id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-002001 | Precalculus: Polynomial Roots | 2 | Exercise: Find all real solutions and justify each step:
$$x^2 + (-3)x + (-70) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-3)=3$ and $uv=-7... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-7,10\\}$.\nFactoring yields roots $r_1=-7$ and $r_2=10$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=289.",
"robustness_analysis": "Robustness note: Factoring is fast when inte... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -7 and 10. (Here the result is $\boxed{\{-7,10\}$.) |
math-002002 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-19) = -318 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificati... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-19)$ from both sides: $23x=-299$.",
"Step 2: Since $23\\neq 0$, divide by $23$: $x=\\frac{-299}{23}$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-13}$.\nBoth methods reduce the equation to $x=\\frac{-299}{23}$ and compute the same integer $x=-13$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=23... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-13$ because $a=23\neq 0$. (Here the result is $\boxed{-13}$.) |
math-002003 | Elementary Algebra: Linear Equations — Verification | 2 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 3x + (-22) = -28 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(-22)$ from both sides: $3x=-6$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{-6}{3}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-6}{3}$ and compute the same integer $x=-2$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $ax... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | 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 $a=3\neq 0$. (Here the result is $\boxed{-2}$.) |
math-002004 | Elementary Algebra: Linear Equations — Verification | 2 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-33) = -271 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-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 $f(x)=17x+(-33)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-238}{17}$ and compute the same integer $x=-14$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works fo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | 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 $a=17\neq 0$. (Here the result is $\boxed{-14}$.) |
math-002005 | Algebra: Quadratics — Vieta Cross-Check | 2 | Indicate where a theorem is used: Find all real solutions and justify each step:
$$x^2 + (8)x + (15) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(8)=-8$ and $uv=15... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-5,-3\\}$.\nFactoring yields roots $r_1=-5$ and $r_2=-3$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=4.",
"robustness_analysis": "Sensitivity analysis: Factoring is fast ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -5 and -3. |
math-002006 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 21x + (14) = 392 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
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 $(14)$ from both sides: $21x=378$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{378}{21}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{18}$.\nBoth methods reduce the equation to $x=\\frac{378}{21}$ and compute the same integer $x=18$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=21... | 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 $a=21\neq 0$. (Here the result is $\boxed{18}$.) |
math-002007 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 2x + (49) = 61 $.
(b) Substitute your value of $x$ 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 $(49)$ from both sides: $2x=12$.",
"Step 2: Since $2\\neq 0$, divide by $2$: $x=\\frac{12}{2}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6}$.\nBoth methods reduce the equation to $x=\\frac{12}{2}$ and compute the same integer $x=6$; substitution confirms equality.",
"robustness_analysis": "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=2\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=6$ because $a=2\neq 0$. (Here the result is $\boxed{6}$.) |
math-002008 | Precalculus: Polynomial Roots | 2 | Warm-up: Find all real solutions and justify each step:
$$x^2 + (10)x + (-24) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(10)x+(-24)=(x-(2))(x-(-12))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-12,2\\}$.\nFactoring yields roots $r_1=2$ and $r_2=-12$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=196.",
"robustness_analysis": "If the problem were perturbed: Factori... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -12 and 2. |
math-002009 | Prealgebra: Solving for a Variable | 2 | Start by stating any domain restrictions: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-47) = -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 ... | [
{
"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: $20x=-100$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{-100}{20}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-100}{20}$ and compute the same integer $x=-5$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works f... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=20... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-5$ because $a=20\neq 0$. (Here the result is $\boxed{-5}$.) |
math-002010 | Algebra: Quadratics — Quadratic Formula | 2 | Checkpoint: Find all real solutions and justify each step:
$$x^2 + (-14)x + (-95) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-14)=14$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-5,19\\}$.\nFactoring yields roots $r_1=19$ and $r_2=-5$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=576.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -5 and 19. |
math-002011 | Prealgebra: Solving for a Variable | 2 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 24x + (57) = 513 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=24x+(57)$. Since the slope $24\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{456}{24}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=24... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=24\neq 0$. |
math-002012 | Prealgebra: Solving for a Variable | 2 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 30x + (-16) = -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 verificati... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-16)$ from both sides: $30x=-30$.",
"Step 2: Since $30\\neq 0$, divide by $30$: $x=\\frac{-30}{30}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-30}{30}$ and compute the same integer $x=-1$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equa... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=30... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=30\neq 0$. (Here the result is $\boxed{-1}$.) |
math-002013 | Prealgebra: Solving for a Variable | 2 | Task: Solve for $x$ and verify your result:
(a) Solve $ 26x + (34) = -590 $.
(b) Substitute your value of $x$ back into the 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 $(34)$ from both sides: $26x=-624$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{-624}{26}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-24}$.\nBoth methods reduce the equation to $x=\\frac{-624}{26}$ and compute the same integer $x=-24$; 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... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-24$ because $a=26\neq 0$. (Here the result is $\boxed{-24}$.) |
math-002014 | Algebra: Quadratics — Vieta Cross-Check | 2 | Make each step logically reversible (or explain if not): Find all real solutions and justify each step:
$$x^2 + (2)x + (-195) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-ch... | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(2)=-2$ and $uv=-1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-15,13\\}$.\nFactoring yields roots $r_1=-15$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=784.",
"robustness_analysis": "If the problem were perturbed: Facto... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -15 and 13. |
math-002015 | Precalculus: Polynomial Roots | 2 | Use two approaches if possible: Find all real solutions and justify each step:
$$x^2 + (-6)x + (-187) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-6)=6$ and $uv=-1... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-11,17\\}$.\nFactoring yields roots $r_1=17$ and $r_2=-11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=784.",
"robustness_analysis": "Generality... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -11 and 17. |
math-002016 | Algebra: Quadratics — Vieta Cross-Check | 2 | Solve with verification: Find all real solutions and justify each step:
$$x^2 + (-1)x + (-156) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-1)=1$ and $uv=-1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-12,13\\}$.\nFactoring yields roots $r_1=-12$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=625.",
"robustness_analysis": "Robustness note: Factoring is fast w... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -12 and 13. |
math-002017 | Prealgebra: Solving for a Variable | 2 | Find the exact value: Solve for $x$ and verify your result:
(a) Solve $ 20x + (76) = 296 $.
(b) Substitute your value of $x$ back into the 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 $(76)$ from both sides: $20x=220$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{220}{20}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{11}$.\nBoth methods reduce the equation to $x=\\frac{220}{20}$ and compute the same integer $x=11$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works fo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=20... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=11$ because $a=20\neq 0$. (Here the result is $\boxed{11}$.) |
math-002018 | Elementary Algebra: Linear Equations — Verification | 2 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-23) = -273 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=25x+(-23)$. Since the slope $25\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-250}{25}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=25\neq 0$. (Here the result is $\boxed{-10}$.) |
math-002019 | Algebra: Quadratics — Factoring + Zero Product | 2 | Derive the result step-by-step: Find all real solutions and justify each step:
$$x^2 + (-29)x + (204) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(-29)x+(204)=(x-(17))(x-(12))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{12,17\\}$.\nFactoring yields roots $r_1=17$ and $r_2=12$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=25.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 12 and 17. (Here the result is $\boxed{\{12,17\}$.) |
math-002020 | Prealgebra: Solving for a Variable | 2 | Provide a rigorous solution: Solve for $x$ and verify your result:
(a) Solve $ 28x + (41) = -127 $.
(b) Substitute your value of $x$ back into the 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 $(41)$ from both sides: $28x=-168$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{-168}{28}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-168}{28}$ and compute the same integer $x=-6$; 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=28... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=28\neq 0$. |
math-002021 | Prealgebra: Solving for a Variable | 2 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-34) = -384 $.
(b) Substitute your value of $x$ back into the 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 $(-34)$ from both sides: $25x=-350$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-350}{25}$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-14}$.\nBoth methods reduce the equation to $x=\\frac{-350}{25}$ and compute the same integer $x=-14$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | 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 $a=25\neq 0$. (Here the result is $\boxed{-14}$.) |
math-002022 | Elementary Algebra: Linear Equations — Verification | 2 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 14x + (27) = 55 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Incl... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(27)$ from both sides: $14x=28$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{28}{14}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{28}{14}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for 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=14... | 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 $a=14\neq 0$. |
math-002023 | Algebra: Quadratics — Factoring + Zero Product | 2 | Question: Find all real solutions and justify each step:
$$x^2 + (22)x + (85) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=22$ and $c=85$,... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-17,-5\\}$.\nFactoring yields roots $r_1=-5$ and $r_2=-17$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=144.",
"robustness_analysis": "Robustness note: Factoring is fast when in... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -17 and -5. (Here the result is $\boxed{\{-17,-5\}$.) |
math-002024 | Algebra: Quadratics — Vieta Cross-Check | 2 | Explain each transformation: Find all real solutions and justify each step:
$$x^2 + (19)x + (18) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(19)=-19$ and $uv=... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-18,-1\\}$.\nFactoring yields roots $r_1=-1$ and $r_2=-18$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=289.",
"robustness_analysis": "Sensitivity analysis: Factoring is fast wh... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -18 and -1. |
math-002025 | Algebra: Quadratics — Quadratic Formula | 2 | Write the solution set clearly: Find all real solutions and justify each step:
$$x^2 + (21)x + (54) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(21)x+(54)=(x-(-3))(x-(-18))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-18,-3\\}$.\nFactoring yields roots $r_1=-3$ and $r_2=-18$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=225.",
"robustness_analysis": "Generality note: Factoring is fast when in... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -18 and -3. (Here the result is $\boxed{\{-18,-3\}$.) |
math-002026 | Precalculus: Polynomial Roots | 2 | Answer using clear logical steps: Find all real solutions and justify each step:
$$x^2 + (16)x + (60) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(16)x+(60)=(x-(-6))(x-(-10))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-10,-6\\}$.\nFactoring yields roots $r_1=-6$ and $r_2=-10$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=16.",
"robustness_analysis": "If the problem were perturbed: Factor... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -10 and -6. (Here the result is $\boxed{\{-10,-6\}$.) |
math-002027 | Prealgebra: Solving for a Variable | 2 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-66) = -484 $.
(b) Substitute your value of $x$ back into the 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 $(-66)$ from both sides: $22x=-418$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{-418}{22}$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-418}{22}$ and compute the same integer $x=-19$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=22... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=22\neq 0$. (Here the result is $\boxed{-19}$.) |
math-002028 | Prealgebra: Solving for a Variable | 2 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-68) = -215 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-68)$ from both sides: $21x=-147$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{-147}{21}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-147}{21}$ and compute the same integer $x=-7$; 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=21... | 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 $a=21\neq 0$. (Here the result is $\boxed{-7}$.) |
math-002029 | Precalculus: Polynomial Roots | 2 | Write the solution set clearly: Find all real solutions and justify each step:
$$x^2 + (21)x + (68) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(21)=-21$ and $uv=... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-17,-4\\}$.\nFactoring yields roots $r_1=-4$ and $r_2=-17$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=169.",
"robustness_analysis": "Robustness note: Factoring is fast w... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -17 and -4. (Here the result is $\boxed{\{-17,-4\}$.) |
math-002030 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Give reasoning, not just computation: Solve for $x$ and verify your result:
(a) Solve $ 26x + (44) = -112 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verif... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=26x+(44)$. Since the slope $26\\neq 0$, $f$ 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 $x=\\frac{-156}{26}$ and compute the same integer $x=-6$; 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=26... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=26\neq 0$. (Here the result is $\boxed{-6}$.) |
math-002031 | Prealgebra: Solving for a Variable | 2 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-9) = -429 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-9)$ from both sides: $21x=-420$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{-420}{21}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-420}{21}$ and compute the same integer $x=-20$; 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=21... | 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 $a=21\neq 0$. |
math-002032 | Precalculus: Polynomial Roots | 2 | Task: Find all real solutions and justify each step:
$$x^2 + (-14)x + (33) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-14)=14$ and $uv=... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{3,11\\}$.\nFactoring yields roots $r_1=11$ and $r_2=3$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=64.",
"robustness_analysis": "If the problem were perturbed: Factoring ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 3 and 11. |
math-002033 | Prealgebra: Solving for a Variable | 2 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 16x + (-32) = -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 verificati... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-32)$ from both sides: $16x=-272$.",
"Step 2: Since $16\\neq 0$, divide by $16$: $x=\\frac{-272}{16}$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-272}{16}$ and compute the same integer $x=-17$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any lin... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-17$ because $a=16\neq 0$. (Here the result is $\boxed{-17}$.) |
math-002034 | Algebra: Quadratics — Factoring + Zero Product | 2 | Problem: Find all real solutions and justify each step:
$$x^2 + (-4)x + (-5) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(-4)x+(-5)=(x-(5))(x-(-1))$ by expansion.",
"Step 2: By the zero-product property, a prod... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-1,5\\}$.\nFactoring yields roots $r_1=5$ and $r_2=-1$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=36.",
"robustness_analysis": "Sensitivity analysis: Factoring is fast w... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -1 and 5. |
math-002035 | Elementary Algebra: Linear Equations — Verification | 2 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-28) = -56 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-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 $(-28)$ from both sides: $28x=-28$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{-28}{28}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-28}{28}$ and compute the same integer $x=-1$; 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=28... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=28\neq 0$. (Here the result is $\boxed{-1}$.) |
math-002036 | Algebra: Affine Functions — Injectivity | 2 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 20x + (18) = -22 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-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 $f(x)=20x+(18)$. Since the slope $20\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-40}{20}$ 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=20... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-2$ because $a=20\neq 0$. (Here the result is $\boxed{-2}$.) |
math-002037 | Algebra: Quadratics — Vieta Cross-Check | 2 | Show all reasoning: Find all real solutions and justify each step:
$$x^2 + (-15)x + (54) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=-15$ and $c=54$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{6,9\\}$.\nFactoring yields roots $r_1=9$ and $r_2=6$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=9.",
"robustness_analysis": "Robustness note: Factoring is fast when integer ro... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 6 and 9. |
math-002038 | Prealgebra: Solving for a Variable | 2 | Checkpoint: Solve for $x$ and verify your result:
(a) Solve $ 26x + (-53) = 207 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-53)$ from both sides: $26x=260$.",
"Step 2: Since $26\\neq 0$, divide by $26$: $x=\\frac{260}{26}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "If the problem were perturbed: Inverse-opera... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=26... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=10$ because $a=26\neq 0$. (Here the result is $\boxed{10}$.) |
math-002039 | Algebra: Quadratics — Factoring + Zero Product | 2 | Keep the final answer in boxed form: Find all real solutions and justify each step:
$$x^2 + (-1)x + (-182) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=-1$ and $c=-182... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-13,14\\}$.\nFactoring yields roots $r_1=-13$ and $r_2=14$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=729.",
"robustness_analysis": "Sensitivity analysis: Factoring is f... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -13 and 14. (Here the result is $\boxed{\{-13,14\}$.) |
math-002040 | Algebra: Affine Functions — Injectivity | 2 | State any required conditions first: Solve for $x$ and verify your result:
(a) Solve $ 5x + (60) = -15 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verifica... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=5x+(60)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-15}$.\nBoth methods reduce the equation to $x=\\frac{-75}{5}$ and compute the same integer $x=-15$; substitution confirms equality.",
"robustness_analysis": "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\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-15$ because $a=5\neq 0$. (Here the result is $\boxed{-15}$.) |
math-002041 | Algebra: Quadratics — Vieta Cross-Check | 2 | Provide a rigorous solution: Find all real solutions and justify each step:
$$x^2 + (17)x + (-18) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(17)x+(-18)=(x-(1))(x-(-18))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-18,1\\}$.\nFactoring yields roots $r_1=1$ and $r_2=-18$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=361.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -18 and 1. |
math-002042 | Algebra: Quadratics — Quadratic Formula | 2 | Show all reasoning: Find all real solutions and justify each step:
$$x^2 + (23)x + (120) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(23)=-23$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-15,-8\\}$.\nFactoring yields roots $r_1=-8$ and $r_2=-15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=49.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -15 and -8. (Here the result is $\boxed{\{-15,-8\}$.) |
math-002043 | Algebra: Quadratics — Vieta Cross-Check | 2 | Carefully track domains: Find all real solutions and justify each step:
$$x^2 + (-18)x + (-40) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-18)=18$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-2,20\\}$.\nFactoring yields roots $r_1=20$ and $r_2=-2$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=484.",
"robustness_analysis": "If the probl... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -2 and 20. |
math-002044 | Elementary Algebra: Linear Equations — Verification | 2 | Solve and then verify: Solve for $x$ and verify your result:
(a) Solve $ 22x + (36) = 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-ch... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(36)$ from both sides: $22x=88$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{88}{22}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods reduce the equation to $x=\\frac{88}{22}$ and compute the same integer $x=4$; 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=22... | 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 $a=22\neq 0$. (Here the result is $\boxed{4}$.) |
math-002045 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 7x + (80) = 234 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(80)$ from both sides: $7x=154$.",
"Step 2: Since $7\\neq 0$, divide by $7$: $x=\\frac{154}{7}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{154}{7}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inver... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=7\neq 0$. (Here the result is $\boxed{22}$.) |
math-002046 | Algebra: Quadratics — Quadratic Formula | 2 | Complete the analysis: Find all real solutions and justify each step:
$$x^2 + (12)x + (-108) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(12)=-12$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-18,6\\}$.\nFactoring yields roots $r_1=6$ and $r_2=-18$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=576.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -18 and 6. (Here the result is $\boxed{\{-18,6\}$.) |
math-002047 | Algebra: Affine Functions — Injectivity | 2 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 29x + (13) = -16 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(13)$ from both sides: $29x=-29$.",
"Step 2: Since $29\\neq 0$, divide by $29$: $x=\\frac{-29}{29}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-29}{29}$ and compute the same integer $x=-1$; 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=29... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=29\neq 0$. (Here the result is $\boxed{-1}$.) |
math-002048 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 29x + (-48) = 677 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verifica... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=29x+(-48)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{25}$.\nBoth methods reduce the equation to $x=\\frac{725}{29}$ and compute the same integer $x=25$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works fo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | 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 $a=29\neq 0$. (Here the result is $\boxed{25}$.) |
math-002049 | Elementary Algebra: Linear Equations — Verification | 2 | Question: Solve for $x$ and verify your result:
(a) Solve $ 23x + (-2) = 573 $.
(b) Substitute your value of $x$ back into the 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 $(-2)$ from both sides: $23x=575$.",
"Step 2: Since $23\\neq 0$, divide by $23$: $x=\\frac{575}{23}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. 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 for any linear equa... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=23... | 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 $a=23\neq 0$. |
math-002050 | Algebra: Quadratics — Quadratic Formula | 2 | Complete the analysis: Find all real solutions and justify each step:
$$x^2 + (4)x + (-12) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(4)=-4$ and $uv=-1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-6,2\\}$.\nFactoring yields roots $r_1=2$ and $r_2=-6$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=64.",
"robustness_analysis": "Sensitivity analysis: Factoring is fast w... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -6 and 2. (Here the result is $\boxed{\{-6,2\}$.) |
math-002051 | Algebra: Affine Functions — Injectivity | 2 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 13x + (-79) = -352 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-79)$ from both sides: $13x=-273$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{-273}{13}$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-21}$.\nBoth methods reduce the equation to $x=\\frac{-273}{13}$ and compute the same integer $x=-21$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=13... | 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 $a=13\neq 0$. (Here the result is $\boxed{-21}$.) |
math-002052 | Algebra: Quadratics — Vieta Cross-Check | 2 | Prompt: Find all real solutions and justify each step:
$$x^2 + (29)x + (198) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(29)x+(198)=(x-(-11))(x-(-18))$ by expansion.",
"Step 2: By the zero-product property, a ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-18,-11\\}$.\nFactoring yields roots $r_1=-11$ and $r_2=-18$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=49.",
"robustness_analysis": "Robustness note: F... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -18 and -11. |
math-002053 | Algebra: Quadratics — Factoring + Zero Product | 2 | Answer with a short justification: Find all real solutions and justify each step:
$$x^2 + (4)x + (-165) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(4)x+(-165)=(x-(-15))(x-(11))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-15,11\\}$.\nFactoring yields roots $r_1=-15$ and $r_2=11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=676.",
"robustness_analysis": "Generality... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -15 and 11. (Here the result is $\boxed{\{-15,11\}$.) |
math-002054 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Complete the analysis: Solve for $x$ and verify your result:
(a) Solve $ 6x + (20) = 152 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=6x+(20)$. Since the slope $6\\neq 0$, $f$ 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 $x=\\frac{132}{6}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operat... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=22$ because $a=6\neq 0$. (Here the result is $\boxed{22}$.) |
math-002055 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-60) = -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/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 $(-60)$ from both sides: $21x=-126$.",
"Step 2: Since $21\\neq 0$, divide by $21$: $x=\\frac{-126}{21}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-6}$.\nBoth methods reduce the equation to $x=\\frac{-126}{21}$ and compute the same integer $x=-6$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works f... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=21... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=21\neq 0$. |
math-002056 | Elementary Algebra: Linear Equations — Verification | 2 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 14x + (60) = 200 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the end... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=14x+(60)$. 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: $\\boxed{10}$.\nBoth methods reduce the equation to $x=\\frac{140}{14}$ and compute the same integer $x=10$; 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=14... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=10$ because $a=14\neq 0$. |
math-002057 | Precalculus: Polynomial Roots | 2 | Exercise: Find all real solutions and justify each step:
$$x^2 + (29)x + (210) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(29)x+(210)=(x-(-15))(x-(-14))$ by expansion.",
"Step 2: By the zero-product property, a ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-15,-14\\}$.\nFactoring yields roots $r_1=-15$ and $r_2=-14$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -15 and -14. (Here the result is $\boxed{\{-15,-14\}$.) |
math-002058 | Prealgebra: Solving for a Variable | 2 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 29x + (-33) = 373 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=29x+(-33)$. Since the slope $29\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{406}{29}$ 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=29... | 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 $a=29\neq 0$. (Here the result is $\boxed{14}$.) |
math-002059 | Prealgebra: Solving for a Variable | 2 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-37) = 419 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief ve... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-37)$ from both sides: $24x=456$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{456}{24}$."... | {
"consistency_check": "Both approaches agree after simplification. 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": "Sensitivity analysis: Inverse-operations works for any linear equat... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=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 $a=24\neq 0$. |
math-002060 | Prealgebra: Solving for a Variable | 2 | Make each step logically reversible (or explain if not): Solve for $x$ and verify your result:
(a) Solve $ 18x + (-73) = 287 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
In... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-73)$ from both sides: $18x=360$.",
"Step 2: Since $18\\neq 0$, divide by $18$: $x=\\frac{360}{18}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{360}{18}$ and compute the same integer $x=20$; substitution confirms equality.",
"robustness_analysis": "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 $x=20$ because $a=18\neq 0$. |
math-002061 | Elementary Algebra: Linear Equations — Verification | 2 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 2x + (58) = 20 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/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 $f(x)=2x+(58)$. 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: $\\boxed{-19}$.\nBoth methods reduce the equation to $x=\\frac{-38}{2}$ and compute the same integer $x=-19$; 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\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-19$ because $a=2\neq 0$. (Here the result is $\boxed{-19}$.) |
math-002062 | Prealgebra: Solving for a Variable | 2 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 14x + (53) = -171 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross... | [
{
"method_name": "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 $f(x)=14x+(53)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"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 $a=14\neq 0$. |
math-002063 | Algebra: Quadratics — Quadratic Formula | 2 | Proceed methodically: Find all real solutions and justify each step:
$$x^2 + (32)x + (255) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(32)=-32$ and $uv=... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-17,-15\\}$.\nFactoring yields roots $r_1=-15$ and $r_2=-17$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=4.",
"robustness_analysis": "Robustness note: Factoring is fast when in... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -17 and -15. (Here the result is $\boxed{\{-17,-15\}$.) |
math-002064 | Prealgebra: Solving for a Variable | 2 | Proceed methodically: Solve for $x$ and verify your result:
(a) Solve $ 28x + (37) = 429 $.
(b) Substitute your value of $x$ back into the 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 $(37)$ from both sides: $28x=392$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{392}{28}$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{392}{28}$ and compute the same integer $x=14$; 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 $x=14$ because $a=28\neq 0$. (Here the result is $\boxed{14}$.) |
math-002065 | Algebra: Quadratics — Quadratic Formula | 2 | Provide both a computational and a conceptual explanation: Find all real solutions and justify each step:
$$x^2 + (-2)x + (-195) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross... | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=-2$ and $c=-195... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-13,15\\}$.\nFactoring yields roots $r_1=-13$ and $r_2=15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=784.",
"robustness_analysis": "If the problem were perturbed: Facto... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -13 and 15. (Here the result is $\boxed{\{-13,15\}$.) |
math-002066 | Algebra: Quadratics — Factoring + Zero Product | 2 | Provide a rigorous solution: Find all real solutions and justify each step:
$$x^2 + (-15)x + (0) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=-15$ and $c=0$,... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{0,15\\}$.\nFactoring yields roots $r_1=0$ and $r_2=15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=225.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 0 and 15. (Here the result is $\boxed{\{0,15\}$.) |
math-002067 | Prealgebra: Solving for a Variable | 2 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 29x + (11) = -221 $.
(b) Substitute your value of $x$ back into the 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 $(11)$ from both sides: $29x=-232$.",
"Step 2: Since $29\\neq 0$, divide by $29$: $x=\\frac{-232}{29}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-8}$.\nBoth methods reduce the equation to $x=\\frac{-232}{29}$ and compute the same integer $x=-8$; substitution confirms equality.",
"robustness_analysis": "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=29... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-8$ because $a=29\neq 0$. (Here the result is $\boxed{-8}$.) |
math-002068 | Prealgebra: Solving for a Variable | 2 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 19x + (65) = -410 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=19x+(65)$. Since the slope $19\\neq 0$, $f$ 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 $x=\\frac{-475}{19}$ 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=19... | 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 $a=19\neq 0$. (Here the result is $\boxed{-25}$.) |
math-002069 | Algebra: Quadratics — Quadratic Formula | 2 | Give a theorem-based solution: Find all real solutions and justify each step:
$$x^2 + (-9)x + (8) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=-9$ and $c=8$, ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{1,8\\}$.\nFactoring yields roots $r_1=8$ and $r_2=1$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=49.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 1 and 8. |
math-002070 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 14x + (-76) = -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 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 $f(x)=14x+(-76)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{14}{14}$ 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=14... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=14\neq 0$. (Here the result is $\boxed{1}$.) |
math-002071 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Start by stating any domain restrictions: Solve for $x$ and verify your result:
(a) Solve $ 4x + (11) = -81 $.
(b) Substitute your value of $x$ back into the 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 $(11)$ from both sides: $4x=-92$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{-92}{4}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-23}$.\nBoth methods reduce the equation to $x=\\frac{-92}{4}$ and compute the same integer $x=-23$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-oper... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=4\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-23$ because $a=4\neq 0$. (Here the result is $\boxed{-23}$.) |
math-002072 | Precalculus: Polynomial Roots | 2 | Write the solution set clearly: Find all real solutions and justify each step:
$$x^2 + (-7)x + (-260) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-7)=7$ and $uv=-2... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-13,20\\}$.\nFactoring yields roots $r_1=-13$ and $r_2=20$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1089.",
"robustness_analysis": "Robustnes... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -13 and 20. |
math-002073 | Algebra: Quadratics — Factoring + Zero Product | 2 | Explain why your operations are valid: Find all real solutions and justify each step:
$$x^2 + (-14)x + (-32) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(-14)x+(-32)=(x-(16))(x-(-2))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-2,16\\}$.\nFactoring yields roots $r_1=16$ and $r_2=-2$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=324.",
"robustness_analysis": "Generality note: Factoring is fast whe... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -2 and 16. |
math-002074 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-3) = 53 $.
(b) Substitute your value of $x$ back into the 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 $(-3)$ from both sides: $28x=56$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{56}{28}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{56}{28}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | 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 $a=28\neq 0$. |
math-002075 | Precalculus: Polynomial Roots | 2 | Explain what is being counted/optimized: Find all real solutions and justify each step:
$$x^2 + (-32)x + (247) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-32)=32$ and $uv=... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{13,19\\}$.\nFactoring yields roots $r_1=19$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=36.",
"robustness_analysis": "Sensitivity analysis: Factoring is fast when ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 13 and 19. |
math-002076 | Algebra: Affine Functions — Injectivity | 2 | Provide a rigorous solution: Solve for $x$ and verify your result:
(a) Solve $ 26x + (-22) = 342 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=26x+(-22)$. Since the slope $26\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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-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... | 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 $a=26\neq 0$. (Here the result is $\boxed{14}$.) |
math-002077 | Elementary Algebra: Linear Equations — Verification | 2 | Task: Solve for $x$ and verify your result:
(a) Solve $ 7x + (22) = 155 $.
(b) Substitute your value of $x$ back into the 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 $(22)$ from both sides: $7x=133$.",
"Step 2: Since $7\\neq 0$, divide by $7$: $x=\\frac{133}{7}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{19}$.\nBoth methods reduce the equation to $x=\\frac{133}{7}$ and compute the same integer $x=19$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=7\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=7\neq 0$. |
math-002078 | Precalculus: Polynomial Roots | 2 | Explain each transformation: Find all real solutions and justify each step:
$$x^2 + (10)x + (-56) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=10$ and $c=-56$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-14,4\\}$.\nFactoring yields roots $r_1=-14$ and $r_2=4$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=324.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -14 and 4. (Here the result is $\boxed{\{-14,4\}$.) |
math-002079 | Algebra: Affine Functions — Injectivity | 2 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 18x + (-65) = -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 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 $f(x)=18x+(-65)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nBoth methods reduce the equation to $x=\\frac{18}{18}$ and compute the same integer $x=1$; 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=18... | 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 $a=18\neq 0$. |
math-002080 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 10x + (9) = 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 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 $f(x)=10x+(9)$. Since the slope $10\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nBoth methods reduce the equation to $x=\\frac{70}{10}$ and compute the same integer $x=7$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=10... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=7$ because $a=10\neq 0$. |
math-002081 | Algebra: Quadratics — Quadratic Formula | 2 | Where appropriate, name the theorem you use: Find all real solutions and justify each step:
$$x^2 + (1)x + (-342) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the e... | [
{
"method_name": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(1)x+(-342)=(x-(18))(x-(-19))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-19,18\\}$.\nFactoring yields roots $r_1=18$ and $r_2=-19$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1369.",
"robustness_analysis": "Generality note: F... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -19 and 18. |
math-002082 | Precalculus: Polynomial Roots | 2 | Solve and sanity-check: Find all real solutions and justify each step:
$$x^2 + (-1)x + (-72) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-1)=1$ and $uv=-7... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-8,9\\}$.\nFactoring yields roots $r_1=-8$ and $r_2=9$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=289.",
"robustness_analysis": "Robustness note: Factor... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Takeaway: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -8 and 9. (Here the result is $\boxed{\{-8,9\}$.) |
math-002083 | Algebra: Quadratics — Factoring + Zero Product | 2 | Track units/moduli carefully: Find all real solutions and justify each step:
$$x^2 + (7)x + (-198) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(7)=-7$ and $uv=-1... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-18,11\\}$.\nFactoring yields roots $r_1=11$ and $r_2=-18$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=841.",
"robustness_analysis": "Generality note: Factoring is fast when in... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -18 and 11. (Here the result is $\boxed{\{-18,11\}$.) |
math-002084 | Prealgebra: Solving for a Variable | 2 | Give reasoning, not just computation: Solve for $x$ and verify your result:
(a) Solve $ 17x + (-74) = -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 veri... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=17x+(-74)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-85}{17}$ and compute the same integer $x=-5$; 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=17... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-5$ because $a=17\neq 0$. |
math-002085 | Elementary Algebra: Linear Equations — Verification | 2 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 10x + (16) = -154 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(16)$ from both sides: $10x=-170$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{-170}{10}$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-17}$.\nBoth methods reduce the equation to $x=\\frac{-170}{10}$ and compute the same integer $x=-17$; 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 $x=-17$ because $a=10\neq 0$. (Here the result is $\boxed{-17}$.) |
math-002086 | Elementary Algebra: Linear Equations — Verification | 2 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-24) = -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 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 $f(x)=22x+(-24)$. Since the slope $22\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-154}{22}$ and compute the same integer $x=-7$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=22... | 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 $a=22\neq 0$. (Here the result is $\boxed{-7}$.) |
math-002087 | Prealgebra: Solving for a Variable | 2 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-74) = -317 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificati... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-74)$ from both sides: $27x=-243$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-243}{27}$... | {
"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-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=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{-9}$.) |
math-002088 | Elementary Algebra: Linear Equations — Verification | 2 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-56) = -296 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificati... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-56)$ from both sides: $15x=-240$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{-240}{15}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-16}$.\nBoth methods reduce the equation to $x=\\frac{-240}{15}$ and compute the same integer $x=-16$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=15... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-16$ because $a=15\neq 0$. |
math-002089 | Precalculus: Polynomial Roots | 2 | Solve (and briefly cross-validate): Find all real solutions and justify each step:
$$x^2 + (22)x + (57) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=22$ and $c=57$,... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-19,-3\\}$.\nFactoring yields roots $r_1=-19$ and $r_2=-3$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=256.",
"robustness_analysis": "Sensitivity analysi... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Key idea: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -19 and -3. (Here the result is $\boxed{\{-19,-3\}$.) |
math-002090 | Elementary Algebra: Linear Equations — Verification | 2 | Be explicit about assumptions: Solve for $x$ and verify your result:
(a) Solve $ 6x + (37) = 127 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=6x+(37)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{90}{6}$ and compute the same integer $x=15$; 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=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=6\neq 0$. |
math-002091 | Elementary Algebra: Linear Equations — Verification | 2 | Carefully track domains: Solve for $x$ and verify your result:
(a) Solve $ 9x + (-58) = 41 $.
(b) Substitute your value of $x$ back into the 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 $(-58)$ from both sides: $9x=99$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{99}{9}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{11}$.\nBoth methods reduce the equation to $x=\\frac{99}{9}$ and compute the same integer $x=11$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=9\... | 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 $a=9\neq 0$. (Here the result is $\boxed{11}$.) |
math-002092 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 5x + (-6) = -56 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $f(x)=5x+(-6)$. Since the slope $5\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-10}$.\nBoth methods reduce the equation to $x=\\frac{-50}{5}$ and compute the same integer $x=-10$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inv... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=5\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-10$ because $a=5\neq 0$. (Here the result is $\boxed{-10}$.) |
math-002093 | Algebra: Quadratics — Vieta Cross-Check | 2 | Be explicit about assumptions: Find all real solutions and justify each step:
$$x^2 + (-12)x + (-160) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Vieta's Relations (Cross-Check)",
"approach": "Use Vieta: for monic $x^2+bx+c$, roots satisfy $r_1+r_2=-b$ and $r_1r_2=c$; verify the candidate roots and conclude they are the solutions.",
"steps": [
"Step 1: For a monic quadratic, any roots $u,v$ satisfy $u+v=-(-12)=12$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-8,20\\}$.\nFactoring yields roots $r_1=-8$ and $r_2=20$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=784.",
"robustness_analysis": "Sensitivity ... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -8 and 20. |
math-002094 | Algebra: Quadratics — Vieta Cross-Check | 2 | Indicate where a theorem is used: Find all real solutions and justify each step:
$$x^2 + (-21)x + (110) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=-21$ and $c=110... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{10,11\\}$.\nFactoring yields roots $r_1=10$ and $r_2=11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1.",
"robustness_analysis": "Robustness note: Factoring is fast when intege... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Remember: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly 10 and 11. (Here the result is $\boxed{\{10,11\}$.) |
math-002095 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 3x + (-70) = -61 $.
(b) Substitute your value of $x$ back into the 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 $(-70)$ from both sides: $3x=9$.",
"Step 2: Since $3\\neq 0$, divide by $3$: $x=\\frac{9}{3}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "Sensitivity analysis: 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=3\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=3$ because $a=3\neq 0$. |
math-002096 | Algebra: Quadratics — Factoring + Zero Product | 2 | Compute the requested quantity: Find all real solutions and justify each step:
$$x^2 + (12)x + (-45) = 0.$$
If you use a theorem (e.g., quadratic formula), name it explicitly.
Also check your solutions by direct substitution into the original polynomial.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quadratic Formula",
"approach": "Use the quadratic formula and simplify; the discriminant being a perfect square yields integer roots.",
"steps": [
"Step 1: For $x^2+bx+c=0$, the quadratic formula gives $x=\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$.",
"Step 2: Here $b=12$ and $c=-45$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-15,3\\}$.\nFactoring yields roots $r_1=-15$ and $r_2=3$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=324.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Treated $uv=0$ as requiring $u=0$ AND $v=0$.",
"why_plausible": "Beginners may interpret 'product is zero' as 'both are zero'.",
"why_wrong": "The correct logical statement is OR: at least one factor must be zero; requiring both can discard valid solutions.",
"which_method_ca... | Core principle: Quadratics can be solved by factoring or the quadratic formula; using Vieta or substitution provides a reliable cross-check. Here the roots are exactly -15 and 3. (Here the result is $\boxed{\{-15,3\}$.) |
math-002097 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Derive the result step-by-step: Solve for $x$ and verify your result:
(a) Solve $ 12x + (37) = -191 $.
(b) Substitute your value of $x$ back into the 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 $(37)$ from both sides: $12x=-228$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{-228}{12}$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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": "If the problem were perturbed: I... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=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 $a=12\neq 0$. |
math-002098 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 12x + (10) = -242 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verifica... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=12x+(10)$. Since the slope $12\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-21}$.\nBoth methods reduce the equation to $x=\\frac{-252}{12}$ and compute the same integer $x=-21$; 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=12... | 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 $a=12\neq 0$. (Here the result is $\boxed{-21}$.) |
math-002099 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 23x + (29) = 604 $.
(b) Substitute your value of $x$ back into the 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 $(29)$ from both sides: $23x=575$.",
"Step 2: Since $23\\neq 0$, divide by $23$: $x=\\frac{575}{23}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly 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 $a=23\neq 0$. |
math-002100 | Algebra: Affine Functions — Injectivity | 2 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 18x + (60) = 60 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=18x+(60)$. Since the slope $18\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. 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": "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=18... | 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 $a=18\neq 0$. (Here the result is $\boxed{0}$.) |
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