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 $\boxed{-15}$.) |
math-000058 | Elementary Algebra: Linear Equations — Verification | 1 | Find the exact value: Solve for $x$ and verify your result:
(a) Solve $ 11x + (71) = -50 $.
(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 $f(x)=11x+(71)$. Since the slope $11\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-11}$.\nBoth methods reduce the equation to $x=\\frac{-121}{11}$ and compute the same integer $x=-11$; 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=11... | 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 $a=11\neq 0$. (Here the result is $\boxed{-11}$.) |
math-000059 | Prealgebra: Solving for a Variable | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 24x + (12) = 468 $.
(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 $(12)$ from both sides: $24x=456$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{456}{24}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "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=24... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=24\neq 0$. (Here the result is $\boxed{19}$.) |
math-000060 | Elementary Algebra: Linear Equations — Verification | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 30x + (53) = 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 verification/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 $(53)$ from both sides: $30x=-30$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{-30}{30}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-30}{30}$ and compute the same integer $x=-1$; 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=30... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=30\neq 0$. (Here the result is $\boxed{-1}$.) |
math-000061 | Prealgebra: Solving for a Variable | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 10x + (-9) = -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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-9)$ from both sides: $10x=-140$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{-140}{10}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-140}{10}$ and compute the same integer $x=-14$; 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=10... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-14$ because $a=10\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000062 | Prealgebra: Solving for a Variable | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 30x + (70) = -230 $.
(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 $f(x)=30x+(70)$. Since the slope $30\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-300}{30}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=30... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=30\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000063 | Elementary Algebra: Linear Equations — Verification | 1 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 6x + (-4) = 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).
Incl... | [
{
"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 $f(x)=6x+(-4)$. 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: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{126}{6}$ and compute the same integer $x=21$; 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=6\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=6\neq 0$. (Here the result is $\boxed{21}$.) |
math-000064 | Prealgebra: Solving for a Variable | 1 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 21x + (29) = -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).
In... | [
{
"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 $f(x)=21x+(29)$. Since the slope $21\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-525}{21}$ and compute the same integer $x=-25$; 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=21... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-25$ because $a=21\neq 0$. |
math-000065 | Elementary Algebra: Linear Equations — Verification | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 23x + (47) = 346 $.
(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 $f(x)=23x+(47)$. Since the slope $23\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{299}{23}$ and compute the same integer $x=13$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equa... | [
{
"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=23... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=23\neq 0$. (Here the result is $\boxed{13}$.) |
math-000066 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Try to avoid pattern-matching; explain why: Solve for $x$ and verify your result:
(a) Solve $ 21x + (76) = -50 $.
(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 $(76)$ from both sides: $21x=-126$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{-126}{21}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-126}{21}$ and compute the same integer $x=-6$; 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=21... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=21\neq 0$. |
math-000067 | Prealgebra: Solving for a Variable | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-68) = -383 $.
(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).
... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-68)$ from both sides: $21x=-315$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{-315}{21}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-315}{21}$ 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=21... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=21\neq 0$. (Here the result is $\boxed{-15}$.) |
math-000068 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 25x + (76) = -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/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 $(76)$ from both sides: $25x=-225$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-225}{25}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-9}$.\nBoth methods reduce the equation to $x=\\frac{-225}{25}$ and compute the same integer $x=-9$; 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=25... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-9$ because $a=25\neq 0$. (Here the result is $\boxed{-9}$.) |
math-000069 | Algebra: Affine Functions — Injectivity | 1 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 28x + (46) = 382 $.
(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": "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 $f(x)=28x+(46)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. 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": "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... | Remember: 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-000070 | Elementary Algebra: Linear Equations — Verification | 1 | Keep the final answer in boxed form: Solve for $x$ and verify your result:
(a) Solve $ 28x + (60) = 424 $.
(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 $f(x)=28x+(60)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "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=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-000071 | Algebra: Affine Functions — Injectivity | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-29) = 277 $.
(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 $(-29)$ from both sides: $17x=306$.",
"Step 2: Since $17\\neq 0$, divide by $17$: $x=\\frac{306}{17}$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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": "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=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-000072 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 18x + (68) = 32 $.
(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-chec... | [
{
"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 $f(x)=18x+(68)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-36}{18}$ and compute the same integer $x=-2$; 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=18... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=18\neq 0$. (Here the result is $\boxed{-2}$.) |
math-000073 | Prealgebra: Solving for a Variable | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 29x + (26) = 548 $.
(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 $f(x)=29x+(26)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{522}{29}$ and compute the same integer $x=18$; 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=29... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=29\neq 0$. (Here the result is $\boxed{18}$.) |
math-000074 | Elementary Algebra: Linear Equations — Verification | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 22x + (3) = -19 $.
(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 $f(x)=22x+(3)$. Since the slope $22\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-22}{22}$ and compute the same integer $x=-1$; substitution confirms equality.",
"robustness_analysis": "Generality 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=22... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=22\neq 0$. (Here the result is $\boxed{-1}$.) |
math-000075 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 7x + (-65) = -212 $.
(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 $f(x)=7x+(-65)$. 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: $\\boxed{-21}$.\nBoth methods reduce the equation to $x=\\frac{-147}{7}$ and compute the same integer $x=-21$; substitution confirms equality.",
"robustness_analysis": "Robustness note: 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=7\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-21$ because $a=7\neq 0$. |
math-000076 | Algebra: Affine Functions — Injectivity | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 15x + (7) = -293 $.
(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 $f(x)=15x+(7)$. Since the slope $15\\neq 0$, $f$ 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 $x=\\frac{-300}{15}$ and compute the same integer $x=-20$; 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... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-20$ because $a=15\neq 0$. (Here the result is $\boxed{-20}$.) |
math-000077 | Prealgebra: Solving for a Variable | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 15x + (3) = -222 $.
(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 $f(x)=15x+(3)$. Since the slope $15\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-225}{15}$ and compute the same integer $x=-15$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: 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=15... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=15\neq 0$. |
math-000078 | Prealgebra: Solving for a Variable | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 25x + (44) = -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/cross-che... | [
{
"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: $25x=-100$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-100}{25}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-100}{25}$ and compute the same integer $x=-4$; 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=25... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-4$ because $a=25\neq 0$. (Here the result is $\boxed{-4}$.) |
math-000079 | Elementary Algebra: Linear Equations — Verification | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 11x + (1) = -241 $.
(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": "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 $f(x)=11x+(1)$. Since the slope $11\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-22}$.\nBoth methods reduce the equation to $x=\\frac{-242}{11}$ and compute the same integer $x=-22$; substitution confirms equality.",
"robustness_analysis": "Robustness 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=11... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-22$ because $a=11\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000080 | Elementary Algebra: Linear Equations — Verification | 1 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 11x + (-69) = -146 $.
(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 $(-69)$ from both sides: $11x=-77$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{-77}{11}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-77}{11}$ and compute the same integer $x=-7$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: 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=11... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-7$ because $a=11\neq 0$. |
math-000081 | Prealgebra: Solving for a Variable | 1 | Explain each transformation: Solve for $x$ and verify your result:
(a) Solve $ 18x + (14) = -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/cr... | [
{
"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 $f(x)=18x+(14)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-72}{18}$ and compute the same integer $x=-4$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: 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=18... | 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 $a=18\neq 0$. (Here the result is $\boxed{-4}$.) |
math-000082 | Elementary Algebra: Linear Equations — Verification | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 20x + (52) = 392 $.
(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 $(52)$ from both sides: $20x=340$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{340}{20}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{340}{20}$ and compute the same integer $x=17$; 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=20... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=20\neq 0$. |
math-000083 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Keep the final answer in boxed form: Solve for $x$ and verify your result:
(a) Solve $ 16x + (-26) = -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 verif... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-26)$ from both sides: $16x=-128$.",
"Step 2: Since $16\\neq 0$, divide by $16$: $x=\\frac{-128}{16}$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-128}{16}$ and compute the same integer $x=-8$; 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=16... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-8$ because $a=16\neq 0$. (Here the result is $\boxed{-8}$.) |
math-000084 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 11x + (-67) = 208 $.
(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).
In... | [
{
"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: $11x=275$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{275}{11}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{25}$.\nBoth methods reduce the equation to $x=\\frac{275}{11}$ and compute the same integer $x=25$; 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=11... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=25$ because $a=11\neq 0$. (Here the result is $\boxed{25}$.) |
math-000085 | Prealgebra: Solving for a Variable | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 2x + (68) = 54 $.
(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 $f(x)=2x+(68)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-14}{2}$ and compute the same integer $x=-7$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear equati... | [
{
"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\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-7$ because $a=2\neq 0$. (Here the result is $\boxed{-7}$.) |
math-000086 | Algebra: Affine Functions — Injectivity | 1 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 19x + (-56) = 115 $.
(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": "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: $19x=171$.",
"Step 2: Since $19\\neq 0$, divide by $19$: $x=\\frac{171}{19}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9}$.\nBoth methods reduce the equation to $x=\\frac{171}{19}$ and compute the same integer $x=9$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any linea... | [
{
"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=19... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=9$ because $a=19\neq 0$. |
math-000087 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 15x + (31) = 271 $.
(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 $(31)$ from both sides: $15x=240$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{240}{15}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nBoth methods reduce the equation to $x=\\frac{240}{15}$ and compute the same integer $x=16$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equa... | [
{
"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... | 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 $a=15\neq 0$. (Here the result is $\boxed{16}$.) |
math-000088 | Prealgebra: Solving for a Variable | 1 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-11) = 583 $.
(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 $f(x)=27x+(-11)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{594}{27}$ and compute the same integer $x=22$; 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=27... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=27\neq 0$. (Here the result is $\boxed{22}$.) |
math-000089 | Prealgebra: Solving for a Variable | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-6) = -40 $.
(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 $f(x)=2x+(-6)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-34}{2}$ and compute the same integer $x=-17$; 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=2\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-17$ because $a=2\neq 0$. (Here the result is $\boxed{-17}$.) |
math-000090 | Algebra: Affine Functions — Injectivity | 1 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 8x + (60) = -100 $.
(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 $f(x)=8x+(60)$. Since the slope $8\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-160}{8}$ and compute the same integer $x=-20$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations w... | [
{
"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 $x=-20$ because $a=8\neq 0$. |
math-000091 | Algebra: Affine Functions — Injectivity | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 29x + (26) = -264 $.
(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": "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 $f(x)=29x+(26)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-290}{29}$ and compute the same integer $x=-10$; 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=29... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=29\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000092 | Algebra: Affine Functions — Injectivity | 1 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-18) = -118 $.
(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": "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 $f(x)=25x+(-18)$. Since the slope $25\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-100}{25}$ and compute the same integer $x=-4$; 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=25... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-4$ because $a=25\neq 0$. (Here the result is $\boxed{-4}$.) |
math-000093 | Elementary Algebra: Linear Equations — Verification | 1 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-1) = -505 $.
(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 $(-1)$ from both sides: $28x=-504$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{-504}{28}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-504}{28}$ and compute the same integer $x=-18$; 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=28... | 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 $a=28\neq 0$. (Here the result is $\boxed{-18}$.) |
math-000094 | Elementary Algebra: Linear Equations — Verification | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-36) = 49 $.
(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 $f(x)=5x+(-36)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{85}{5}$ and compute the same integer $x=17$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equati... | [
{
"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\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=5\neq 0$. (Here the result is $\boxed{17}$.) |
math-000095 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 30x + (-33) = -603 $.
(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).
... | [
{
"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 $f(x)=30x+(-33)$. Since the slope $30\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-570}{30}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Robustness note: 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=30... | 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 $a=30\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000096 | Elementary Algebra: Linear Equations — Verification | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-7) = -307 $.
(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 $(-7)$ from both sides: $15x=-300$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{-300}{15}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-300}{15}$ and compute the same integer $x=-20$; 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=15... | 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 $a=15\neq 0$. |
math-000097 | Algebra: Affine Functions — Injectivity | 1 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 11x + (8) = -234 $.
(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).
Inc... | [
{
"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: $11x=-242$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{-242}{11}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-22}$.\nBoth methods reduce the equation to $x=\\frac{-242}{11}$ and compute the same integer $x=-22$; substitution confirms equality.",
"robustness_analysis": "Robustness note: 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=11... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-22$ because $a=11\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000098 | Algebra: Affine Functions — Injectivity | 1 | Provide a rigorous solution: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-8) = -128 $.
(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 $f(x)=15x+(-8)$. Since the slope $15\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-120}{15}$ and compute the same integer $x=-8$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equ... | [
{
"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... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-8$ because $a=15\neq 0$. (Here the result is $\boxed{-8}$.) |
math-000099 | Algebra: Affine Functions — Injectivity | 1 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-72) = -28 $.
(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 $(-72)$ from both sides: $22x=44$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{44}{22}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{44}{22}$ and compute the same integer $x=2$; 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=22... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=22\neq 0$. |
math-000100 | Prealgebra: Solving for a Variable | 1 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 15x + (58) = -212 $.
(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": "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 $f(x)=15x+(58)$. Since the slope $15\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-270}{15}$ and compute the same integer $x=-18$; 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=15... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-18$ because $a=15\neq 0$. (Here the result is $\boxed{-18}$.) |
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