Datasets:
id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-000001 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Task: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-42) = -498 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the end. | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-42)$ from both sides: $24x=-456$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{-456}{24}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-456}{24}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=24... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=24\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000002 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-18) = -438 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=28x+(-18)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-420}{28}$ and compute the same integer $x=-15$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=28\neq 0$. (Here the result is $\boxed{-15}$.) |
math-000003 | Prealgebra: Solving for a Variable | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 4x + (-5) = -77 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-c... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-5)$ from both sides: $4x=-72$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{-72}{4}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-72}{4}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=4\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-18$ because $a=4\neq 0$. (Here the result is $\boxed{-18}$.) |
math-000004 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 16x + (-80) = 16 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the end... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=16x+(-80)$. Since the slope $16\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6}$.\nBoth methods reduce the equation to $x=\\frac{96}{16}$ and compute the same integer $x=6$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear eq... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=6$ because $a=16\neq 0$. |
math-000005 | Elementary Algebra: Linear Equations — Verification | 1 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 14x + (-39) = 45 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the ... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-39)$ from both sides: $14x=84$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{84}{14}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6}$.\nBoth methods reduce the equation to $x=\\frac{84}{14}$ and compute the same integer $x=6$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=6$ because $a=14\neq 0$. (Here the result is $\boxed{6}$.) |
math-000006 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 28x + (76) = 412 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-ch... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=28x+(76)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12}$.\nBoth methods reduce the equation to $x=\\frac{336}{28}$ and compute the same integer $x=12$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=12$ because $a=28\neq 0$. (Here the result is $\boxed{12}$.) |
math-000007 | Algebra: Affine Functions — Injectivity | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 3x + (78) = 15 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=3x+(78)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-21}$.\nBoth methods reduce the equation to $x=\\frac{-63}{3}$ and compute the same integer $x=-21$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-oper... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-21$ because $a=3\neq 0$. (Here the result is $\boxed{-21}$.) |
math-000008 | Algebra: Affine Functions — Injectivity | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 2x + (23) = 35 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=2x+(23)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6}$.\nBoth methods reduce the equation to $x=\\frac{12}{2}$ and compute the same integer $x=6$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any l... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=2\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=6$ because $a=2\neq 0$. |
math-000009 | Algebra: Affine Functions — Injectivity | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 7x + (-34) = 92 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=7x+(-34)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{126}{7}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=7\neq 0$. |
math-000010 | Elementary Algebra: Linear Equations — Verification | 1 | Explain why your operations are valid: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-76) = -92 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verif... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=2x+(-76)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-16}{2}$ and compute the same integer $x=-8$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equat... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=2\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-8$ because $a=2\neq 0$. (Here the result is $\boxed{-8}$.) |
math-000011 | Algebra: Affine Functions — Injectivity | 1 | Track units/moduli carefully: Solve for $x$ and verify your result:
(a) Solve $ 30x + (-38) = 352 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-38)$ from both sides: $30x=390$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{390}{30}$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{390}{30}$ and compute the same integer $x=13$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=30... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=30\neq 0$. (Here the result is $\boxed{13}$.) |
math-000012 | Algebra: Affine Functions — Injectivity | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-71) = -29 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-che... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=2x+(-71)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{42}{2}$ and compute the same integer $x=21$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Invers... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=2\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=2\neq 0$. |
math-000013 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 4x + (-42) = 38 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the en... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=4x+(-42)$. Since the slope $4\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{80}{4}$ and compute the same integer $x=20$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=4\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=20$ because $a=4\neq 0$. (Here the result is $\boxed{20}$.) |
math-000014 | Elementary Algebra: Linear Equations — Verification | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 7x + (-47) = 65 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the en... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=7x+(-47)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\nBoth methods reduce the equation to $x=\\frac{112}{7}$ and compute the same integer $x=16$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations work... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=16$ because $a=7\neq 0$. (Here the result is $\boxed{16}$.) |
math-000015 | Algebra: Affine Functions — Injectivity | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 8x + (-69) = -149 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=8x+(-69)$. Since the slope $8\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-80}{8}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works f... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=8\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=8\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000016 | Elementary Algebra: Linear Equations — Verification | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 12x + (66) = 90 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=12x+(66)$. Since the slope $12\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{24}{12}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax+... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=12... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=12\neq 0$. (Here the result is $\boxed{2}$.) |
math-000017 | Algebra: Affine Functions — Injectivity | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-72) = -112 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief ve... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(-72)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-40}{5}$ and compute the same integer $x=-8$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-8$ because $a=5\neq 0$. (Here the result is $\boxed{-8}$.) |
math-000018 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 7x + (72) = -26 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verifica... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(72)$ from both sides: $7x=-98$.",
"Step 2: Since $7\\neq 0$, divide by $7$: $x=\\frac{-98}{7}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-98}{7}$ and compute the same integer $x=-14$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works f... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-14$ because $a=7\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000019 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-76) = -576 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=25x+(-76)$. Since the slope $25\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-500}{25}$ and compute the same integer $x=-20$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-20$ because $a=25\neq 0$. (Here the result is $\boxed{-20}$.) |
math-000020 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 9x + (-78) = -150 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=9x+(-78)$. Since the slope $9\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-72}{9}$ and compute the same integer $x=-8$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for an... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=9\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-8$ because $a=9\neq 0$. (Here the result is $\boxed{-8}$.) |
math-000021 | Algebra: Affine Functions — Injectivity | 1 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 11x + (45) = 23 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(45)$ from both sides: $11x=-22$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{-22}{11}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-22}{11}$ and compute the same integer $x=-2$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=11... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=11\neq 0$. (Here the result is $\boxed{-2}$.) |
math-000022 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 29x + (38) = 357 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(38)$ from both sides: $29x=319$.",
"Step 2: Since $29\\neq 0$, divide by $29$: $x=\\frac{319}{29}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{11}$.\nBoth methods reduce the equation to $x=\\frac{319}{29}$ and compute the same integer $x=11$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=11$ because $a=29\neq 0$. (Here the result is $\boxed{11}$.) |
math-000023 | Prealgebra: Solving for a Variable | 1 | Keep the final answer in boxed form: Solve for $x$ and verify your result:
(a) Solve $ 30x + (18) = -192 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verifi... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(18)$ from both sides: $30x=-210$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{-210}{30}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-210}{30}$ and compute the same integer $x=-7$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-oper... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=30... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-7$ because $a=30\neq 0$. (Here the result is $\boxed{-7}$.) |
math-000024 | Prealgebra: Solving for a Variable | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 3x + (76) = 145 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificati... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(76)$ from both sides: $3x=69$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{69}{3}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{69}{3}$ and compute the same integer $x=23$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=23$ because $a=3\neq 0$. |
math-000025 | Elementary Algebra: Linear Equations — Verification | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 3x + (-80) = -17 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the en... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=3x+(-80)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{63}{3}$ and compute the same integer $x=21$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $ax... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=3\neq 0$. |
math-000026 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 14x + (59) = 3 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the end. | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=14x+(59)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-56}{14}$ and compute the same integer $x=-4$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works fo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-4$ because $a=14\neq 0$. (Here the result is $\boxed{-4}$.) |
math-000027 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 7x + (44) = 149 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=7x+(44)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{105}{7}$ and compute the same integer $x=15$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=7\neq 0$. |
math-000028 | Elementary Algebra: Linear Equations — Verification | 1 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 22x + (44) = 528 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the en... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(44)$ from both sides: $22x=484$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{484}{22}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{484}{22}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=22... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=22\neq 0$. |
math-000029 | Prealgebra: Solving for a Variable | 1 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 8x + (3) = 123 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(3)$ from both sides: $8x=120$.",
"Step 2: Since $8\\neq 0$, divide by $8$: $x=\\frac{120}{8}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{120}{8}$ and compute the same integer $x=15$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=8\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=8\neq 0$. (Here the result is $\boxed{15}$.) |
math-000030 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give reasoning, not just computation: Solve for $x$ and verify your result:
(a) Solve $ 7x + (-23) = -114 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verif... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=7x+(-23)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-13}$.\nBoth methods reduce the equation to $x=\\frac{-91}{7}$ and compute the same integer $x=-13$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any li... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-13$ because $a=7\neq 0$. (Here the result is $\boxed{-13}$.) |
math-000031 | Prealgebra: Solving for a Variable | 1 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-31) = 360 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=17x+(-31)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{391}{17}$ and compute the same integer $x=23$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=23$ because $a=17\neq 0$. (Here the result is $\boxed{23}$.) |
math-000032 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 6x + (43) = -59 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief veri... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=6x+(43)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-102}{6}$ and compute the same integer $x=-17$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any line... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-17$ because $a=6\neq 0$. |
math-000033 | Algebra: Affine Functions — Injectivity | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 17x + (34) = 102 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the e... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=17x+(34)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{68}{17}$ and compute the same integer $x=4$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax+... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=4$ because $a=17\neq 0$. |
math-000034 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Start by stating any domain restrictions: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-21) = 343 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief v... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=28x+(-21)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{364}{28}$ and compute the same integer $x=13$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=28\neq 0$. (Here the result is $\boxed{13}$.) |
math-000035 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 30x + (74) = 734 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the e... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=30x+(74)$. Since the slope $30\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{660}{30}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works fo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=30... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=30\neq 0$. (Here the result is $\boxed{22}$.) |
math-000036 | Elementary Algebra: Linear Equations — Verification | 1 | Solve (and briefly cross-validate): Solve for $x$ and verify your result:
(a) Solve $ 14x + (-63) = 133 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verific... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=14x+(-63)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{196}{14}$ and compute the same integer $x=14$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations wor... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=14\neq 0$. (Here the result is $\boxed{14}$.) |
math-000037 | Elementary Algebra: Linear Equations — Verification | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 26x + (-55) = 543 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-55)$ from both sides: $26x=598$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{598}{26}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{598}{26}$ and compute the same integer $x=23$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=26... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=23$ because $a=26\neq 0$. |
math-000038 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 30x + (23) = 533 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the e... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(23)$ from both sides: $30x=510$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{510}{30}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{510}{30}$ and compute the same integer $x=17$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=30... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=30\neq 0$. |
math-000039 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 11x + (48) = 158 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(48)$ from both sides: $11x=110$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{110}{11}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{10}$.\nBoth methods reduce the equation to $x=\\frac{110}{11}$ and compute the same integer $x=10$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inve... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=11... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=10$ because $a=11\neq 0$. (Here the result is $\boxed{10}$.) |
math-000040 | Elementary Algebra: Linear Equations — Verification | 1 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 29x + (-34) = -295 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at th... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-34)$ from both sides: $29x=-261$.",
"Step 2: Since $29\\neq 0$, divide by $29$: $x=\\frac{-261}{29}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-9}$.\nBoth methods reduce the equation to $x=\\frac{-261}{29}$ and compute the same integer $x=-9$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operation... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-9$ because $a=29\neq 0$. |
math-000041 | Algebra: Affine Functions — Injectivity | 1 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 4x + (8) = -92 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the en... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(8)$ from both sides: $4x=-100$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{-100}{4}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-100}{4}$ and compute the same integer $x=-25$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-ope... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=4\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-25$ because $a=4\neq 0$. (Here the result is $\boxed{-25}$.) |
math-000042 | Elementary Algebra: Linear Equations — Verification | 1 | Try to avoid pattern-matching; explain why: Solve for $x$ and verify your result:
(a) Solve $ 4x + (-61) = -57 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief ... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=4x+(-61)$. Since the slope $4\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{4}{4}$ and compute the same integer $x=1$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any li... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=4\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=4\neq 0$. |
math-000043 | Algebra: Affine Functions — Injectivity | 1 | Derive the result step-by-step: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-4) = -310 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-4)$ from both sides: $17x=-306$.",
"Step 2: Since $17\\neq 0$, divide by $17$: $x=\\frac{-306}{17}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-306}{17}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any lin... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-18$ because $a=17\neq 0$. |
math-000044 | Algebra: Affine Functions — Injectivity | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 11x + (-43) = 56 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cr... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-43)$ from both sides: $11x=99$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{99}{11}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{9}$.\nBoth methods reduce the equation to $x=\\frac{99}{11}$ and compute the same integer $x=9$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=11... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=9$ because $a=11\neq 0$. |
math-000045 | Algebra: Affine Functions — Injectivity | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 14x + (59) = 171 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(59)$ from both sides: $14x=112$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{112}{14}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{112}{14}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=14\neq 0$. (Here the result is $\boxed{8}$.) |
math-000046 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 6x + (39) = -21 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the end. | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=6x+(39)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-60}{6}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-oper... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=6\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000047 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and sanity-check: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-51) = -442 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=17x+(-51)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-23}$.\nBoth methods reduce the equation to $x=\\frac{-391}{17}$ and compute the same integer $x=-23$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-23$ because $a=17\neq 0$. (Here the result is $\boxed{-23}$.) |
math-000048 | Elementary Algebra: Linear Equations — Verification | 1 | Solve (and briefly cross-validate): Solve for $x$ and verify your result:
(a) Solve $ 18x + (-74) = -434 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verifi... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-74)$ from both sides: $18x=-360$.",
"Step 2: Since $18\\neq 0$, divide by $18$: $x=\\frac{-360}{18}$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-360}{18}$ and compute the same integer $x=-20$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any lin... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=18... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-20$ because $a=18\neq 0$. |
math-000049 | Algebra: Affine Functions — Injectivity | 1 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-78) = -58 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the end... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(-78)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{20}{5}$ and compute the same integer $x=4$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operation... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=4$ because $a=5\neq 0$. (Here the result is $\boxed{4}$.) |
math-000050 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 8x + (-37) = 107 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-ch... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-37)$ from both sides: $8x=144$.",
"Step 2: Since $8\\neq 0$, divide by $8$: $x=\\frac{144}{8}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{144}{8}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=8\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=8\neq 0$. (Here the result is $\boxed{18}$.) |
math-000051 | Algebra: Affine Functions — Injectivity | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 3x + (65) = 122 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=3x+(65)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{57}{3}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=3\neq 0$. (Here the result is $\boxed{19}$.) |
math-000052 | Algebra: Affine Functions — Injectivity | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 6x + (-77) = 37 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=6x+(-77)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{114}{6}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=6\neq 0$. |
math-000053 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 25x + (61) = -414 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(61)$ from both sides: $25x=-475$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-475}{25}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-475}{25}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=25\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000054 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Solve (and briefly cross-validate): Solve for $x$ and verify your result:
(a) Solve $ 29x + (-46) = 389 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verific... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-46)$ from both sides: $29x=435$.",
"Step 2: Since $29\\neq 0$, divide by $29$: $x=\\frac{435}{29}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{435}{29}$ and compute the same integer $x=15$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear equat... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=29\neq 0$. (Here the result is $\boxed{15}$.) |
math-000055 | Prealgebra: Solving for a Variable | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 15x + (56) = -154 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(56)$ from both sides: $15x=-210$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{-210}{15}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-210}{15}$ and compute the same integer $x=-14$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=15... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-14$ because $a=15\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000056 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 24x + (32) = -496 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-ch... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=24x+(32)$. Since the slope $24\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-22}$.\nBoth methods reduce the equation to $x=\\frac{-528}{24}$ and compute the same integer $x=-22$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=24... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-22$ because $a=24\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000057 | Elementary Algebra: Linear Equations — Verification | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 10x + (67) = -83 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(67)$ from both sides: $10x=-150$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{-150}{10}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-150}{10}$ and compute the same integer $x=-15$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=10... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=10\neq 0$. (Here the result is $\boxed{-15}$.) |
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Math-HQ-20k
A curated synthetic dataset of 20,000 English-language math reasoning examples designed for supervised fine-tuning, reasoning evaluation, and error-analysis workflows.
Each record contains a problem statement, one or more solution paths, a consistency reconciliation, a catalogue of plausible mistakes, and a short conceptual takeaway.
Dataset summary
- Records: 20,000
- Topics: 120
- Difficulty levels: 1–10
- Format: JSONL
- Language: English
- License: MIT
- File size: ~72.2 MB
What is inside
Each example includes:
id— unique row identifiertopic— topic labeldifficulty— integer difficulty rating from 1 to 10problem_statement— the task promptsolution_paths— multiple worked solution methodsreconciliation— cross-check and robustness noteserror_catalogue— plausible mistakes and why they are wrongconceptual_takeaway— short summary of the key idea
Technical specifications
| Metric | Value |
|---|---|
| Total examples | 20,000 |
| Unique topics | 120 |
| Difficulty range | 1 to 10 |
| Difficulty distribution | 2,000 examples per level |
| JSONL validity | 100% parse success |
| Top-level field completeness | 100% |
| Exact prompt duplicates | 0 |
| Average problem length | 63.1 words |
| Median problem length | 62 words |
| Average solution content | 88.5 words |
| Average reconciliation content | 68.1 words |
| Average solution paths per item | 2.04 |
| Items with 2 solution paths | 19,298 |
| Items with 3 solution paths | 702 |
| Average error entries per item | 2.88 |
Schema
{
"id": "math-000001",
"topic": "Elementary Algebra: Linear Equations — Inverse Operations",
"difficulty": 1,
"problem_statement": "...",
"solution_paths": [
{
"method_name": "Inverse Operations",
"approach": "...",
"steps": ["...", "..."],
"final_answer": "..."
}
],
"reconciliation": {
"consistency_check": "...",
"robustness_analysis": "..."
},
"error_catalogue": [
{
"error_description": "...",
"why_plausible": "...",
"why_wrong": "...",
"which_method_catches_it": "..."
}
],
"conceptual_takeaway": "..."
}
Intended use
This dataset is suitable for:
- supervised fine-tuning on step-by-step mathematical reasoning
- multi-solution reasoning behavior
- answer verification and self-check training
- error detection and correction tasks
- curriculum-style difficulty experiments
Notes
The dataset is synthetic and intentionally structured for reasoning quality.
Most examples contain more than one solution path to support comparison and verification.
The error catalogue is designed to model common student and model mistakes.
The prompts are intentionally consistent in style to make reasoning supervision easier.
Limitations
The dataset is not a collection of real-world student work.
The writing style is intentionally template-like, which may reduce natural-language diversity.
The dataset focuses on mathematical reasoning and does not aim to cover general open-domain QA.
Citation
If you use this dataset, please cite it as:
Math-HQ-20k. Synthetic JSONL dataset for supervised mathematical reasoning and verification.
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