id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-000201 | Elementary Algebra: Linear Equations — Verification | 1 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 11x + (-48) = 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 verification/cross-check... | [
{
"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 $f(x)=11x+(-48)$. Since the slope $11\\neq 0$, $f$ 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 $x=\\frac{66}{11}$ 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=11... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=6$ because $a=11\neq 0$. |
math-000202 | Algebra: Affine Functions — Injectivity | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 29x + (7) = 558 $.
(b) Substitute your value of $x$ 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 $f(x)=29x+(7)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{551}{29}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=29\neq 0$. |
math-000203 | Algebra: Affine Functions — Injectivity | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 30x + (-15) = -435 $.
(b) Substitute your value of $x$ 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 $(-15)$ from both sides: $30x=-420$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{-420}{30}$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-420}{30}$ 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=30... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-14$ because $a=30\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000204 | Prealgebra: Solving for a Variable | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 3x + (3) = -63 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=3x+(3)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Step... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-22}$.\nBoth methods reduce the equation to $x=\\frac{-66}{3}$ and compute the same integer $x=-22$; 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=3\... | 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 $a=3\neq 0$. |
math-000205 | Elementary Algebra: Linear Equations — Verification | 1 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 4x + (58) = 130 $.
(b) Substitute your value of $x$ 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 $(58)$ from both sides: $4x=72$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{72}{4}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{72}{4}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "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=4\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=4\neq 0$. (Here the result is $\boxed{18}$.) |
math-000206 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 11x + (41) = 283 $.
(b) Substitute your value of $x$ 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 $(41)$ from both sides: $11x=242$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{242}{11}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{242}{11}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "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=11... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=11\neq 0$. (Here the result is $\boxed{22}$.) |
math-000207 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 9x + (38) = 47 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=9x+(38)$. Since the slope $9\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{9}{9}$ and compute the same integer $x=1$; substitution confirms equality.",
"robustness_analysis": "Robustness note: 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=9\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=9\neq 0$. (Here the result is $\boxed{1}$.) |
math-000208 | Prealgebra: Solving for a Variable | 1 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 19x + (35) = 130 $.
(b) Substitute your value of $x$ 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 $(35)$ from both sides: $19x=95$.",
"Step 2: Since $19\\neq 0$, divide by $19$: $x=\\frac{95}{19}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5}$.\nBoth methods reduce the equation to $x=\\frac{95}{19}$ and compute the same integer $x=5$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=19... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=5$ because $a=19\neq 0$. |
math-000209 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 12x + (-1) = -229 $.
(b) Substitute your value of $x$ 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 $f(x)=12x+(-1)$. Since the slope $12\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-228}{12}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=12... | 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 $a=12\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000210 | Algebra: Affine Functions — Injectivity | 1 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 11x + (53) = -222 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-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 $f(x)=11x+(53)$. 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: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-275}{11}$ and compute the same integer $x=-25$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=11... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-25$ because $a=11\neq 0$. (Here the result is $\boxed{-25}$.) |
math-000211 | Prealgebra: Solving for a Variable | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 10x + (-7) = -147 $.
(b) Substitute your value of $x$ 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 $f(x)=10x+(-7)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-140}{10}$ and compute the same integer $x=-14$; substitution confirms equality.",
"robustness_analysis": "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=10... | 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 $a=10\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000212 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 26x + (25) = 389 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(25)$ from both sides: $26x=364$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{364}{26}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{364}{26}$ and compute the same integer $x=14$; 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=26... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=26\neq 0$. |
math-000213 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-56) = -581 $.
(b) Substitute your value of $x$ 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 $(-56)$ from both sides: $21x=-525$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{-525}{21}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-525}{21}$ and compute the same integer $x=-25$; substitution confirms equality.",
"robustness_analysis": "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=21... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-25$ because $a=21\neq 0$. |
math-000214 | Elementary Algebra: Linear Equations — Verification | 1 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 18x + (-70) = -70 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-chec... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=18x+(-70)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods reduce the equation to $x=\\frac{0}{18}$ and compute the same integer $x=0$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operation... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=18... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=0$ because $a=18\neq 0$. |
math-000215 | Algebra: Affine Functions — Injectivity | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-9) = -119 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verific... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(-9)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-22}$.\nBoth methods reduce the equation to $x=\\frac{-110}{5}$ and compute the same integer $x=-22$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations w... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | 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 $a=5\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000216 | Algebra: Affine Functions — Injectivity | 1 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 13x + (-23) = 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/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 $(-23)$ from both sides: $13x=182$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{182}{13}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{182}{13}$ and compute the same integer $x=14$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations wor... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=13... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=13\neq 0$. (Here the result is $\boxed{14}$.) |
math-000217 | Prealgebra: Solving for a Variable | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 9x + (-46) = -46 $.
(b) Substitute your value of $x$ 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 $(-46)$ from both sides: $9x=0$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{0}{9}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{0}$.\nBoth methods reduce the equation to $x=\\frac{0}{9}$ and compute the same integer $x=0$; 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=9\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=0$ because $a=9\neq 0$. |
math-000218 | Elementary Algebra: Linear Equations — Verification | 1 | Task: Solve for $x$ and verify your result:
(a) Solve $ 23x + (50) = 625 $.
(b) Substitute your value of $x$ 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 $(50)$ from both sides: $23x=575$.",
"Step 2: Since $23\\neq 0$, divide by $23$: $x=\\frac{575}{23}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{25}$.\nBoth methods reduce the equation to $x=\\frac{575}{23}$ and compute the same integer $x=25$; 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=23... | 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 $a=23\neq 0$. (Here the result is $\boxed{25}$.) |
math-000219 | Prealgebra: Solving for a Variable | 1 | Find the exact value: Solve for $x$ and verify your result:
(a) Solve $ 18x + (47) = 353 $.
(b) Substitute your value of $x$ 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 $f(x)=18x+(47)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{306}{18}$ 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=18... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=18\neq 0$. |
math-000220 | Elementary Algebra: Linear Equations — Verification | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 9x + (43) = 124 $.
(b) Substitute your value of $x$ 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 $(43)$ from both sides: $9x=81$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{81}{9}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods reduce the equation to $x=\\frac{81}{9}$ and compute the same integer $x=9$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly 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 $x=9$ because $a=9\neq 0$. (Here the result is $\boxed{9}$.) |
math-000221 | Elementary Algebra: Linear Equations — Verification | 1 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 9x + (72) = 54 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-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 $f(x)=9x+(72)$. Since the slope $9\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "Robustness note: Inverse-operations works for... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=9\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=9\neq 0$. (Here the result is $\boxed{-2}$.) |
math-000222 | Prealgebra: Solving for a Variable | 1 | Keep the final answer in boxed form: Solve for $x$ and verify your result:
(a) Solve $ 16x + (39) = -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 verifi... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=16x+(39)$. Since the slope $16\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-304}{16}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear eq... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=16\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000223 | Prealgebra: Solving for a Variable | 1 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 14x + (37) = -173 $.
(b) Substitute your value of $x$ 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 $f(x)=14x+(37)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-210}{14}$ and compute the same integer $x=-15$; 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=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=14\neq 0$. (Here the result is $\boxed{-15}$.) |
math-000224 | Algebra: Affine Functions — Injectivity | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 6x + (74) = 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 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 $(74)$ from both sides: $6x=132$.",
"Step 2: Since $6\\neq 0$, divide by $6$: $x=\\frac{132}{6}$.",
... | {
"consistency_check": "Both approaches agree after simplification. 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": "If the problem were perturbed: Inverse-operations works for any line... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=6\neq 0$. (Here the result is $\boxed{22}$.) |
math-000225 | Algebra: Affine Functions — Injectivity | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 30x + (47) = 467 $.
(b) Substitute your value of $x$ 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 $(47)$ from both sides: $30x=420$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{420}{30}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{420}{30}$ and compute the same integer $x=14$; 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=30... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=30\neq 0$. (Here the result is $\boxed{14}$.) |
math-000226 | Prealgebra: Solving for a Variable | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 26x + (33) = -227 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=26x+(33)$. Since the slope $26\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-260}{26}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=26... | 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 $a=26\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000227 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-40) = -283 $.
(b) Substitute your value of $x$ 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 $f(x)=27x+(-40)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-9}$.\nBoth methods reduce the equation to $x=\\frac{-243}{27}$ and compute the same integer $x=-9$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{-9}$.) |
math-000228 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-38) = -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-check at the end. | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=2x+(-38)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-28}{2}$ and compute the same integer $x=-14$; 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=2\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-14$ because $a=2\neq 0$. (Here the result is $\boxed{-14}$.) |
math-000229 | Algebra: Affine Functions — Injectivity | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 19x + (-61) = -99 $.
(b) Substitute your value of $x$ 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 $f(x)=19x+(-61)$. Since the slope $19\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-38}{19}$ and compute the same integer $x=-2$; 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 $x=-2$ because $a=19\neq 0$. (Here the result is $\boxed{-2}$.) |
math-000230 | Elementary Algebra: Linear Equations — Verification | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 30x + (-45) = 75 $.
(b) Substitute your value of $x$ 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 $(-45)$ from both sides: $30x=120$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{120}{30}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{120}{30}$ 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=30... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=4$ because $a=30\neq 0$. |
math-000231 | Elementary Algebra: Linear Equations — Verification | 1 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 20x + (13) = 413 $.
(b) Substitute your value of $x$ 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 $f(x)=20x+(13)$. Since the slope $20\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{400}{20}$ and compute the same integer $x=20$; 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=20... | 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 $a=20\neq 0$. (Here the result is $\boxed{20}$.) |
math-000232 | Elementary Algebra: Linear Equations — Verification | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 10x + (-44) = -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 verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=10x+(-44)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. 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 equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=10... | 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 $a=10\neq 0$. |
math-000233 | Prealgebra: Solving for a Variable | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 16x + (62) = -226 $.
(b) Substitute your value of $x$ 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 $(62)$ from both sides: $16x=-288$.",
"Step 2: Since $16\\neq 0$, divide by $16$: $x=\\frac{-288}{16}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-288}{16}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | 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 $a=16\neq 0$. |
math-000234 | Prealgebra: Solving for a Variable | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 25x + (76) = 526 $.
(b) Substitute your value of $x$ 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 $(76)$ from both sides: $25x=450$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{450}{25}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{450}{25}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=25\neq 0$. |
math-000235 | Algebra: Affine Functions — Injectivity | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-11) = -35 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(-11)$ from both sides: $24x=-24$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{-24}{24}$."... | {
"consistency_check": "The two methods are consistent and must coincide. 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": "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=24... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=24\neq 0$. (Here the result is $\boxed{-1}$.) |
math-000236 | Elementary Algebra: Linear Equations — Verification | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 8x + (58) = 114 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $f(x)=8x+(58)$. Since the slope $8\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nBoth methods reduce the equation to $x=\\frac{56}{8}$ and compute the same integer $x=7$; 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=8\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=7$ because $a=8\neq 0$. (Here the result is $\boxed{7}$.) |
math-000237 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 5x + (70) = 170 $.
(b) Substitute your value of $x$ 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 $f(x)=5x+(70)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{100}{5}$ and compute the same integer $x=20$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations work... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | 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 $a=5\neq 0$. |
math-000238 | Elementary Algebra: Linear Equations — Verification | 1 | Solve (and briefly cross-validate): Solve for $x$ and verify your result:
(a) Solve $ 9x + (-70) = 137 $.
(b) Substitute your value of $x$ 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 $(-70)$ from both sides: $9x=207$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{207}{9}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{207}{9}$ and compute the same integer $x=23$; 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\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=23$ because $a=9\neq 0$. (Here the result is $\boxed{23}$.) |
math-000239 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and sanity-check: Solve for $x$ and verify your result:
(a) Solve $ 16x + (-3) = -307 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-3)$ from both sides: $16x=-304$.",
"Step 2: Since $16\\neq 0$, divide by $16$: $x=\\frac{-304}{16}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-304}{16}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=16\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000240 | Algebra: Affine Functions — Injectivity | 1 | Start by stating any domain restrictions: Solve for $x$ and verify your result:
(a) Solve $ 9x + (48) = 156 $.
(b) Substitute your value of $x$ 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": "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: $9x=108$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{108}{9}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12}$.\nBoth methods reduce the equation to $x=\\frac{108}{9}$ and compute the same integer $x=12$; 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=9\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=12$ because $a=9\neq 0$. (Here the result is $\boxed{12}$.) |
math-000241 | Elementary Algebra: Linear Equations — Verification | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 6x + (-52) = 98 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief ... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=6x+(-52)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{25}$.\nBoth methods reduce the equation to $x=\\frac{150}{6}$ and compute the same integer $x=25$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=25$ because $a=6\neq 0$. (Here the result is $\boxed{25}$.) |
math-000242 | Prealgebra: Solving for a Variable | 1 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 4x + (78) = 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/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 $(78)$ from both sides: $4x=8$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{8}{4}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{8}{4}$ and compute the same integer $x=2$; 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=4\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=4\neq 0$. (Here the result is $\boxed{2}$.) |
math-000243 | Algebra: Affine Functions — Injectivity | 1 | Question: Solve for $x$ and verify your result:
(a) Solve $ 13x + (77) = -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-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 $(77)$ from both sides: $13x=-143$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{-143}{13}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-11}$.\nBoth methods reduce the equation to $x=\\frac{-143}{13}$ and compute the same integer $x=-11$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=13... | 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 $a=13\neq 0$. |
math-000244 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Explain each transformation: Solve for $x$ and verify your result:
(a) Solve $ 10x + (73) = 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/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 $f(x)=10x+(73)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods reduce the equation to $x=\\frac{90}{10}$ and compute the same integer $x=9$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly 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 $x=9$ because $a=10\neq 0$. (Here the result is $\boxed{9}$.) |
math-000245 | Elementary Algebra: Linear Equations — Verification | 1 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 14x + (-6) = -230 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(-6)$ from both sides: $14x=-224$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{-224}{14}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-16}$.\nBoth methods reduce the equation to $x=\\frac{-224}{14}$ and compute the same integer $x=-16$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-16$ because $a=14\neq 0$. (Here the result is $\boxed{-16}$.) |
math-000246 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 12x + (-18) = 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 $(-18)$ from both sides: $12x=96$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{96}{12}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{96}{12}$ and compute the same integer $x=8$; 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=12... | 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 $a=12\neq 0$. (Here the result is $\boxed{8}$.) |
math-000247 | Algebra: Affine Functions — Injectivity | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-32) = 388 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=28x+(-32)$. Since the slope $28\\neq 0$, $f$ 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 $x=\\frac{420}{28}$ and compute the same integer $x=15$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | 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 $a=28\neq 0$. |
math-000248 | Algebra: Affine Functions — Injectivity | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 26x + (39) = 585 $.
(b) Substitute your value of $x$ 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": "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: $26x=546$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{546}{26}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{546}{26}$ and compute the same integer $x=21$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=26... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=26\neq 0$. (Here the result is $\boxed{21}$.) |
math-000249 | Elementary Algebra: Linear Equations — Verification | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-16) = 32 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-16)$ from both sides: $24x=48$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{48}{24}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{48}{24}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Generality 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=24... | 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 $a=24\neq 0$. |
math-000250 | Prealgebra: Solving for a Variable | 1 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 12x + (-23) = 181 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=12x+(-23)$. Since the slope $12\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{204}{12}$ 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=12... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=12\neq 0$. (Here the result is $\boxed{17}$.) |
math-000251 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-64) = -88 $.
(b) Substitute your value of $x$ 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 $f(x)=2x+(-64)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-12}$.\nBoth methods reduce the equation to $x=\\frac{-24}{2}$ and compute the same integer $x=-12$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=2\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-12$ because $a=2\neq 0$. (Here the result is $\boxed{-12}$.) |
math-000252 | Prealgebra: Solving for a Variable | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 5x + (55) = 65 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(55)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{10}{5}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: 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=5\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=5\neq 0$. |
math-000253 | Elementary Algebra: Linear Equations — Verification | 1 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 24x + (72) = 432 $.
(b) Substitute your value of $x$ 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 $f(x)=24x+(72)$. Since the slope $24\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{360}{24}$ and compute the same integer $x=15$; 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=24... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=24\neq 0$. (Here the result is $\boxed{15}$.) |
math-000254 | Algebra: Affine Functions — Injectivity | 1 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 3x + (-36) = -45 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
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 $(-36)$ from both sides: $3x=-9$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{-9}{3}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-3}$.\nBoth methods reduce the equation to $x=\\frac{-9}{3}$ and compute the same integer $x=-3$; 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=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-3$ because $a=3\neq 0$. (Here the result is $\boxed{-3}$.) |
math-000255 | Prealgebra: Solving for a Variable | 1 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-6) = 219 $.
(b) Substitute your value of $x$ 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 $(-6)$ from both sides: $15x=225$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{225}{15}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{225}{15}$ and compute the same integer $x=15$; substitution confirms equality.",
"robustness_analysis": "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 $x=15$ because $a=15\neq 0$. |
math-000256 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 13x + (46) = 150 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=13x+(46)$. Since the slope $13\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{104}{13}$ and compute the same integer $x=8$; 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=13... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=13\neq 0$. (Here the result is $\boxed{8}$.) |
math-000257 | Prealgebra: Solving for a Variable | 1 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 28x + (30) = 646 $.
(b) Substitute your value of $x$ 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 $(30)$ 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": "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=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 $a=28\neq 0$. |
math-000258 | Elementary Algebra: Linear Equations — Verification | 1 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-73) = -529 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-che... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-73)$ from both sides: $24x=-456$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{-456}{24}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-456}{24}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "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=24... | 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 $a=24\neq 0$. (Here the result is $\boxed{-19}$.) |
math-000259 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 12x + (28) = 88 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-che... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(28)$ from both sides: $12x=60$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{60}{12}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5}$.\nBoth methods reduce the equation to $x=\\frac{60}{12}$ and compute the same integer $x=5$; 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=12... | 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 $a=12\neq 0$. (Here the result is $\boxed{5}$.) |
math-000260 | Algebra: Affine Functions — Injectivity | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 2x + (73) = 49 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/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 $f(x)=2x+(73)$. Since the slope $2\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-12}$.\nBoth methods reduce the equation to $x=\\frac{-24}{2}$ and compute the same integer $x=-12$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=2\... | 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 $a=2\neq 0$. (Here the result is $\boxed{-12}$.) |
math-000261 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-25) = -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/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 $f(x)=5x+(-25)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-5}{5}$ and compute the same integer $x=-1$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=5\neq 0$. (Here the result is $\boxed{-1}$.) |
math-000262 | Algebra: Affine Functions — Injectivity | 1 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 29x + (28) = -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/cross-check at the en... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=29x+(28)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-58}{29}$ and compute the same integer $x=-2$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=29\neq 0$. (Here the result is $\boxed{-2}$.) |
math-000263 | Prealgebra: Solving for a Variable | 1 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-19) = -9 $.
(b) Substitute your value of $x$ 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 ... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(-19)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{10}{5}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | 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 $a=5\neq 0$. |
math-000264 | Elementary Algebra: Linear Equations — Verification | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 30x + (68) = -52 $.
(b) Substitute your value of $x$ 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 $(68)$ from both sides: $30x=-120$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{-120}{30}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-120}{30}$ and compute the same integer $x=-4$; substitution confirms equality.",
"robustness_analysis": "Generality 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=30... | 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 $a=30\neq 0$. (Here the result is $\boxed{-4}$.) |
math-000265 | Elementary Algebra: Linear Equations — Verification | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 14x + (-50) = -120 $.
(b) Substitute your value of $x$ 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": "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 $f(x)=14x+(-50)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-70}{14}$ and compute the same integer $x=-5$; 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=14... | 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 $a=14\neq 0$. |
math-000266 | Elementary Algebra: Linear Equations — Verification | 1 | Task: Solve for $x$ and verify your result:
(a) Solve $ 28x + (74) = -430 $.
(b) Substitute your value of $x$ 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 $f(x)=28x+(74)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-504}{28}$ and compute the same integer $x=-18$; 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=28... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-18$ because $a=28\neq 0$. (Here the result is $\boxed{-18}$.) |
math-000267 | Elementary Algebra: Linear Equations — Verification | 1 | Derive the result step-by-step: Solve for $x$ and verify your result:
(a) Solve $ 11x + (42) = -68 $.
(b) Substitute your value of $x$ 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 $f(x)=11x+(42)$. Since the slope $11\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-110}{11}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=11... | 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 $a=11\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000268 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Try to avoid pattern-matching; explain why: Solve for $x$ and verify your result:
(a) Solve $ 5x + (59) = 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 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 $(59)$ from both sides: $5x=5$.",
"Step 2: Since $5\\neq 0$, divide by $5$: $x=\\frac{5}{5}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{5}{5}$ and compute the same integer $x=1$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations work... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | 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 $a=5\neq 0$. (Here the result is $\boxed{1}$.) |
math-000269 | Elementary Algebra: Linear Equations — Verification | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 16x + (-80) = 48 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cr... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-80)$ from both sides: $16x=128$.",
"Step 2: Since $16\\neq 0$, divide by $16$: $x=\\frac{128}{16}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{128}{16}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "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=16... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=16\neq 0$. |
math-000270 | Algebra: Affine Functions — Injectivity | 1 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 8x + (-73) = -49 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/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 $(-73)$ from both sides: $8x=24$.",
"Step 2: Since $8\\neq 0$, divide by $8$: $x=\\frac{24}{8}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3}$.\nBoth methods reduce the equation to $x=\\frac{24}{8}$ and compute the same integer $x=3$; substitution confirms equality.",
"robustness_analysis": "Generality 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=8\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=3$ because $a=8\neq 0$. (Here the result is $\boxed{3}$.) |
math-000271 | Algebra: Affine Functions — Injectivity | 1 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 13x + (48) = 282 $.
(b) Substitute your value of $x$ 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: $13x=234$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{234}{13}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{234}{13}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=13... | 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 $a=13\neq 0$. (Here the result is $\boxed{18}$.) |
math-000272 | Prealgebra: Solving for a Variable | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 10x + (-20) = -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).
... | [
{
"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: $10x=-10$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{-10}{10}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-10}{10}$ and compute the same integer $x=-1$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=10... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=10\neq 0$. (Here the result is $\boxed{-1}$.) |
math-000273 | Elementary Algebra: Linear Equations — Verification | 1 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-2) = -38 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(-2)$ from both sides: $2x=-36$.",
"Step 2: Since $2\\neq 0$, divide by $2$: $x=\\frac{-36}{2}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-36}{2}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any li... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=2\... | 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 $a=2\neq 0$. (Here the result is $\boxed{-18}$.) |
math-000274 | Prealgebra: Solving for a Variable | 1 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-17) = -97 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(-17)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-16}$.\nBoth methods reduce the equation to $x=\\frac{-80}{5}$ 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=5\... | 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 $a=5\neq 0$. |
math-000275 | Elementary Algebra: Linear Equations — Verification | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 26x + (7) = 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 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 $f(x)=26x+(7)$. Since the slope $26\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{26}{26}$ and compute the same integer $x=1$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=26... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=26\neq 0$. (Here the result is $\boxed{1}$.) |
math-000276 | Prealgebra: Solving for a Variable | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 27x + (61) = 304 $.
(b) Substitute your value of $x$ 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 $f(x)=27x+(61)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods reduce the equation to $x=\\frac{243}{27}$ and compute the same integer $x=9$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{9}$.) |
math-000277 | Algebra: Affine Functions — Injectivity | 1 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 13x + (-14) = -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 verifi... | [
{
"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 $f(x)=13x+(-14)$. Since the slope $13\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-65}{13}$ and compute the same integer $x=-5$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inve... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=13... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-5$ because $a=13\neq 0$. (Here the result is $\boxed{-5}$.) |
math-000278 | Elementary Algebra: Linear Equations — Verification | 1 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-70) = 91 $.
(b) Substitute your value of $x$ 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 $(-70)$ from both sides: $23x=161$.",
"Step 2: Since $23\\neq 0$, divide by $23$: $x=\\frac{161}{23}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nBoth methods reduce the equation to $x=\\frac{161}{23}$ and compute the same integer $x=7$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=23... | 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 $a=23\neq 0$. |
math-000279 | Algebra: Affine Functions — Injectivity | 1 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 14x + (41) = 307 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cr... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(41)$ from both sides: $14x=266$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{266}{14}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{266}{14}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | 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 $a=14\neq 0$. |
math-000280 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-6) = -108 $.
(b) Substitute your value of $x$ 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 $f(x)=17x+(-6)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-102}{17}$ and compute the same integer $x=-6$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=17\neq 0$. (Here the result is $\boxed{-6}$.) |
math-000281 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Explain each transformation: Solve for $x$ and verify your result:
(a) Solve $ 7x + (-3) = -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/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 $f(x)=7x+(-3)$. Since the slope $7\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-70}{7}$ and compute the same integer $x=-10$; 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=7\... | 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 $a=7\neq 0$. (Here the result is $\boxed{-10}$.) |
math-000282 | Algebra: Affine Functions — Injectivity | 1 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 18x + (42) = 366 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=18x+(42)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{324}{18}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=18... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=18$ because $a=18\neq 0$. (Here the result is $\boxed{18}$.) |
math-000283 | Prealgebra: Solving for a Variable | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 13x + (-33) = -254 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-33)$ from both sides: $13x=-221$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{-221}{13}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-221}{13}$ and compute the same integer $x=-17$; 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=13... | 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 $a=13\neq 0$. |
math-000284 | Prealgebra: Solving for a Variable | 1 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 3x + (32) = 2 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=3x+(32)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"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": "If the problem were perturbed: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=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 $a=3\neq 0$. |
math-000285 | Elementary Algebra: Linear Equations — Verification | 1 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 3x + (45) = -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... | [
{
"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 $f(x)=3x+(45)$. 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: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-57}{3}$ and compute the same integer $x=-19$; 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=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=3\neq 0$. |
math-000286 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 15x + (13) = -77 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-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 $f(x)=15x+(13)$. 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: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-90}{15}$ and compute the same integer $x=-6$; 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=15... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=15\neq 0$. (Here the result is $\boxed{-6}$.) |
math-000287 | Elementary Algebra: Linear Equations — Verification | 1 | Solve and sanity-check: Solve for $x$ and verify your result:
(a) Solve $ 14x + (46) = -178 $.
(b) Substitute your value of $x$ 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 $f(x)=14x+(46)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-16}$.\nBoth methods reduce the equation to $x=\\frac{-224}{14}$ and compute the same integer $x=-16$; 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=14... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-16$ because $a=14\neq 0$. |
math-000288 | Elementary Algebra: Linear Equations — Verification | 1 | Where appropriate, name the theorem you use: Solve for $x$ and verify your result:
(a) Solve $ 26x + (-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 brie... | [
{
"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: $26x=0$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{0}{26}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{0}$.\nBoth methods reduce the equation to $x=\\frac{0}{26}$ and compute the same integer $x=0$; substitution confirms equality.",
"robustness_analysis": "Robustness 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=26... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=0$ because $a=26\neq 0$. |
math-000289 | Algebra: Affine Functions — Injectivity | 1 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 16x + (34) = -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-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 $f(x)=16x+(34)$. Since the slope $16\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-23}$.\nBoth methods reduce the equation to $x=\\frac{-368}{16}$ and compute the same integer $x=-23$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-23$ because $a=16\neq 0$. (Here the result is $\boxed{-23}$.) |
math-000290 | Algebra: Affine Functions — Injectivity | 1 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 18x + (-70) = -268 $.
(b) Substitute your value of $x$ 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 $f(x)=18x+(-70)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-11}$.\nBoth methods reduce the equation to $x=\\frac{-198}{18}$ and compute the same integer $x=-11$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=18... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-11$ because $a=18\neq 0$. (Here the result is $\boxed{-11}$.) |
math-000291 | Prealgebra: Solving for a Variable | 1 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-3) = -141 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=23x+(-3)$. Since the slope $23\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-138}{23}$ and compute the same integer $x=-6$; 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=23... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=23\neq 0$. |
math-000292 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 10x + (76) = 286 $.
(b) Substitute your value of $x$ 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 $f(x)=10x+(76)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"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: Inve... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=10... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=10\neq 0$. |
math-000293 | Algebra: Affine Functions — Injectivity | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-51) = -591 $.
(b) Substitute your value of $x$ 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 $f(x)=27x+(-51)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"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": "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... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-20$ because $a=27\neq 0$. (Here the result is $\boxed{-20}$.) |
math-000294 | Elementary Algebra: Linear Equations — Inverse Operations | 1 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 3x + (51) = 0 $.
(b) Substitute your value of $x$ 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 $f(x)=3x+(51)$. 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: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-51}{3}$ and compute the same integer $x=-17$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-oper... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-17$ because $a=3\neq 0$. |
math-000295 | Elementary Algebra: Linear Equations — Verification | 1 | Challenge: Solve for $x$ and verify your result:
(a) Solve $ 13x + (20) = 124 $.
(b) Substitute your value of $x$ 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 $(20)$ from both sides: $13x=104$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{104}{13}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{104}{13}$ and compute the same integer $x=8$; 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=13... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=13\neq 0$. |
math-000296 | Prealgebra: Solving for a Variable | 1 | Start by stating any domain restrictions: Solve for $x$ and verify your result:
(a) Solve $ 6x + (70) = -62 $.
(b) Substitute your value of $x$ 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 $f(x)=6x+(70)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. 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-operations works for any linear equ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-22$ because $a=6\neq 0$. (Here the result is $\boxed{-22}$.) |
math-000297 | Algebra: Affine Functions — Injectivity | 1 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 28x + (38) = -438 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $f(x)=28x+(38)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-476}{28}$ and compute the same integer $x=-17$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-17$ because $a=28\neq 0$. |
math-000298 | Algebra: Affine Functions — Injectivity | 1 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 19x + (60) = -301 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=19x+(60)$. Since the slope $19\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-361}{19}$ and compute the same integer $x=-19$; 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=19... | 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 $a=19\neq 0$. |
math-000299 | Prealgebra: Solving for a Variable | 1 | Provide a rigorous solution: Solve for $x$ and verify your result:
(a) Solve $ 27x + (53) = -568 $.
(b) Substitute your value of $x$ 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 $f(x)=27x+(53)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-23}$.\nBoth methods reduce the equation to $x=\\frac{-621}{27}$ and compute the same integer $x=-23$; 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=27... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-23$ because $a=27\neq 0$. (Here the result is $\boxed{-23}$.) |
math-000300 | Prealgebra: Solving for a Variable | 1 | Derive the result step-by-step: Solve for $x$ and verify your result:
(a) Solve $ 11x + (-1) = 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... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-1)$ from both sides: $11x=187$.",
"Step 2: Since $11\\neq 0$, divide by $11$: $x=\\frac{187}{11}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{187}{11}$ and compute the same integer $x=17$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=11... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=17$ because $a=11\neq 0$. |
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