x+y=645$.
(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 $y=645-x$ with $x\\in(0,645)$. Then $P(x)=xy=x(645... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{416025}$.\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 $S^2/4=\\frac{416025}{4}$.",
"robustness_analysis": "Robustness n... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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 $x=y=322.5$. |
math-012318 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Provide a rigorous 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:
$$\frac{x-(-5)}{x-(5)}=\frac{-7}{3}.$$
Your final response must include (i) the solution set ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-7}{3}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=2}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-7}{3}$), giving the unique solution $x=2$. The domain check $x\\neq 5$ is satisfied here.",
"robustness_analysis": "Generality 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'.",
... | 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 $\boxed{x=2}$.) |
math-012319 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Prompt: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-10)}{x-(9)}=\frac{30}{11}.$$
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 $x\\neq 9$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{30}{11}$), giving the unique solution $x=20$. The domain check $x\\neq 9$ is satisfied here.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=9$). |
math-012320 | Inequalities: Product Given Sum | 7 | Solve and sanity-check: Let $x,y>0$ satisfy $x+y=510$.
(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 $y=510-x$ with $x\\in(0,510)$. Then $P(x)=xy=x(510... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{65025}$.\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 $S^2/4=65025$.",
"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 $x=y=255.0$. |
math-012321 | Inequalities: AM–GM — Equality Conditions | 7 | Give an answer and a quick verification: Let $x,y>0$ satisfy $x+y=408$.
(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 $x+y=408$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{41616}$.\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 $S^2/4=41616$.",
"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 $x=y=204.0$. |
math-012322 | Optimization: Two Variables — Concavity | 7 | Explain each transformation: Let $x,y>0$ satisfy $x+y=32$.
(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 $y=32-x$ with $x\\in(0,32)$. Then $P(x)=xy=x(32-x)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{256}$.\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 $S^2/4=256$.",
"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 $x=y=16.0$. |
math-012323 | Algebra: Rational Equations — Verification by Substitution | 7 | Answer with a short justification: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(9)}=\frac{17}{9}.$$
Your final response must include (i) the solution... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 9$ so that $x-(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{17}{9}$), giving the unique solution $x=18$. The domain check $x\\neq 9$ is satisfied here.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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-012324 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Warm-up: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-14)}{x-(4)}=\frac{16}{7}.$$
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 $x\\neq 4$ so that $x-(... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{16}{7}$), giving the unique solution $x=18$. 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'.",
... | 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 $\boxed{x=18}$.) |
math-012325 | Algebra: Rational Equations — Domain Restrictions | 7 | Solve and include a self-check: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-11)}{x-(-7)}=\frac{7}{11}.$$
Your final response mus... | [
{
"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: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{11}$), giving the unique solution $x=-18$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). (Here the result is $\boxed{x=-18}$.) |
math-012326 | Inequalities: AM–GM — Equality Conditions | 7 | Solve with verification: Let $x,y>0$ satisfy $x+y=500$.
(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 $x+y=500$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{62500}$.\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 $S^2/4=62500$.",
"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_... | 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 $x=y=250.0$. |
math-012327 | Inequalities: Product Given Sum | 7 | Solve (and briefly cross-validate): Let $x,y>0$ satisfy $x+y=545$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theore... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=545-x$ with $x\\in(0,545)$. Then $P(x)=xy=x(545... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{297025}$.\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 $S^2/4=\\frac{297025}{4}$.",
"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 $x=y=272.5$. |
math-012328 | Algebra: Rational Equations — Domain Restrictions | 7 | Answer with a short justification: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(7)}{x-(-15)}=\frac{13}{2}.$$
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 $x\\neq -15$ so that $x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{2}$), giving the unique solution $x=-19$. The domain check $x\\neq -15$ is satisfied here.",
"robustness_analysis... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-15$). (Here the result is $\boxed{x=-19}$.) |
math-012329 | Algebra: Rational Equations — Clearing Denominators | 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:
$$\frac{x-(13)}{x-(-4)}=\frac{-12}{5}.$$
Your final response must include (i) the solution set and (ii) a... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-12}{5}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-12}{5}$), giving the unique solution $x=1$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-4$). (Here the result is $\boxed{x=1}$.) |
math-012330 | Algebra: Rational Equations — Verification by Substitution | 7 | Compute the requested quantity: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(15)}=\frac{13}{27}.$$
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 $(x-a) - \\frac{13}{27}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-12}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{27}$), giving the unique solution $x=-12$. The domain check $x\\neq 15$ 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 $x=15$). (Here the result is $\boxed{x=-12}$.) |
math-012331 | Algebra: Rational Equations — Clearing Denominators | 7 | Work this out carefully: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(10)}{x-(0)}=\frac{17}{7}.$$
Your final response must include (i) the solution set and ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 0$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{17}{7}$), giving the unique solution $x=-7$. The domain check $x\\neq 0$ is satisfied here.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=0$). (Here the result is $\boxed{x=-7}$.) |
math-012332 | Algebra: Rational Equations — Verification by Substitution | 7 | Carefully track domains: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(2)}=\frac{13}{2}.$$
Your final response must include (i) the soluti... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 2$ 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 $k=\\frac{13}{2}$), giving the unique solution $x=4$. The domain check $x\\neq 2$ 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'.",
... | 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-012333 | Inequalities: AM–GM — Equality Conditions | 7 | Solve with verification: Let $x,y>0$ satisfy $x+y=559$.
(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 $y=559-x$ with $x\\in(0,559)$. Then $P(x)=xy=x(559... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{312481}$.\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 $S^2/4=\\frac{312481}{4}$.",
"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 $x=y=279.5$. |
math-012334 | Inequalities: Product Given Sum | 7 | Derive the result step-by-step: Let $x,y>0$ satisfy $x+y=533$.
(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 $x+y=533$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{284089}$.\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 $S^2/4=\\frac{284089}{4}$.",
"robustness_analysis": "Robust... | [
{
"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 $x=y=266.5$. (Here the result is $\boxed{\frac{284089}$.) |
math-012335 | Algebra: Extremal Values — Global Bounds | 7 | Answer using clear logical steps: Let $x,y>0$ satisfy $x+y=342$.
(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 $y=342-x$ with $x\\in(0,342)$. Then $P(x)=xy=x(342... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{29241}$.\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 $S^2/4=29241$.",
"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 $x=y=171.0$. |
math-012336 | Algebra: Rational Equations — Domain Restrictions | 7 | Make each step logically reversible (or explain if not): 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:
$$\frac{x-(-1)}{x-(12)}=\frac{15}{28}.$$
Your f... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 12$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{15}{28}$), giving the unique solution $x=-16$. The domain check $x\\neq 12$ 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 $x=12$). (Here the result is $\boxed{x=-16}$.) |
math-012337 | Inequalities: AM–GM — Equality Conditions | 7 | Checkpoint: Let $x,y>0$ satisfy $x+y=436$.
(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 $y=436-x$ with $x\\in(0,436)$. Then $P(x)=xy=x(436... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{47524}$.\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 $S^2/4=47524$.",
"robustness_analysis": "Robustness note: AM–GM generaliz... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=218.0$. (Here the result is $\boxed{47524}$.) |
math-012338 | Algebra: Rational Equations — Domain Restrictions | 7 | Give a fully justified 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:
$$\frac{x-(-14)}{x-(14)}=0.$$
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 $(x-a) - 0(x-b)=0$ with the domain restriction $x\\neq 14$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-14}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=0$), giving the unique solution $x=-14$. The domain check $x\\neq 14$ 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'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=14$). (Here the result is $\boxed{x=-14}$.) |
math-012339 | Precalculus: Rational Expressions — Valid Cancellation | 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:
$$\frac{x-(-13)}{x-(-2)}=\frac{16}{5}.$$
Your final... | [
{
"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": "Both approaches agree after simplification. Final answer: $\\boxed{x=3}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{16}{5}$), giving the unique solution $x=3$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis": "Generality note: Clearin... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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-012340 | Inequalities: Product Given Sum | 7 | Solve with verification: Let $x,y>0$ satisfy $x+y=888$.
(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 $y=888-x$ with $x\\in(0,888)$. Then $P(x)=xy=x(888... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{197136}$.\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 $S^2/4=197136$.",
"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 $x=y=444.0$. |
math-012341 | Algebra: Extremal Values — Global Bounds | 7 | Solve and then verify: Let $x,y>0$ satisfy $x+y=154$.
(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 $y=154-x$ with $x\\in(0,154)$. Then $P(x)=xy=x(154... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5929}$.\nAM–GM gives $xy\\le (S/2)^2$ and calculus finds the maximizer $x=y=S/2$ for the concave quadratic $x(S-x)$. Both yield the same maximum value $S^2/4=5929$.",
"robustness_analysis": "If the problem were perturbed: 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 $x=y=77.0$. |
math-012342 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Where appropriate, name the theorem you use: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(1)}{x-(13)}=\frac... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 13$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{31}$), giving the unique solution $x=-18$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). |
math-012343 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Prompt: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(-6)}=\frac{1}{4}.$$
Your final response must include (i) the solution set and (ii) a ... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{1}{4}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{4}$), giving the unique solution $x=10$. The domain check $x\\neq -6$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-6$). (Here the result is $\boxed{x=10}$.) |
math-012344 | Algebra: Rational Equations — Clearing Denominators | 7 | Use two approaches if possible: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(10)}=\frac{-10}{7}.$$
Your final response must include (i) t... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 10$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=3}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-10}{7}$), giving the unique solution $x=3$. The domain check $x\\neq 10$ is satisfied here.",
"robustness_ana... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=10$). |
math-012345 | Algebra: Rational Equations — Verification by Substitution | 7 | Solve (and briefly cross-validate): You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(2)}{x-(-15)}=\frac{-1}{16}.$$
Your final response must include (... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -15$ so that $x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-1}{16}$), giving the unique solution $x=1$. The domain check $x\\neq -15$ is satisfied here.",
"robustness_analysis": "Robustness note:... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-15$). (Here the result is $\boxed{x=1}$.) |
math-012346 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Complete the analysis: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(7)}{x-(13)}=\frac{13}{16}.$$
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=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{16}$), giving the unique solution $x=-19$. The domain check $x\\neq 13$ is satisfied here.",
"robustness... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). (Here the result is $\boxed{x=-19}$.) |
math-012347 | Algebra: Extremal Values — Global Bounds | 7 | Answer with a short justification: Let $x,y>0$ satisfy $x+y=423$.
(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=423-x$ with $x\\in(0,423)$. Then $P(x)=xy=x(423... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{178929}$.\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{178929}{4}$.",
"robustness_analysis": "Robust... | [
{
"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=211.5$. (Here the result is $\boxed{\frac{178929}$.) |
math-012348 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | State any required conditions first: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(3)}{x-(9)}=\frac{11}{14}.$$
Your final response must include... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{11}{14}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{11}{14}$), giving the unique solution $x=-19$. The domain check $x\\neq 9$ is satisfied here.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=9$). (Here the result is $\boxed{x=-19}$.) |
math-012349 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Prompt: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(11)}{x-(7)}=\frac{27}{23}.$$
Your final response must include (i) the solutio... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{27}{23}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{27}{23}$), giving the unique solution $x=-16$. The domain check $x\\neq 7$ is satisfied here.",
"robustness_analysis"... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=7$). (Here the result is $\boxed{x=-16}$.) |
math-012350 | Algebra: Rational Equations — Clearing Denominators | 7 | Explain what is being counted/optimized: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-6)}{x-(-5)}=\frac{19}{18}.$$
Your final response must include (i) the... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -5$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{18}$), giving the unique solution $x=13$. The domain check $x\\neq -5$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-5$). (Here the result is $\boxed{x=13}$.) |
math-012351 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Solve (and briefly cross-validate): Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(15)}=\frac{1}{3}.$$
Your final response m... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 15$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-15}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{3}$), giving the unique solution $x=-15$. The domain check $x\\neq 15$ is satisfied here.",
"robustness_a... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=15$). |
math-012352 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Find the exact value: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(11)}{x-(-11)}=\frac{14}{3}.$$
Your final response must include (i) the solutio... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{14}{3}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-17}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{14}{3}$), giving the unique solution $x=-17$. The domain check $x\\neq -11$ is satisfied here.",
"robustness_analysis": "Robustness 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'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-11$). (Here the result is $\boxed{x=-17}$.) |
math-012353 | Inequalities: Product Given Sum | 7 | Give a fully justified solution: Let $x,y>0$ satisfy $x+y=624$.
(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=624-x$ with $x\\in(0,624)$. Then $P(x)=xy=x(624... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{97344}$.\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=97344$.",
"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_... | 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=312.0$. |
math-012354 | Inequalities: Product Given Sum | 7 | Complete the analysis: Let $x,y>0$ satisfy $x+y=171$.
(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=171-x$ with $x\\in(0,171)$. Then $P(x)=xy=x(171... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{29241}$.\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{29241}{4}$.",
"robustness_ana... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Remember: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=85.5$. (Here the result is $\boxed{\frac{29241}$.) |
math-012355 | Inequalities: AM–GM — Equality Conditions | 7 | Derive the result step-by-step: Let $x,y>0$ satisfy $x+y=651$.
(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=651-x$ with $x\\in(0,651)$. Then $P(x)=xy=x(651... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{423801}$.\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{423801}{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=325.5$. (Here the result is $\boxed{\frac{423801}$.) |
math-012356 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | State any required conditions first: 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-(12)}{x-(14)}=\frac{13}{15}.$$
Your final ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 14$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{15}$), giving the unique solution $x=-1$. The domain check $x\\neq 14$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=14$). (Here the result is $\boxed{x=-1}$.) |
math-012357 | Algebra: Rational Equations — Verification by Substitution | 7 | Keep the final answer in boxed form: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(13)}=\frac{7}{29}.$$
Your final response must include (... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{7}{29}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{29}$), giving the unique solution $x=-16$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis"... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). |
math-012358 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Show all reasoning: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(2)}{x-(-7)}=-8.$$
Your final response must include (i) the soluti... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -7$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=-8$), giving the unique solution $x=-6$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analysis": "Generality note: Clearing denomin... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). |
math-012359 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Give an answer and a quick verification: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-2)}{x-(-1)}=\frac{4}{3}.$$
Your final respo... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -1$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=2}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{3}$), giving the unique solution $x=2$. The domain check $x\\neq -1$ is satisfied here.",
"robustness_analysis": "Robustness note: Cl... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-1$). |
math-012360 | Algebra: Rational Equations — Verification by Substitution | 7 | Warm-up: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(5)}=\frac{22}{21}.$$
Your final response must include (i) the solution set an... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 5$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{22}{21}$), giving the unique solution $x=-16$. The domain check $x\\neq 5$ is satisfied here.",
"robustness_analysis": "Robustness note: Cle... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=5$). |
math-012361 | Algebra: Rational Equations — Verification by Substitution | 7 | Explain why your operations are valid: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(7)}{x-(-9)}=17.$$
Your final response must include (i) the... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -9$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=17$), giving the unique solution $x=-10$. The domain check $x\\neq -9$ is satisfied here.",
"robustness_analysis": ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-9$). |
math-012362 | Algebra: Rational Equations — Domain Restrictions | 7 | Explain each transformation: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(14)}{x-(1)}=\frac{25}{12}.$$
Your final response must in... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 1$ so that $x-(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{25}{12}$), giving the unique solution $x=-11$. The domain check $x\\neq 1$ is satisfied here.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=1$). |
math-012363 | Algebra: Rational Equations — Clearing Denominators | 7 | Exercise: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(12)}{x-(4)}=\frac{1}{2}.$$
Your final response must include (i) the solution set and (ii) ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 4$ so that $x-(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=20}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{2}$), giving the unique solution $x=20$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_anal... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=4$). (Here the result is $\boxed{x=20}$.) |
math-012364 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Challenge: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(12)}=\frac{-4}{13}.$$
Your final response must include (i) the solution se... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{-4}{13}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-1}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{-4}{13}$), giving the unique solution $x=-1$. The domain check $x\\neq 12$ is satisfied here.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=12$). (Here the result is $\boxed{x=-1}$.) |
math-012365 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Show all reasoning: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-7)}{x-(-5)}=\frac{1}{2}.$$
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 -5$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{2}$), giving the unique solution $x=-9$. The domain check $x\\neq -5$ 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'.",
... | 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$). (Here the result is $\boxed{x=-9}$.) |
math-012366 | Inequalities: Product Given Sum | 7 | Write the solution set clearly: Let $x,y>0$ satisfy $x+y=397$.
(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=397$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{157609}$.\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{157609}{4}$.",
"robustness_analysis": "Robustness n... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | 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=198.5$. (Here the result is $\boxed{\frac{157609}$.) |
math-012367 | Inequalities: AM–GM — Equality Conditions | 7 | Explain each transformation: Let $x,y>0$ satisfy $x+y=582$.
(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=582$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{84681}$.\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=84681$.",
"robustness_analysis": "Generalit... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Takeaway: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=291.0$. (Here the result is $\boxed{84681}$.) |
math-012368 | Inequalities: Product Given Sum | 7 | Explain why your operations are valid: Let $x,y>0$ satisfy $x+y=788$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus the... | [
{
"method_name": "AM–GM",
"approach": "Apply AM–GM to $x$ and $y$ to bound $\\sqrt{xy}$ in terms of $x+y$; equality characterizes the maximizer.",
"steps": [
"Step 1: By AM–GM, for $x,y>0$ we have $\\frac{x+y}{2}\\ge \\sqrt{xy}$.",
"Step 2: Substitute $x+y=788$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{155236}$.\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=155236$.",
"robustness_analysis": "Robustness 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_... | 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=394.0$. |
math-012369 | Inequalities: AM–GM — Equality Conditions | 7 | Exercise: Let $x,y>0$ satisfy $x+y=503$.
(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=503-x$ with $x\\in(0,503)$. Then $P(x)=xy=x(503... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{253009}$.\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{253009}{4}$.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=251.5$. (Here the result is $\boxed{\frac{253009}$.) |
math-012370 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Solve and then verify: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-15)}{x-(-7)}=\frac{5}{3}.$$
Your final response must include (i) the solutio... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{5}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=5}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{5}{3}$), giving the unique solution $x=5$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analysis": "Robustness note: Cl... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). (Here the result is $\boxed{x=5}$.) |
math-012371 | Algebra: Rational Equations — Clearing Denominators | 7 | Keep the final answer in boxed form: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-9)}{x-(-13)}=\frac{4}{5}.$$
Your final r... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{4}{5}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=7}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{4}{5}$), giving the unique solution $x=7$. The domain check $x\\neq -13$ is satisfied here.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-13$). (Here the result is $\boxed{x=7}$.) |
math-012372 | Optimization: Two Variables — Concavity | 7 | Answer using clear logical steps: Let $x,y>0$ satisfy $x+y=587$.
(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=587$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{344569}$.\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{344569}{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_... | 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=293.5$. |
math-012373 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Where appropriate, name the theorem you use: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(8)}{x-(-7)}=\frac{26}{11}.$$
Your final response must i... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{26}{11}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{26}{11}$), giving the unique solution $x=-18$. The domain check $x\\neq -7$ is satisfied here.",
"robustness_analysis... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-7$). (Here the result is $\boxed{x=-18}$.) |
math-012374 | Algebra: Rational Equations — Domain Restrictions | 7 | Question: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(12)}{x-(-15)}=\frac{7}{34}.$$
Your final response must include (i) t... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{7}{34}(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=19}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{7}{34}$), giving the unique solution $x=19$. The domain check $x\\neq -15$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-15$). |
math-012375 | Inequalities: AM–GM — Equality Conditions | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=402$.
(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=402-x$ with $x\\in(0,402)$. Then $P(x)=xy=x(402... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{40401}$.\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=40401$.",
"robustness_analysis": "If the problem were perturbed: A... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Key idea: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=201.0$. (Here the result is $\boxed{40401}$.) |
math-012376 | Algebra: Rational Equations — Verification by Substitution | 7 | Find the exact value: 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-(-14)}{x-(-4)}=\frac{3}{2}.$$
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 -4$ so that $x-... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{3}{2}$), giving the unique solution $x=16$. The domain check $x\\neq -4$ 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'.",
... | 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=16}$.) |
math-012377 | Inequalities: Product Given Sum | 7 | Use two approaches if possible: Let $x,y>0$ satisfy $x+y=313$.
(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=313$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{97969}$.\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{97969}{4}$.",
"robustness_analysis": "Robustness 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=156.5$. |
math-012378 | Algebra: Extremal Values — Global Bounds | 7 | Provide a rigorous solution: Let $x,y>0$ satisfy $x+y=792$.
(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=792$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{156816}$.\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=156816$.",
"robustness_analysis": "Robustness 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=396.0$. |
math-012379 | Algebra: Rational Equations — Domain Restrictions | 7 | Challenge: 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{-8}{19}.$$
Your final response must include (i) the so... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 14$ 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=\\frac{-8}{19}$), giving the unique solution $x=-5$. The domain check $x\\neq 14$ is satisfied here.",
"robustness_analysis":... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=14$). |
math-012380 | Inequalities: Product Given Sum | 7 | Exercise: Let $x,y>0$ satisfy $x+y=628$.
(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=628$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{98596}$.\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=98596$.",
"robustness_analysis": "Robustness note: AM–GM gen... | [
{
"error_description": "Squared an inequality without stating nonnegativity.",
"why_plausible": "Squaring is common and seems automatically valid.",
"why_wrong": "Squaring reverses inequalities when negative quantities are involved; the justification is that $\\sqrt{xy}\\ge 0$ and $S/2>0$.",
"which_... | Core principle: With a fixed positive sum, the product is maximized when the numbers are equal; AM–GM gives a global bound and identifies equality at $x=y=314.0$. (Here the result is $\boxed{98596}$.) |
math-012381 | Algebra: Rational Equations — Clearing Denominators | 7 | Give a theorem-based solution: 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-(-4)}=\frac{23}{10}.$$
Your final response must include (i) ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -4$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-14}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{23}{10}$), giving the unique solution $x=-14$. The domain check $x\\neq -4$ is satisfied here.",
"robustness_analysis... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Remember: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-4$). |
math-012382 | Algebra: Rational Equations — Clearing Denominators | 7 | Give a fully justified solution: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(2)}{x-(1)}=\frac{18}{17}.$$
Your final response must... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 1$ so that $x-(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-16}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{18}{17}$), giving the unique solution $x=-16$. The domain check $x\\neq 1$ is satisfied here.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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=-16}$.) |
math-012383 | Inequalities: AM–GM — Equality Conditions | 7 | Explain what is being counted/optimized: Let $x,y>0$ satisfy $x+y=427$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus t... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=427-x$ with $x\\in(0,427)$. Then $P(x)=xy=x(427... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{182329}$.\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{182329}{4}$.",
"robustness_analysis": "Robust... | [
{
"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=213.5$. |
math-012384 | Algebra: Rational Equations — Domain Restrictions | 7 | Challenge: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-2)}{x-(6)}=\frac{2}{3}.$$
Your final response must include (i) the soluti... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{2}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-18}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{3}$), giving the unique solution $x=-18$. The domain check $x\\neq 6$ 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'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=6$). |
math-012385 | Optimization: Two Variables — Concavity | 7 | Give reasoning, not just computation: Let $x,y>0$ satisfy $x+y=672$.
(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": "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=672$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{112896}$.\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=112896$.",
"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_... | 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=336.0$. |
math-012386 | Algebra: Rational Equations — Verification by Substitution | 7 | Answer using clear logical steps: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-5)}{x-(13)}=\frac{1}{19}.$$
Your final response must inclu... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 13$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{1}{19}$), giving the unique solution $x=-6$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). |
math-012387 | Algebra: Rational Equations — Verification by Substitution | 7 | Carefully track domains: Find the solution set of the rational equation. Your answer must include a short 'domain + check' section:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-1)}{x-(-12)}=\frac{13}{2}.$$
Your final response must include (i) the s... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -12$ so that $x... | {
"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{13}{2}$), giving the unique solution $x=-14$. The domain check $x\\neq -12$ is satisfied here.",
"robustness... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-12$). (Here the result is $\boxed{x=-14}$.) |
math-012388 | Optimization: Two Variables — Concavity | 7 | Start by stating any domain restrictions: Let $x,y>0$ satisfy $x+y=482$.
(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=482$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{58081}$.\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=58081$.",
"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=241.0$. (Here the result is $\boxed{58081}$.) |
math-012389 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Carefully track domains: Rational equation (watch for extraneous roots). Solve and then give a one-line verification:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-8)}{x-(5)}=0.$$
Your final response must include (i) the solution set and (ii) a brie... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 5$ so that $x-(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-8}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=0$), giving the unique solution $x=-8$. The domain check $x\\neq 5$ is satisfied here.",
"robustness_analysis": "If the problem were perturbed: Clea... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=5$). (Here the result is $\boxed{x=-8}$.) |
math-012390 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Give a theorem-based solution: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(6)}{x-(4)}=\frac{2}{3}.$$
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{2}{3}(x-b)=0$ with the domain restriction $x\\neq ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{3}$), giving the unique solution $x=10$. The domain check $x\\neq 4$ is satisfied here.",
"robustness_anal... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | 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=10}$.) |
math-012391 | Optimization: Two Variables — Concavity | 7 | Solve and then verify: Let $x,y>0$ satisfy $x+y=425$.
(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=425$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{180625}$.\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{180625}{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=212.5$. |
math-012392 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Be explicit about assumptions: 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-(-3)}{x-(-14)}=\frac{9}{20}.$$
Your final response must include (i) th... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{9}{20}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{9}{20}$), giving the unique solution $x=6$. The domain check $x\\neq -14$ is satisfied here.",
"robustness_analysis": "... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-14$). (Here the result is $\boxed{x=6}$.) |
math-012393 | Algebra: Equations — Conditions for Valid Multiplication | 7 | Indicate where a theorem is used: Solve the equation and explain why clearing denominators is logically valid **only after** excluding forbidden values:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(5)}{x-(13)}=\frac{2}{3}.$$
Your final response must... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 13$ so that $x-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{x=-11}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{2}{3}$), giving the unique solution $x=-11$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "Generality note: Clea... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). (Here the result is $\boxed{x=-11}$.) |
math-012394 | Inequalities: Product Given Sum | 7 | Track quantifiers carefully: Let $x,y>0$ satisfy $x+y=468$.
(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=468$ to get $\\sqrt{xy}\\le \\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{54756}$.\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=54756$.",
"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_... | 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=234.0$. |
math-012395 | Algebra: Rational Equations — Domain Restrictions | 7 | Solve and sanity-check: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-6)}{x-(-11)}=\frac{12}{17}.$$
Your f... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{12}{17}(x-b)=0$ with the domain restriction $x\\ne... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=6}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{12}{17}$), giving the unique solution $x=6$. The domain check $x\\neq -11$ is satisfied here.",
"robustness_an... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Key idea: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-11$). (Here the result is $\boxed{x=6}$.) |
math-012396 | Precalculus: Rational Expressions — Valid Cancellation | 7 | Proceed methodically: Compute the real solutions. (i) state the domain restriction, (ii) clear denominators, (iii) check the result:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(-10)}{x-(2)}=\frac{19}{7}.$$
Your final response must include (i) the s... | [
{
"method_name": "Affine Transformation View",
"approach": "Rewrite the equation as equality of two Möbius/affine transforms and use uniqueness of solutions for linear equations after rearrangement.",
"steps": [
"Step 1: Rewrite as $(x-a) - \\frac{19}{7}(x-b)=0$ with the domain restriction $x\\neq... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{x=9}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{19}{7}$), giving the unique solution $x=9$. The domain check $x\\neq 2$ is satisfied here.",
"robustness_analy... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=2$). |
math-012397 | Algebra: Rational Equations — Domain Restrictions | 7 | Complete the analysis: Determine all real $x$ satisfying the equation below, and **explicitly list excluded values** from the domain before solving:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(13)}{x-(-2)}=\frac{26}{11}.$$
Your final response must ... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq -2$ so that $x-... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{x=-13}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{26}{11}$), giving the unique solution $x=-13$. The domain check $x\\neq -2$ is satisfied here.",
"robustness_analysis... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=-2$). |
math-012398 | Algebra: Rational Equations — Extraneous Roots Detection | 7 | Derive the result step-by-step: Solve over $\mathbb{R}$ **with domain bookkeeping**. Start by stating when denominators vanish, then solve and check for extraneous solutions:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(3)}{x-(11)}=\frac{13}{21}.$$
... | [
{
"method_name": "Clear Denominators + Domain Check",
"approach": "Multiply both sides by the (nonzero) denominator expression, solve the resulting linear equation, then reject any solution that violates the original domain.",
"steps": [
"Step 1: Domain restriction: require $x\\neq 11$ so that $x-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{x=-10}$.\nBoth methods derive the same linear equation $(1-k)x=a-kb$ (with $k=\\frac{13}{21}$), giving the unique solution $x=-10$. The domain check $x\\neq 11$ is satisfied here.",
"robustness_analysis": "Sensitivity a... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Core principle: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=11$). (Here the result is $\boxed{x=-10}$.) |
math-012399 | Algebra: Rational Equations — Linear-Fractional Forms | 7 | Solve and justify each step: You are given a linear-fractional equation. Solve it and justify any cancellation/multiplication steps:
Solve the rational equation over the reals, and state any values that must be excluded from the domain:
$$\frac{x-(11)}{x-(13)}=\frac{9}{11}.$$
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 $(x-a) - \\frac{9}{11}(x-b)=0$ with the domain restriction $x\\neq... | {
"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 $k=\\frac{9}{11}$), giving the unique solution $x=2$. The domain check $x\\neq 13$ is satisfied here.",
"robustness_analysis": "Generality note: Clearin... | [
{
"error_description": "Forgot to exclude $x=b$ before multiplying both sides by $(x-b)$.",
"why_plausible": "Clearing denominators is often taught as a mechanical step.",
"why_wrong": "If $x=b$, the original equation is undefined, but after multiplying you may accidentally admit it as a 'solution'.",
... | Takeaway: Rational equations require domain bookkeeping: exclude values that make denominators zero, then solve and finally substitute back to avoid extraneous solutions (here exclude $x=13$). (Here the result is $\boxed{x=2}$.) |
math-012400 | Algebra: Extremal Values — Global Bounds | 7 | Warm-up: Let $x,y>0$ satisfy $x+y=132$.
(a) Find the maximum possible value of $xy$.
(b) State precisely when equality (the maximum) occurs.
(c) Explain why your argument guarantees a global maximum (not just a local one).
Your solution must explicitly cite either AM–GM or a calculus theorem (e.g., concavity/second d... | [
{
"method_name": "Calculus (Concave Quadratic)",
"approach": "Use the constraint to write $xy$ as a concave quadratic in one variable; a concave function on an interval has a unique global maximum at its critical point.",
"steps": [
"Step 1: Write $y=132-x$ with $x\\in(0,132)$. Then $P(x)=xy=x(132... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4356}$.\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 $S^2/4=4356$.",
"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_... | 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 $x=y=66.0$. (Here the result is $\boxed{4356}$.) |
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