id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-000301 | Algebra: Affine Functions — Injectivity | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 27x + (10) = 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-... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(10)$ from both sides: $27x=108$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{108}{27}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{108}{27}$ and compute the same integer $x=4$; 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=27... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=4$ because $a=27\neq 0$. |
math-000302 | Elementary Algebra: Linear Equations — Verification | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 18x + (-32) = 4 $.
(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 $f(x)=18x+(-32)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"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": "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=18... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=18\neq 0$. |
math-000303 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 26x + (-46) = -566 $.
(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 $(-46)$ from both sides: $26x=-520$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{-520}{26}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-520}{26}$ and compute the same integer $x=-20$; 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=26... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-20$ because $a=26\neq 0$. (Here the result is $\boxed{-20}$.) |
math-000304 | Algebra: Affine Functions — Injectivity | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-79) = -159 $.
(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/cros... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-79)$ from both sides: $20x=-80$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{-80}{20}$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-80}{20}$ and compute the same integer $x=-4$; 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=20... | 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 $a=20\neq 0$. |
math-000305 | Prealgebra: Solving for a Variable | 1 | Find the exact value: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-47) = -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).
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 $f(x)=17x+(-47)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-11}$.\nBoth methods reduce the equation to $x=\\frac{-187}{17}$ and compute the same integer $x=-11$; 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=17... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-11$ because $a=17\neq 0$. (Here the result is $\boxed{-11}$.) |
math-000306 | Algebra: Affine Functions — Injectivity | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 6x + (39) = -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 veri... | [
{
"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: $6x=-54$.",
"Step 2: Since $6\\neq 0$, divide by $6$: $x=\\frac{-54}{6}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-9}$.\nBoth methods reduce the equation to $x=\\frac{-54}{6}$ and compute the same integer $x=-9$; 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=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-9$ because $a=6\neq 0$. (Here the result is $\boxed{-9}$.) |
math-000307 | Prealgebra: Solving for a Variable | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-37) = -312 $.
(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 $f(x)=25x+(-37)$. Since the slope $25\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-11}$.\nBoth methods reduce the equation to $x=\\frac{-275}{25}$ and compute the same integer $x=-11$; 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=25... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-11$ because $a=25\neq 0$. (Here the result is $\boxed{-11}$.) |
math-000308 | Algebra: Affine Functions — Injectivity | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-19) = 116 $.
(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 $f(x)=27x+(-19)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5}$.\nBoth methods reduce the equation to $x=\\frac{135}{27}$ and compute the same integer $x=5$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=27... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=5$ because $a=27\neq 0$. (Here the result is $\boxed{5}$.) |
math-000309 | Prealgebra: Solving for a Variable | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 9x + (-15) = -33 $.
(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 $f(x)=9x+(-15)$. 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: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-18}{9}$ and compute the same integer $x=-2$; substitution confirms equality.",
"robustness_analysis": "Generality 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=9\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=9\neq 0$. |
math-000310 | Prealgebra: Solving for a Variable | 1 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-12) = -131 $.
(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 $(-12)$ from both sides: $17x=-119$.",
"Step 2: Since $17\\neq 0$, divide by $17$: $x=\\frac{-119}{17}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-119}{17}$ and compute the same integer $x=-7$; 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=17... | 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 $a=17\neq 0$. |
math-000311 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-71) = 209 $.
(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 $f(x)=20x+(-71)$. Since the slope $20\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{280}{20}$ and compute the same integer $x=14$; 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=20... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=20\neq 0$. |
math-000312 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-29) = 457 $.
(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 $f(x)=27x+(-29)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{486}{27}$ and compute the same integer $x=18$; 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=27... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=27\neq 0$. (Here the result is $\boxed{18}$.) |
math-000313 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 3x + (-11) = 22 $.
(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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-11)$ from both sides: $3x=33$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{33}{3}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{11}$.\nBoth methods reduce the equation to $x=\\frac{33}{3}$ and compute the same integer $x=11$; substitution confirms equality.",
"robustness_analysis": "Generality 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=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=11$ because $a=3\neq 0$. |
math-000314 | Elementary Algebra: Linear Equations — Verification | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 3x + (20) = 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 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 $(20)$ from both sides: $3x=72$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{72}{3}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{72}{3}$ and compute the same integer $x=24$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=3\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=3\neq 0$. (Here the result is $\boxed{24}$.) |
math-000315 | Prealgebra: Solving for a Variable | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 6x + (28) = 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 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 $(28)$ from both sides: $6x=84$.",
"Step 2: Since $6\\neq 0$, divide by $6$: $x=\\frac{84}{6}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{84}{6}$ and compute the same integer $x=14$; 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=6\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=6\neq 0$. (Here the result is $\boxed{14}$.) |
math-000316 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 24x + (4) = 580 $.
(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 $f(x)=24x+(4)$. Since the slope $24\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{576}{24}$ and compute the same integer $x=24$; 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... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=24\neq 0$. (Here the result is $\boxed{24}$.) |
math-000317 | Algebra: Affine Functions — Injectivity | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 19x + (-3) = 339 $.
(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 $(-3)$ from both sides: $19x=342$.",
"Step 2: Since $19\\neq 0$, divide by $19$: $x=\\frac{342}{19}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{342}{19}$ and compute the same integer $x=18$; 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=19... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=19\neq 0$. (Here the result is $\boxed{18}$.) |
math-000318 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Track units/moduli carefully: Solve for $x$ and verify your result:
(a) Solve $ 10x + (-57) = -267 $.
(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 $(-57)$ from both sides: $10x=-210$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{-210}{10}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-21}$.\nBoth methods reduce the equation to $x=\\frac{-210}{10}$ and compute the same integer $x=-21$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: I... | [
{
"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 $x=-21$ because $a=10\neq 0$. (Here the result is $\boxed{-21}$.) |
math-000319 | Elementary Algebra: Linear Equations — Verification | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-6) = 474 $.
(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 $(-6)$ from both sides: $20x=480$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{480}{20}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{480}{20}$ and compute the same integer $x=24$; 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=20... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=20\neq 0$. |
math-000320 | Elementary Algebra: Linear Equations — Verification | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 12x + (-42) = -78 $.
(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 $(-42)$ from both sides: $12x=-36$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{-36}{12}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-3}$.\nBoth methods reduce the equation to $x=\\frac{-36}{12}$ and compute the same integer $x=-3$; 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=12... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-3$ because $a=12\neq 0$. |
math-000321 | Algebra: Affine Functions — Injectivity | 1 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 17x + (37) = 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/cro... | [
{
"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 $f(x)=17x+(37)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods reduce the equation to $x=\\frac{0}{17}$ and compute the same integer $x=0$; substitution confirms equality.",
"robustness_analysis": "Generality note: 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=17... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=0$ because $a=17\neq 0$. |
math-000322 | Prealgebra: Solving for a Variable | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-11) = -161 $.
(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 bri... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-11)$ from both sides: $25x=-150$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-150}{25}$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-150}{25}$ and compute the same integer $x=-6$; 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=25... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=25\neq 0$. (Here the result is $\boxed{-6}$.) |
math-000323 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-16) = 329 $.
(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 $f(x)=23x+(-16)$. Since the slope $23\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{345}{23}$ and compute the same integer $x=15$; 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=23... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=23\neq 0$. |
math-000324 | Prealgebra: Solving for a Variable | 1 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-9) = -515 $.
(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: $23x=-506$.",
"Step 2: Since $23\\neq 0$, divide by $23$: $x=\\frac{-506}{23}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-22}$.\nBoth methods reduce the equation to $x=\\frac{-506}{23}$ and compute the same integer $x=-22$; 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=23... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-22$ because $a=23\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000325 | Algebra: Affine Functions — Injectivity | 1 | Carefully track domains: Solve for $x$ and verify your result:
(a) Solve $ 24x + (43) = -245 $.
(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 $f(x)=24x+(43)$. Since the slope $24\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-12}$.\nBoth methods reduce the equation to $x=\\frac{-288}{24}$ and compute the same integer $x=-12$; 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=24... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-12$ because $a=24\neq 0$. (Here the result is $\boxed{-12}$.) |
math-000326 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 4x + (64) = -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 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 $(64)$ from both sides: $4x=-92$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{-92}{4}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-23}$.\nBoth methods reduce the equation to $x=\\frac{-92}{4}$ and compute the same integer $x=-23$; 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=4\... | 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 $a=4\neq 0$. (Here the result is $\boxed{-23}$.) |
math-000327 | Prealgebra: Solving for a Variable | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 11x + (64) = 86 $.
(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 $f(x)=11x+(64)$. 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: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{22}{11}$ and compute the same integer $x=2$; 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=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 $a=11\neq 0$. |
math-000328 | Algebra: Affine Functions — Injectivity | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 20x + (25) = 525 $.
(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 $f(x)=20x+(25)$. Since the slope $20\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{25}$.\nBoth methods reduce the equation to $x=\\frac{500}{20}$ and compute the same integer $x=25$; 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=20... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=25$ because $a=20\neq 0$. |
math-000329 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 27x + (67) = 175 $.
(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: $27x=108$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{108}{27}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{108}{27}$ and compute the same integer $x=4$; 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=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{4}$.) |
math-000330 | Prealgebra: Solving for a Variable | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 9x + (36) = 153 $.
(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 $f(x)=9x+(36)$. Since the slope $9\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{117}{9}$ and compute the same integer $x=13$; 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=9\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=9\neq 0$. (Here the result is $\boxed{13}$.) |
math-000331 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 21x + (22) = 379 $.
(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": "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 $f(x)=21x+(22)$. Since the slope $21\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{357}{21}$ and compute the same integer $x=17$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=21... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=21\neq 0$. (Here the result is $\boxed{17}$.) |
math-000332 | Elementary Algebra: Linear Equations — Verification | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-60) = 255 $.
(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 $f(x)=21x+(-60)$. Since the slope $21\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. 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-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=21... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=21\neq 0$. (Here the result is $\boxed{15}$.) |
math-000333 | Elementary Algebra: Linear Equations — Verification | 1 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 13x + (-40) = 207 $.
(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 $(-40)$ from both sides: $13x=247$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{247}{13}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{247}{13}$ and compute the same integer $x=19$; 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=13... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=13\neq 0$. (Here the result is $\boxed{19}$.) |
math-000334 | Elementary Algebra: Linear Equations — Verification | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 3x + (39) = 12 $.
(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 $f(x)=3x+(39)$. 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: $\\boxed{-9}$.\nBoth methods reduce the equation to $x=\\frac{-27}{3}$ and compute the same integer $x=-9$; 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=3\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-9$ because $a=3\neq 0$. |
math-000335 | Prealgebra: Solving for a Variable | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-72) = -477 $.
(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 bri... | [
{
"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: $27x=-405$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-405}{27}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-405}{27}$ and compute the same integer $x=-15$; substitution confirms equality.",
"robustness_analysis": "Generality 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=27... | 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 $a=27\neq 0$. |
math-000336 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Explain why your operations are valid: Solve for $x$ and verify your result:
(a) Solve $ 30x + (-15) = 345 $.
(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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-15)$ from both sides: $30x=360$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{360}{30}$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12}$.\nBoth methods reduce the equation to $x=\\frac{360}{30}$ 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=30... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=12$ because $a=30\neq 0$. (Here the result is $\boxed{12}$.) |
math-000337 | Prealgebra: Solving for a Variable | 1 | Task: Solve for $x$ and verify your result:
(a) Solve $ 3x + (-77) = -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 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 $(-77)$ from both sides: $3x=-15$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{-15}{3}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-15}{3}$ and compute the same integer $x=-5$; 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=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-5$ because $a=3\neq 0$. |
math-000338 | Algebra: Affine Functions — Injectivity | 1 | Provide a rigorous solution: Solve for $x$ and verify your result:
(a) Solve $ 29x + (46) = -186 $.
(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 $f(x)=29x+(46)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-232}{29}$ and compute the same integer $x=-8$; 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=29... | 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 $a=29\neq 0$. (Here the result is $\boxed{-8}$.) |
math-000339 | Algebra: Affine Functions — Injectivity | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 8x + (31) = -161 $.
(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 $(31)$ from both sides: $8x=-192$.",
"Step 2: Since $8\\neq 0$, divide by $8$: $x=\\frac{-192}{8}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-24}$.\nBoth methods reduce the equation to $x=\\frac{-192}{8}$ and compute the same integer $x=-24$; substitution confirms equality.",
"robustness_analysis": "Generality 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=8\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-24$ because $a=8\neq 0$. (Here the result is $\boxed{-24}$.) |
math-000340 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 27x + (23) = -517 $.
(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 $(23)$ from both sides: $27x=-540$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-540}{27}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-540}{27}$ and compute the same integer $x=-20$; substitution confirms equality.",
"robustness_analysis": "Generality 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=27... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-20$ because $a=27\neq 0$. |
math-000341 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give reasoning, not just computation: Solve for $x$ and verify your result:
(a) Solve $ 6x + (-60) = 18 $.
(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 $(-60)$ from both sides: $6x=78$.",
"Step 2: Since $6\\neq 0$, divide by $6$: $x=\\frac{78}{6}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{78}{6}$ and compute the same integer $x=13$; substitution confirms equality.",
"robustness_analysis": "Generality 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=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=6\neq 0$. (Here the result is $\boxed{13}$.) |
math-000342 | Algebra: Affine Functions — Injectivity | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 5x + (2) = -73 $.
(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 $(2)$ from both sides: $5x=-75$.",
"Step 2: Since $5\\neq 0$, divide by $5$: $x=\\frac{-75}{5}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-75}{5}$ and compute the same integer $x=-15$; 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=5\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=5\neq 0$. (Here the result is $\boxed{-15}$.) |
math-000343 | Elementary Algebra: Linear Equations — Verification | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 15x + (38) = 248 $.
(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 $f(x)=15x+(38)$. 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: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{210}{15}$ and compute the same integer $x=14$; 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=15... | Key idea: 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-000344 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 3x + (57) = 69 $.
(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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(57)$ from both sides: $3x=12$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{12}{3}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{12}{3}$ and compute the same integer $x=4$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: 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=3\... | 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 $a=3\neq 0$. (Here the result is $\boxed{4}$.) |
math-000345 | Elementary Algebra: Linear Equations — Verification | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-40) = 334 $.
(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 $(-40)$ from both sides: $22x=374$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{374}{22}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{374}{22}$ and compute the same integer $x=17$; 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=22... | 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 $a=22\neq 0$. (Here the result is $\boxed{17}$.) |
math-000346 | 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 $ 24x + (-46) = 530 $.
(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 $(-46)$ from both sides: $24x=576$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{576}{24}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{576}{24}$ and compute the same integer $x=24$; 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=24... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=24\neq 0$. (Here the result is $\boxed{24}$.) |
math-000347 | Prealgebra: Solving for a Variable | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 4x + (-12) = -64 $.
(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 $(-12)$ from both sides: $4x=-52$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{-52}{4}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-13}$.\nBoth methods reduce the equation to $x=\\frac{-52}{4}$ and compute the same integer $x=-13$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations wo... | [
{
"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 $x=-13$ because $a=4\neq 0$. |
math-000348 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 19x + (66) = 522 $.
(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 ver... | [
{
"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 $f(x)=19x+(66)$. Since the slope $19\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{456}{19}$ and compute the same integer $x=24$; 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=19... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=19\neq 0$. (Here the result is $\boxed{24}$.) |
math-000349 | Prealgebra: Solving for a Variable | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 6x + (31) = 163 $.
(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": "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: $6x=132$.",
"Step 2: Since $6\\neq 0$, divide by $6$: $x=\\frac{132}{6}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{132}{6}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operat... | [
{
"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\... | 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 $a=6\neq 0$. (Here the result is $\boxed{22}$.) |
math-000350 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 19x + (-28) = -256 $.
(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 $f(x)=19x+(-28)$. Since the slope $19\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-12}$.\nBoth methods reduce the equation to $x=\\frac{-228}{19}$ and compute the same integer $x=-12$; 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=19... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-12$ because $a=19\neq 0$. (Here the result is $\boxed{-12}$.) |
math-000351 | Algebra: Affine Functions — Injectivity | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-55) = 561 $.
(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 $(-55)$ from both sides: $28x=616$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{616}{28}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{616}{28}$ 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=28... | 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 $a=28\neq 0$. (Here the result is $\boxed{22}$.) |
math-000352 | Algebra: Affine Functions — Injectivity | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 7x + (51) = 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 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 $f(x)=7x+(51)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{11}$.\nBoth methods reduce the equation to $x=\\frac{77}{7}$ and compute the same integer $x=11$; substitution confirms equality.",
"robustness_analysis": "Robustness note: 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=7\... | 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 $a=7\neq 0$. |
math-000353 | Prealgebra: Solving for a Variable | 1 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-10) = -490 $.
(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 $(-10)$ from both sides: $24x=-480$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{-480}{24}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-480}{24}$ 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=24... | 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 $a=24\neq 0$. (Here the result is $\boxed{-20}$.) |
math-000354 | Algebra: Affine Functions — Injectivity | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 7x + (30) = 79 $.
(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 $(30)$ from both sides: $7x=49$.",
"Step 2: Since $7\\neq 0$, divide by $7$: $x=\\frac{49}{7}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nBoth methods reduce the equation to $x=\\frac{49}{7}$ and compute the same integer $x=7$; substitution confirms equality.",
"robustness_analysis": "Robustness note: 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=7\... | 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 $a=7\neq 0$. (Here the result is $\boxed{7}$.) |
math-000355 | Elementary Algebra: Linear Equations — Verification | 1 | Explain why your operations are valid: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-40) = -265 $.
(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 ver... | [
{
"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 $f(x)=15x+(-40)$. Since the slope $15\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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: I... | [
{
"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... | 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 $a=15\neq 0$. (Here the result is $\boxed{-15}$.) |
math-000356 | Prealgebra: Solving for a Variable | 1 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 21x + (42) = 546 $.
(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 $(42)$ from both sides: $21x=504$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{504}{21}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{504}{21}$ and compute the same integer $x=24$; 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=21... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=21\neq 0$. (Here the result is $\boxed{24}$.) |
math-000357 | Algebra: Affine Functions — Injectivity | 1 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-43) = -263 $.
(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 $f(x)=22x+(-43)$. Since the slope $22\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-220}{22}$ and compute the same integer $x=-10$; 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=22... | 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 $a=22\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000358 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 11x + (37) = 257 $.
(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 $(37)$ from both sides: $11x=220$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{220}{11}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{220}{11}$ and compute the same integer $x=20$; 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=11... | 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 $a=11\neq 0$. (Here the result is $\boxed{20}$.) |
math-000359 | Prealgebra: Solving for a Variable | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-48) = 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... | [
{
"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: $22x=440$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{440}{22}$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{440}{22}$ and compute the same integer $x=20$; 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=22... | 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 $a=22\neq 0$. (Here the result is $\boxed{20}$.) |
math-000360 | Prealgebra: Solving for a Variable | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 16x + (34) = 194 $.
(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 $(34)$ from both sides: $16x=160$.",
"Step 2: Since $16\\neq 0$, divide by $16$: $x=\\frac{160}{16}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{10}$.\nBoth methods reduce the equation to $x=\\frac{160}{16}$ and compute the same integer $x=10$; 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=16... | 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 $a=16\neq 0$. (Here the result is $\boxed{10}$.) |
math-000361 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and sanity-check: Solve for $x$ and verify your result:
(a) Solve $ 20x + (34) = 34 $.
(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 $(34)$ from both sides: $20x=0$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{0}{20}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{0}$.\nBoth methods reduce the equation to $x=\\frac{0}{20}$ and compute the same integer $x=0$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax+b... | [
{
"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... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=0$ because $a=20\neq 0$. |
math-000362 | Prealgebra: Solving for a Variable | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 17x + (2) = -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).
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 $f(x)=17x+(2)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-34}{17}$ and compute the same integer $x=-2$; 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=17... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=17\neq 0$. (Here the result is $\boxed{-2}$.) |
math-000363 | Elementary Algebra: Linear Equations — Verification | 1 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 8x + (46) = 206 $.
(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 $f(x)=8x+(46)$. Since the slope $8\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. 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 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=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 $a=8\neq 0$. (Here the result is $\boxed{20}$.) |
math-000364 | Prealgebra: Solving for a Variable | 1 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-74) = -560 $.
(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 $(-74)$ from both sides: $27x=-486$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-486}{27}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-486}{27}$ 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=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{-18}$.) |
math-000365 | Algebra: Affine Functions — Injectivity | 1 | Explain why your operations are valid: Solve for $x$ and verify your result:
(a) Solve $ 10x + (66) = 216 $.
(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 $(66)$ from both sides: $10x=150$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{150}{10}$.",... | {
"consistency_check": "Both approaches agree after simplification. 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": "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=10... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=10\neq 0$. |
math-000366 | Algebra: Affine Functions — Injectivity | 1 | Keep the final answer in boxed form: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-36) = -316 $.
(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 $f(x)=20x+(-36)$. Since the slope $20\\neq 0$, $f$ 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 $x=\\frac{-280}{20}$ and compute the same integer $x=-14$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=20... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-14$ because $a=20\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000367 | Prealgebra: Solving for a Variable | 1 | Explain why your operations are valid: Solve for $x$ and verify your result:
(a) Solve $ 27x + (32) = -562 $.
(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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(32)$ from both sides: $27x=-594$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-594}{27}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "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=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000368 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 3x + (-40) = 20 $.
(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 $f(x)=3x+(-40)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{60}{3}$ and compute the same integer $x=20$; 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=3\... | 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 $a=3\neq 0$. |
math-000369 | Algebra: Affine Functions — Injectivity | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 20x + (29) = 409 $.
(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 $(29)$ from both sides: $20x=380$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{380}{20}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{380}{20}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "Generality 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=20... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=20\neq 0$. (Here the result is $\boxed{19}$.) |
math-000370 | Prealgebra: Solving for a Variable | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 14x + (71) = -69 $.
(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 $(71)$ from both sides: $14x=-140$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{-140}{14}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-140}{14}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: I... | [
{
"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... | 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 $a=14\neq 0$. |
math-000371 | Prealgebra: Solving for a Variable | 1 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 24x + (7) = 511 $.
(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 $f(x)=24x+(7)$. Since the slope $24\\neq 0$, $f$ 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 $x=\\frac{504}{24}$ and compute the same integer $x=21$; 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=24... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=24\neq 0$. (Here the result is $\boxed{21}$.) |
math-000372 | Prealgebra: Solving for a Variable | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 22x + (7) = 381 $.
(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 $(7)$ from both sides: $22x=374$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{374}{22}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{374}{22}$ and compute the same integer $x=17$; substitution confirms equality.",
"robustness_analysis": "Generality 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=22... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=22\neq 0$. (Here the result is $\boxed{17}$.) |
math-000373 | Elementary Algebra: Linear Equations — Verification | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 17x + (31) = -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/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 $(31)$ from both sides: $17x=-272$.",
"Step 2: Since $17\\neq 0$, divide by $17$: $x=\\frac{-272}{17}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-16}$.\nBoth methods reduce the equation to $x=\\frac{-272}{17}$ and compute the same integer $x=-16$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: I... | [
{
"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 $x=-16$ because $a=17\neq 0$. (Here the result is $\boxed{-16}$.) |
math-000374 | Algebra: Affine Functions — Injectivity | 1 | Keep the final answer in boxed form: Solve for $x$ and verify your result:
(a) Solve $ 3x + (52) = 22 $.
(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 $(52)$ from both sides: $3x=-30$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{-30}{3}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-30}{3}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=3\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000375 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 28x + (55) = 671 $.
(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 $(55)$ from both sides: $28x=616$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{616}{28}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{616}{28}$ and compute the same integer $x=22$; 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... | 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 $a=28\neq 0$. (Here the result is $\boxed{22}$.) |
math-000376 | Elementary Algebra: Linear Equations — Verification | 1 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-4) = 30 $.
(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 $f(x)=2x+(-4)$. 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": "Generality 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=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 $a=2\neq 0$. |
math-000377 | Prealgebra: Solving for a Variable | 1 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-12) = 328 $.
(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 $f(x)=17x+(-12)$. 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: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{340}{17}$ and compute the same integer $x=20$; 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... | 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 $a=17\neq 0$. |
math-000378 | Elementary Algebra: Linear Equations — Verification | 1 | Carefully track domains: Solve for $x$ and verify your result:
(a) Solve $ 6x + (4) = 142 $.
(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 $f(x)=6x+(4)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Step... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{138}{6}$ and compute the same integer $x=23$; 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=6\... | 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 $a=6\neq 0$. |
math-000379 | Prealgebra: Solving for a Variable | 1 | Give reasoning, not just computation: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-17) = 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 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 $f(x)=22x+(-17)$. Since the slope $22\\neq 0$, $f$ 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 $x=\\frac{132}{22}$ and compute the same integer $x=6$; 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=22... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=6$ because $a=22\neq 0$. |
math-000380 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-22) = 24 $.
(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 $f(x)=23x+(-22)$. Since the slope $23\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{46}{23}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=23... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=23\neq 0$. |
math-000381 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 14x + (-72) = -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 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 $f(x)=14x+(-72)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-28}{14}$ and compute the same integer $x=-2$; 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=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=14\neq 0$. |
math-000382 | Elementary Algebra: Linear Equations — Verification | 1 | Solve (and briefly cross-validate): Solve for $x$ and verify your result:
(a) Solve $ 10x + (9) = 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 verificati... | [
{
"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 $f(x)=10x+(9)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{40}{10}$ and compute the same integer $x=4$; 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=10... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=4$ because $a=10\neq 0$. |
math-000383 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 11x + (55) = 308 $.
(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 $(55)$ from both sides: $11x=253$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{253}{11}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{253}{11}$ and compute the same integer $x=23$; 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=11... | 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 $a=11\neq 0$. (Here the result is $\boxed{23}$.) |
math-000384 | Algebra: Affine Functions — Injectivity | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 2x + (52) = 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/... | [
{
"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 $f(x)=2x+(52)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{48}{2}$ and compute the same integer $x=24$; 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=2\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=2\neq 0$. (Here the result is $\boxed{24}$.) |
math-000385 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 7x + (-13) = -76 $.
(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 $(-13)$ from both sides: $7x=-63$.",
"Step 2: Since $7\\neq 0$, divide by $7$: $x=\\frac{-63}{7}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-9}$.\nBoth methods reduce the equation to $x=\\frac{-63}{7}$ and compute the same integer $x=-9$; substitution confirms equality.",
"robustness_analysis": "Generality 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=7\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-9$ because $a=7\neq 0$. |
math-000386 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 12x + (4) = 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/... | [
{
"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: $12x=12$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{12}{12}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{12}{12}$ and compute the same integer $x=1$; 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=12... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=12\neq 0$. |
math-000387 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 14x + (40) = 390 $.
(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 $f(x)=14x+(40)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{25}$.\nBoth methods reduce the equation to $x=\\frac{350}{14}$ and compute the same integer $x=25$; 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=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=25$ because $a=14\neq 0$. (Here the result is $\boxed{25}$.) |
math-000388 | Algebra: Affine Functions — Injectivity | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 11x + (-15) = -180 $.
(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 $(-15)$ from both sides: $11x=-165$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{-165}{11}$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-165}{11}$ and compute the same integer $x=-15$; 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... | 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 $a=11\neq 0$. |
math-000389 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 17x + (38) = 344 $.
(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 $(38)$ 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-000390 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 10x + (3) = 233 $.
(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 $f(x)=10x+(3)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{230}{10}$ 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=10... | 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 $a=10\neq 0$. |
math-000391 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 27x + (27) = 594 $.
(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 $f(x)=27x+(27)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{567}{27}$ and compute the same integer $x=21$; 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=27... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=27\neq 0$. (Here the result is $\boxed{21}$.) |
math-000392 | Prealgebra: Solving for a Variable | 1 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 9x + (-38) = -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).
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 $(-38)$ from both sides: $9x=-45$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{-45}{9}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-45}{9}$ and compute the same integer $x=-5$; 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=9\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-5$ because $a=9\neq 0$. (Here the result is $\boxed{-5}$.) |
math-000393 | Elementary Algebra: Linear Equations — Verification | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 8x + (77) = -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 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 $(77)$ from both sides: $8x=-200$.",
"Step 2: Since $8\\neq 0$, divide by $8$: $x=\\frac{-200}{8}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-200}{8}$ and compute the same integer $x=-25$; substitution confirms equality.",
"robustness_analysis": "Generality 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=8\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-25$ because $a=8\neq 0$. (Here the result is $\boxed{-25}$.) |
math-000394 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-37) = -13 $.
(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": "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: $24x=24$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{24}{24}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{24}{24}$ and compute the same integer $x=1$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operatio... | [
{
"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... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=24\neq 0$. |
math-000395 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 6x + (23) = 11 $.
(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 $f(x)=6x+(23)$. 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: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-12}{6}$ and compute the same integer $x=-2$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operat... | [
{
"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\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=6\neq 0$. |
math-000396 | Prealgebra: Solving for a Variable | 1 | Solve and sanity-check: Solve for $x$ and verify your result:
(a) Solve $ 7x + (10) = 66 $.
(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 $f(x)=7x+(10)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{56}{7}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax+b... | [
{
"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 $x=8$ because $a=7\neq 0$. (Here the result is $\boxed{8}$.) |
math-000397 | Prealgebra: Solving for a Variable | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 10x + (-40) = 190 $.
(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 $(-40)$ from both sides: $10x=230$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{230}{10}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{230}{10}$ and compute the same integer $x=23$; 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=10... | 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 $a=10\neq 0$. |
math-000398 | Prealgebra: Solving for a Variable | 1 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-7) = 265 $.
(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 $(-7)$ from both sides: $17x=272$.",
"Step 2: Since $17\\neq 0$, divide by $17$: $x=\\frac{272}{17}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{16}$.\nBoth methods reduce the equation to $x=\\frac{272}{17}$ and compute the same integer $x=16$; 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=17... | 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 $a=17\neq 0$. |
math-000399 | Elementary Algebra: Linear Equations — Verification | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-57) = -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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-57)$ from both sides: $20x=20$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{20}{20}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{20}{20}$ and compute the same integer $x=1$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations wo... | [
{
"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... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=20\neq 0$. (Here the result is $\boxed{1}$.) |
math-000400 | Elementary Algebra: Linear Equations — Verification | 1 | Task: Solve for $x$ and verify your result:
(a) Solve $ 26x + (61) = 269 $.
(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 $f(x)=26x+(61)$. Since the slope $26\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{208}{26}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: 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=26... | 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 $a=26\neq 0$. (Here the result is $\boxed{8}$.) |
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