id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-002201 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 12x + (34) = 130 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(34)$ from both sides: $12x=96$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{96}{12}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{96}{12}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "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=12... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=12\neq 0$. (Here the result is $\boxed{8}$.) |
math-002202 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 9x + (-80) = -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 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: $9x=27$.",
"Step 2: Since $9\\neq 0$, divide by $9$: $x=\\frac{27}{9}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3}$.\nBoth methods reduce the equation to $x=\\frac{27}{9}$ and compute the same integer $x=3$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operation... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=9\... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=3$ because $a=9\neq 0$. (Here the result is $\boxed{3}$.) |
math-002203 | Algebra: Quadratics — Quadratic Formula | 2 | Problem: Find all real solutions and justify each step:
$$x^2 + (14)x + (40) = 0.$$
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": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(14)x+(40)=(x-(-4))(x-(-10))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-10,-4\\}$.\nFactoring yields roots $r_1=-4$ and $r_2=-10$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=36.",
"robustness_analysis": "If the problem were ... | [
{
"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 -4. (Here the result is $\boxed{\{-10,-4\}$.) |
math-002204 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 27x + (59) = -265 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(59)$ from both sides: $27x=-324$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-324}{27}$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-12}$.\nBoth methods reduce the equation to $x=\\frac{-324}{27}$ and compute the same integer $x=-12$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear e... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=27... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-12$ because $a=27\neq 0$. |
math-002205 | Algebra: Affine Functions — Injectivity | 2 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 13x + (15) = 288 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=13x+(15)$. Since the slope $13\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{273}{13}$ and compute the same integer $x=21$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=13... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=21$ because $a=13\neq 0$. |
math-002206 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 29x + (-56) = 2 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/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 $(-56)$ from both sides: $29x=58$.",
"Step 2: Since $29\\neq 0$, divide by $29$: $x=\\frac{58}{29}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{58}{29}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations wo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=29... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=29\neq 0$. |
math-002207 | Prealgebra: Solving for a Variable | 2 | Problem: Solve for $x$ and verify your result:
(a) Solve $ 16x + (-39) = 169 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=16x+(-39)$. Since the slope $16\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{208}{16}$ and compute the same integer $x=13$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equation $... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=16\neq 0$. |
math-002208 | Algebra: Quadratics — Quadratic Formula | 2 | Exercise: Find all real solutions and justify each step:
$$x^2 + (11)x + (-152) = 0.$$
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 $b=11$ and $c=-152... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-19,8\\}$.\nFactoring yields roots $r_1=8$ and $r_2=-19$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=729.",
"robustness_analysis": "Generality 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 -19 and 8. (Here the result is $\boxed{\{-19,8\}$.) |
math-002209 | Prealgebra: Solving for a Variable | 2 | Give a fully justified solution: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-80) = -107 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=27x+(-80)$. Since the slope $27\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-27}{27}$ and compute the same integer $x=-1$; 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=27... | 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 $a=27\neq 0$. (Here the result is $\boxed{-1}$.) |
math-002210 | Algebra: Quadratics — Vieta Cross-Check | 2 | Be explicit about assumptions: Find all real solutions and justify each step:
$$x^2 + (18)x + (-19) = 0.$$
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 $x^2+(18)x+(-19)=(x-(1))(x-(-19))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-19,1\\}$.\nFactoring yields roots $r_1=1$ and $r_2=-19$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=400.",
"robustness_analysis": "If the problem were perturbed: Factoring is ... | [
{
"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 -19 and 1. (Here the result is $\boxed{\{-19,1\}$.) |
math-002211 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Prompt: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-64) = 216 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(-64)$ from both sides: $20x=280$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{280}{20}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{14}$.\nBoth methods reduce the equation to $x=\\frac{280}{20}$ and compute the same integer $x=14$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equa... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=20... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=14$ because $a=20\neq 0$. |
math-002212 | Precalculus: Polynomial Roots | 2 | Determine the requested value: Find all real solutions and justify each step:
$$x^2 + (-11)x + (0) = 0.$$
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": "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 $u+v=-(-11)=11$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{0,11\\}$.\nFactoring yields roots $r_1=0$ and $r_2=11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=121.",
"robustness_analysis": "Robustness not... | [
{
"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 0 and 11. |
math-002213 | Precalculus: Polynomial Roots | 2 | Question: Find all real solutions and justify each step:
$$x^2 + (-1)x + (-56) = 0.$$
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": "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=-5... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-7,8\\}$.\nFactoring yields roots $r_1=8$ and $r_2=-7$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=225.",
"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... | 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 -7 and 8. |
math-002214 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Track quantifiers carefully: Solve for $x$ and verify your result:
(a) Solve $ 3x + (47) = -7 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cros... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=3x+(47)$. Since the slope $3\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"Ste... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-54}{3}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any li... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=3\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-18$ because $a=3\neq 0$. |
math-002215 | Algebra: Quadratics — Vieta Cross-Check | 2 | Problem: Find all real solutions and justify each step:
$$x^2 + (2)x + (-63) = 0.$$
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 $u+v=-(2)=-2$ and $uv=-6... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-9,7\\}$.\nFactoring yields roots $r_1=7$ and $r_2=-9$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=256.",
"robustness_analysis": "Generality 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... | 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 -9 and 7. |
math-002216 | Algebra: Quadratics — Quadratic Formula | 2 | Proceed methodically: Find all real solutions and justify each step:
$$x^2 + (-6)x + (-72) = 0.$$
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 $u+v=-(-6)=6$ and $uv=-7... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-6,12\\}$.\nFactoring yields roots $r_1=-6$ and $r_2=12$. 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... | 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 -6 and 12. (Here the result is $\boxed{\{-6,12\}$.) |
math-002217 | Algebra: Quadratics — Vieta Cross-Check | 2 | Checkpoint: Find all real solutions and justify each step:
$$x^2 + (5)x + (-176) = 0.$$
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 $x^2+(5)x+(-176)=(x-(-16))(x-(11))$ 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: $\\boxed{\\{-16,11\\}$.\nFactoring yields roots $r_1=-16$ and $r_2=11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=729.",
"robustness_analysis": "Robustness note: Fa... | [
{
"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 -16 and 11. (Here the result is $\boxed{\{-16,11\}$.) |
math-002218 | Prealgebra: Solving for a Variable | 2 | Determine the requested value: Solve for $x$ and verify your result:
(a) Solve $ 27x + (-68) = -122 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(-68)$ from both sides: $27x=-54$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{-54}{27}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-2}$.\nBoth methods reduce the equation to $x=\\frac{-54}{27}$ and compute the same integer $x=-2$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-operations works for any linear... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=27... | 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 $a=27\neq 0$. |
math-002219 | Algebra: Quadratics — Vieta Cross-Check | 2 | Give a theorem-based solution: Find all real solutions and justify each step:
$$x^2 + (11)x + (-12) = 0.$$
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": "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 $b=11$ and $c=-12$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-12,1\\}$.\nFactoring yields roots $r_1=1$ and $r_2=-12$. 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 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 -12 and 1. (Here the result is $\boxed{\{-12,1\}$.) |
math-002220 | Precalculus: Polynomial Roots | 2 | Work carefully and justify each inference: Find all real solutions and justify each step:
$$x^2 + (-24)x + (135) = 0.$$
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 en... | [
{
"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 $b=-24$ and $c=135... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{9,15\\}$.\nFactoring yields roots $r_1=9$ and $r_2=15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=36.",
"robustness_analysis": "Generality note: Factoring is fast when i... | [
{
"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 9 and 15. (Here the result is $\boxed{\{9,15\}$.) |
math-002221 | Precalculus: Polynomial Roots | 2 | Solve and then verify: Find all real solutions and justify each step:
$$x^2 + (21)x + (108) = 0.$$
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": "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 $c=108$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-12,-9\\}$.\nFactoring yields roots $r_1=-12$ and $r_2=-9$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=9.",
"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... | 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 -9. (Here the result is $\boxed{\{-12,-9\}$.) |
math-002222 | Algebra: Affine Functions — Injectivity | 2 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 27x + (59) = 248 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(59)$ from both sides: $27x=189$.",
"Step 2: Since $27\\neq 0$, divide by $27$: $x=\\frac{189}{27}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nBoth methods reduce the equation to $x=\\frac{189}{27}$ and compute the same integer $x=7$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operati... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=27... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=7$ because $a=27\neq 0$. (Here the result is $\boxed{7}$.) |
math-002223 | Algebra: Quadratics — Factoring + Zero Product | 2 | Where appropriate, name the theorem you use: Find all real solutions and justify each step:
$$x^2 + (-10)x + (-11) = 0.$$
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 ... | [
{
"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=-11... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-1,11\\}$.\nFactoring yields roots $r_1=11$ and $r_2=-1$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=144.",
"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... | 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 11. (Here the result is $\boxed{\{-1,11\}$.) |
math-002224 | Algebra: Quadratics — Factoring + Zero Product | 2 | Problem: Find all real solutions and justify each step:
$$x^2 + (14)x + (-51) = 0.$$
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 $b=14$ and $c=-51$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-17,3\\}$.\nFactoring yields roots $r_1=3$ and $r_2=-17$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=400.",
"robustness_analysis": "Sensitivity analysis: Factoring is fas... | [
{
"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 3. (Here the result is $\boxed{\{-17,3\}$.) |
math-002225 | Algebra: Quadratics — Vieta Cross-Check | 2 | Give reasoning, not just computation: Find all real solutions and justify each step:
$$x^2 + (-39)x + (380) = 0.$$
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 $x^2+(-39)x+(380)=(x-(19))(x-(20))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{19,20\\}$.\nFactoring yields roots $r_1=19$ and $r_2=20$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1.",
"robustness_analysis": "Sensitivity analysis: Factoring is fast when i... | [
{
"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 19 and 20. (Here the result is $\boxed{\{19,20\}$.) |
math-002226 | Algebra: Affine Functions — Injectivity | 2 | Provide both a computational and a conceptual explanation: Solve for $x$ and verify your result:
(a) Solve $ 9x + (-47) = -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).
I... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=9x+(-47)$. Since the slope $9\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-9}{9}$ and compute the same integer $x=-1$; 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\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=9\neq 0$. |
math-002227 | Algebra: Quadratics — Quadratic Formula | 2 | Provide a rigorous solution: Find all real solutions and justify each step:
$$x^2 + (0)x + (-121) = 0.$$
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 $x^2+(0)x+(-121)=(x-(11))(x-(-11))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-11,11\\}$.\nFactoring yields roots $r_1=11$ and $r_2=-11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=484.",
"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... | 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 -11 and 11. (Here the result is $\boxed{\{-11,11\}$.) |
math-002228 | Algebra: Affine Functions — Injectivity | 2 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 16x + (43) = 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... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=16x+(43)$. Since the slope $16\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{128}{16}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=16... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=16\neq 0$. |
math-002229 | Precalculus: Polynomial Roots | 2 | Be explicit about assumptions: Find all real solutions and justify each step:
$$x^2 + (18)x + (77) = 0.$$
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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-11,-7\\}$.\nFactoring yields roots $r_1=-7$ and $r_2=-11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=16.",
"robustness_analysis": "Generality note: Fac... | [
{
"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 -11 and -7. |
math-002230 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Give reasoning, not just computation: Solve for $x$ and verify your result:
(a) Solve $ 13x + (-74) = 225 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verif... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-74)$ from both sides: $13x=299$.",
"Step 2: Since $13\\neq 0$, divide by $13$: $x=\\frac{299}{13}$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{23}$.\nBoth methods reduce the equation to $x=\\frac{299}{13}$ and compute the same integer $x=23$; 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=13... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=23$ because $a=13\neq 0$. (Here the result is $\boxed{23}$.) |
math-002231 | Algebra: Quadratics — Vieta Cross-Check | 2 | Solve and include a self-check: Find all real solutions and justify each step:
$$x^2 + (15)x + (-16) = 0.$$
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 $u+v=-(15)=-15$ and $uv=... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-16,1\\}$.\nFactoring yields roots $r_1=-16$ and $r_2=1$. 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 -16 and 1. (Here the result is $\boxed{\{-16,1\}$.) |
math-002232 | Algebra: Quadratics — Quadratic Formula | 2 | Answer using clear logical steps: Find all real solutions and justify each step:
$$x^2 + (24)x + (143) = 0.$$
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 $x^2+(24)x+(143)=(x-(-13))(x-(-11))$ by expansion.",
"Step 2: By the zero-product property, a ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-13,-11\\}$.\nFactoring yields roots $r_1=-13$ and $r_2=-11$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=4.",
"robustness_analysis": "If the problem were perturbed: Factoring i... | [
{
"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 -11. (Here the result is $\boxed{\{-13,-11\}$.) |
math-002233 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 25x + (62) = 387 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=25x+(62)$. Since the slope $25\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{13}$.\nBoth methods reduce the equation to $x=\\frac{325}{25}$ and compute the same integer $x=13$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equa... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=13$ because $a=25\neq 0$. |
math-002234 | Algebra: Quadratics — Vieta Cross-Check | 2 | Determine the requested value: Find all real solutions and justify each step:
$$x^2 + (-12)x + (-64) = 0.$$
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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-4,16\\}$.\nFactoring yields roots $r_1=16$ and $r_2=-4$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=400.",
"robustness_analysis": "Robustness note: Fact... | [
{
"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 -4 and 16. (Here the result is $\boxed{\{-4,16\}$.) |
math-002235 | Algebra: Quadratics — Factoring + Zero Product | 2 | Derive the result step-by-step: Find all real solutions and justify each step:
$$x^2 + (-13)x + (0) = 0.$$
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": "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 $u+v=-(-13)=13$ and $uv=... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{0,13\\}$.\nFactoring yields roots $r_1=0$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=169.",
"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 0 and 13. (Here the result is $\boxed{\{0,13\}$.) |
math-002236 | Precalculus: Polynomial Roots | 2 | Try to avoid pattern-matching; explain why: Find all real solutions and justify each step:
$$x^2 + (15)x + (26) = 0.$$
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 $c=26$,... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-13,-2\\}$.\nFactoring yields roots $r_1=-13$ and $r_2=-2$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=121.",
"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 -13 and -2. |
math-002237 | Algebra: Quadratics — Factoring + Zero Product | 2 | Track units/moduli carefully: Find all real solutions and justify each step:
$$x^2 + (-16)x + (55) = 0.$$
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 $x^2+(-16)x+(55)=(x-(11))(x-(5))$ by expansion.",
"Step 2: By the zero-product property, a pro... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{5,11\\}$.\nFactoring yields roots $r_1=11$ and $r_2=5$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=36.",
"robustness_analysis": "Generality note: 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... | 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 11. (Here the result is $\boxed{\{5,11\}$.) |
math-002238 | Prealgebra: Solving for a Variable | 2 | Give a theorem-based solution: Solve for $x$ and verify your result:
(a) Solve $ 6x + (-77) = -59 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=6x+(-77)$. Since the slope $6\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"St... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3}$.\nBoth methods reduce the equation to $x=\\frac{18}{6}$ and compute the same integer $x=3$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for an... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=3$ because $a=6\neq 0$. |
math-002239 | Algebra: Quadratics — Factoring + Zero Product | 2 | Compute the requested quantity: Find all real solutions and justify each step:
$$x^2 + (5)x + (4) = 0.$$
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 $b=5$ and $c=4$, s... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-4,-1\\}$.\nFactoring yields roots $r_1=-4$ and $r_2=-1$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=9.",
"robustness_analysis": "If the problem... | [
{
"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 -4 and -1. (Here the result is $\boxed{\{-4,-1\}$.) |
math-002240 | Precalculus: Polynomial Roots | 2 | Try to avoid pattern-matching; explain why: Find all real solutions and justify each step:
$$x^2 + (34)x + (280) = 0.$$
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 en... | [
{
"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 $u+v=-(34)=-34$ and $uv=... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-20,-14\\}$.\nFactoring yields roots $r_1=-14$ and $r_2=-20$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=36.",
"robustness_analysis": "Generalit... | [
{
"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 -20 and -14. (Here the result is $\boxed{\{-20,-14\}$.) |
math-002241 | Prealgebra: Solving for a Variable | 2 | Work this out carefully: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-39) = -9 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-39)$ from both sides: $15x=30$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{30}{15}$.",
... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{30}{15}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for any linear equatio... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=15... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=15\neq 0$. |
math-002242 | Algebra: Affine Functions — Injectivity | 2 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 25x + (-41) = -291 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-41)$ from both sides: $25x=-250$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-250}{25}$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "If the problem were perturbed: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=25... | 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 $a=25\neq 0$. (Here the result is $\boxed{-10}$.) |
math-002243 | Algebra: Quadratics — Factoring + Zero Product | 2 | Solve and sanity-check: Find all real solutions and justify each step:
$$x^2 + (-9)x + (-52) = 0.$$
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 $u+v=-(-9)=9$ and $uv=-5... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-4,13\\}$.\nFactoring yields roots $r_1=-4$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=289.",
"robustness_analysis": "If the problem were p... | [
{
"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 -4 and 13. (Here the result is $\boxed{\{-4,13\}$.) |
math-002244 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 6x + (37) = 85 $.
(b) Substitute your value of $x$ back into the 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 $(37)$ from both sides: $6x=48$.",
"Step 2: Since $6\\neq 0$, divide by $6$: $x=\\frac{48}{6}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{8}$.\nBoth methods reduce the equation to $x=\\frac{48}{6}$ and compute the same integer $x=8$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax+b... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=6\... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=8$ because $a=6\neq 0$. |
math-002245 | Algebra: Quadratics — Quadratic Formula | 2 | Determine the requested value: Find all real solutions and justify each step:
$$x^2 + (4)x + (-320) = 0.$$
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 $x^2+(4)x+(-320)=(x-(16))(x-(-20))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-20,16\\}$.\nFactoring yields roots $r_1=16$ and $r_2=-20$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1296.",
"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... | 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 -20 and 16. (Here the result is $\boxed{\{-20,16\}$.) |
math-002246 | Algebra: Quadratics — Vieta Cross-Check | 2 | Track units/moduli carefully: Find all real solutions and justify each step:
$$x^2 + (-18)x + (45) = 0.$$
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": "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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{3,15\\}$.\nFactoring yields roots $r_1=3$ and $r_2=15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=144.",
"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... | 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 3 and 15. |
math-002247 | Algebra: Quadratics — Vieta Cross-Check | 2 | Try to avoid pattern-matching; explain why: Find all real solutions and justify each step:
$$x^2 + (-17)x + (52) = 0.$$
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 en... | [
{
"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 $b=-17$ and $c=52$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{4,13\\}$.\nFactoring yields roots $r_1=4$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=81.",
"robustness_analysis": "Robustness note: 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... | 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 4 and 13. (Here the result is $\boxed{\{4,13\}$.) |
math-002248 | Algebra: Quadratics — Vieta Cross-Check | 2 | Explain why your operations are valid: Find all real solutions and justify each step:
$$x^2 + (24)x + (95) = 0.$$
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": "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 $b=24$ and $c=95$,... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-19,-5\\}$.\nFactoring yields roots $r_1=-5$ and $r_2=-19$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=196.",
"robustness_analysis": "If the problem were... | [
{
"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 -5. |
math-002249 | Algebra: Quadratics — Vieta Cross-Check | 2 | Write the solution set clearly: Find all real solutions and justify each step:
$$x^2 + (-9)x + (0) = 0.$$
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 $b=-9$ and $c=0$, ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{0,9\\}$.\nFactoring yields roots $r_1=0$ and $r_2=9$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=81.",
"robustness_analysis": "Robustness note: Factoring is fast when integer r... | [
{
"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 0 and 9. (Here the result is $\boxed{\{0,9\}$.) |
math-002250 | Algebra: Quadratics — Quadratic Formula | 2 | Work carefully and justify each inference: Find all real solutions and justify each step:
$$x^2 + (-1)x + (-272) = 0.$$
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 en... | [
{
"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=-272... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-16,17\\}$.\nFactoring yields roots $r_1=-16$ and $r_2=17$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1089.",
"robustness_analysis": "Generality note: 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... | 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 -16 and 17. |
math-002251 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Warm-up: Solve for $x$ and verify your result:
(a) Solve $ 12x + (18) = 78 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/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 $f(x)=12x+(18)$. 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: $\\boxed{5}$.\nBoth methods reduce the equation to $x=\\frac{60}{12}$ and compute the same integer $x=5$; 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=12... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=5$ because $a=12\neq 0$. |
math-002252 | Algebra: Affine Functions — Injectivity | 2 | Write the solution set clearly: Solve for $x$ and verify your result:
(a) Solve $ 19x + (29) = -370 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=19x+(29)$. Since the slope $19\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-21}$.\nBoth methods reduce the equation to $x=\\frac{-399}{19}$ and compute the same integer $x=-21$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=19... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-21$ because $a=19\neq 0$. (Here the result is $\boxed{-21}$.) |
math-002253 | Algebra: Affine Functions — Injectivity | 2 | Give an answer and a quick verification: Solve for $x$ and verify your result:
(a) Solve $ 12x + (-43) = 185 $.
(b) Substitute your value of $x$ back into the 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 $(-43)$ from both sides: $12x=228$.",
"Step 2: Since $12\\neq 0$, divide by $12$: $x=\\frac{228}{12}$."... | {
"consistency_check": "The two methods are consistent and must coincide. 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: 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=12... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=19$ because $a=12\neq 0$. |
math-002254 | Algebra: Quadratics — Vieta Cross-Check | 2 | Write the solution set clearly: Find all real solutions and justify each step:
$$x^2 + (1)x + (-20) = 0.$$
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 $x^2+(1)x+(-20)=(x-(-5))(x-(4))$ by expansion.",
"Step 2: By the zero-product property, a prod... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-5,4\\}$.\nFactoring yields roots $r_1=-5$ and $r_2=4$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=81.",
"robustness_analysis": "Generality note: 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... | 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 4. (Here the result is $\boxed{\{-5,4\}$.) |
math-002255 | Prealgebra: Solving for a Variable | 2 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 17x + (76) = -230 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $f(x)=17x+(76)$. Since the slope $17\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-306}{17}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=17... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-18$ because $a=17\neq 0$. (Here the result is $\boxed{-18}$.) |
math-002256 | Elementary Algebra: Linear Equations — Verification | 2 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 15x + (35) = 110 $.
(b) Substitute your value of $x$ back into the 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": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(35)$ from both sides: $15x=75$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{75}{15}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5}$.\nBoth methods reduce the equation to $x=\\frac{75}{15}$ and compute the same integer $x=5$; 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=15... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=5$ because $a=15\neq 0$. (Here the result is $\boxed{5}$.) |
math-002257 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-32) = 544 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-chec... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-32)$ from both sides: $24x=576$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{576}{24}$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{24}$.\nBoth methods reduce the equation to $x=\\frac{576}{24}$ and compute the same integer $x=24$; substitution confirms equality.",
"robustness_analysis": "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=24... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=24$ because $a=24\neq 0$. (Here the result is $\boxed{24}$.) |
math-002258 | Prealgebra: Solving for a Variable | 2 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 2x + (-38) = -34 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-check at the en... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-38)$ from both sides: $2x=4$.",
"Step 2: Since $2\\neq 0$, divide by $2$: $x=\\frac{4}{2}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{4}{2}$ and compute the same integer $x=2$; 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=2\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=2\neq 0$. (Here the result is $\boxed{2}$.) |
math-002259 | Algebra: Quadratics — Quadratic Formula | 2 | Solve and then verify: Find all real solutions and justify each step:
$$x^2 + (0)x + (-361) = 0.$$
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 $b=0$ and $c=-361$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-19,19\\}$.\nFactoring yields roots $r_1=19$ and $r_2=-19$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1444.",
"robustness_analysis": "Robustness note: 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 -19 and 19. |
math-002260 | Elementary Algebra: Linear Equations — Verification | 2 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-19) = -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 verific... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-19)$ from both sides: $28x=-140$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{-140}{28}$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-5}$.\nBoth methods reduce the equation to $x=\\frac{-140}{28}$ and compute the same integer $x=-5$; 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... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-5$ because $a=28\neq 0$. |
math-002261 | Algebra: Affine Functions — Injectivity | 2 | Compute the requested quantity: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-1) = 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 verification... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-1)$ from both sides: $20x=420$.",
"Step 2: Since $20\\neq 0$, divide by $20$: $x=\\frac{420}{20}$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{21}$.\nBoth methods reduce the equation to $x=\\frac{420}{20}$ and compute the same integer $x=21$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works fo... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=20... | 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 $a=20\neq 0$. |
math-002262 | Algebra: Quadratics — Quadratic Formula | 2 | Solve and include a self-check: 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": "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 $b=18$ and $c=-40$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-20,2\\}$.\nFactoring yields roots $r_1=2$ and $r_2=-20$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=484.",
"robustness_analysis": "Robustness note: Fact... | [
{
"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 -20 and 2. (Here the result is $\boxed{\{-20,2\}$.) |
math-002263 | Precalculus: Polynomial Roots | 2 | Show all reasoning: Find all real solutions and justify each step:
$$x^2 + (-3)x + (-180) = 0.$$
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 $b=-3$ and $c=-180... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-12,15\\}$.\nFactoring yields roots $r_1=-12$ and $r_2=15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=729.",
"robustness_analysis": "Generality note: Fa... | [
{
"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 -12 and 15. (Here the result is $\boxed{\{-12,15\}$.) |
math-002264 | Algebra: Quadratics — Quadratic Formula | 2 | Give a fully justified solution: Find all real solutions and justify each step:
$$x^2 + (5)x + (-50) = 0.$$
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 $u+v=-(5)=-5$ and $uv=-5... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-10,5\\}$.\nFactoring yields roots $r_1=5$ and $r_2=-10$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=225.",
"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... | 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 5. |
math-002265 | Prealgebra: Solving for a Variable | 2 | Answer using clear logical steps: Solve for $x$ and verify your result:
(a) Solve $ 15x + (-29) = 1 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificatio... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-29)$ from both sides: $15x=30$.",
"Step 2: Since $15\\neq 0$, divide by $15$: $x=\\frac{30}{15}$.",
... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods reduce the equation to $x=\\frac{30}{15}$ and compute the same integer $x=2$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works for any linear equation $ax+... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=15... | Core principle: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=2$ because $a=15\neq 0$. (Here the result is $\boxed{2}$.) |
math-002266 | Precalculus: Polynomial Roots | 2 | Warm-up: Find all real solutions and justify each step:
$$x^2 + (2)x + (-35) = 0.$$
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 $x^2+(2)x+(-35)=(x-(-7))(x-(5))$ by expansion.",
"Step 2: By the zero-product property, a prod... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-7,5\\}$.\nFactoring yields roots $r_1=-7$ and $r_2=5$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=144.",
"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... | 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 -7 and 5. |
math-002267 | Algebra: Quadratics — Quadratic Formula | 2 | Be explicit about assumptions: Find all real solutions and justify each step:
$$x^2 + (-8)x + (-240) = 0.$$
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=-2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-12,20\\}$.\nFactoring yields roots $r_1=20$ and $r_2=-12$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1024.",
"robustness_analysis": "Robustness note: 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... | 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 20. |
math-002268 | Algebra: Quadratics — Vieta Cross-Check | 2 | Challenge: Find all real solutions and justify each step:
$$x^2 + (18)x + (0) = 0.$$
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": "Factoring + Zero-Product",
"approach": "Construct factors from the integer roots (or by matching coefficients), then apply the zero-product property.",
"steps": [
"Step 1: Observe $x^2+(18)x+(0)=(x-(0))(x-(-18))$ by expansion.",
"Step 2: By the zero-product property, a prod... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-18,0\\}$.\nFactoring yields roots $r_1=0$ and $r_2=-18$. 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 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... | 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 0. |
math-002269 | Prealgebra: Solving for a Variable | 2 | Explain what is being counted/optimized: Solve for $x$ and verify your result:
(a) Solve $ 13x + (73) = -187 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief ve... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=13x+(73)$. Since the slope $13\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-20}$.\nBoth methods reduce the equation to $x=\\frac{-260}{13}$ and compute the same integer $x=-20$; 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=13... | 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 $a=13\neq 0$. (Here the result is $\boxed{-20}$.) |
math-002270 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 10x + (21) = -229 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verificat... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(21)$ from both sides: $10x=-250$.",
"Step 2: Since $10\\neq 0$, divide by $10$: $x=\\frac{-250}{10}$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-250}{10}$ and compute the same integer $x=-25$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=10... | 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 $a=10\neq 0$. (Here the result is $\boxed{-25}$.) |
math-002271 | Algebra: Quadratics — Quadratic Formula | 2 | Indicate where a theorem is used: Find all real solutions and justify each step:
$$x^2 + (-30)x + (221) = 0.$$
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 $u+v=-(-30)=30$ and $uv=... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{13,17\\}$.\nFactoring yields roots $r_1=17$ and $r_2=13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=16.",
"robustness_analysis": "Generality note: 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... | 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 13 and 17. |
math-002272 | Algebra: Quadratics — Factoring + Zero Product | 2 | Track units/moduli carefully: Find all real solutions and justify each step:
$$x^2 + (-10)x + (-75) = 0.$$
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 $x^2+(-10)x+(-75)=(x-(-5))(x-(15))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-5,15\\}$.\nFactoring yields roots $r_1=-5$ and $r_2=15$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=400.",
"robustness_analysis": "Sensitivity analysis: Factoring is fas... | [
{
"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 15. |
math-002273 | Algebra: Affine Functions — Injectivity | 2 | Carefully track domains: Solve for $x$ and verify your result:
(a) Solve $ 14x + (-64) = -386 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cros... | [
{
"method_name": "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 $f(x)=14x+(-64)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-23}$.\nBoth methods reduce the equation to $x=\\frac{-322}{14}$ and compute the same integer $x=-23$; substitution confirms equality.",
"robustness_analysis": "Generality note: Inverse-operations works... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-23$ because $a=14\neq 0$. |
math-002274 | Elementary Algebra: Linear Equations — Verification | 2 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 14x + (68) = 278 $.
(b) Substitute your value of $x$ back into the 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 $(68)$ from both sides: $14x=210$.",
"Step 2: Since $14\\neq 0$, divide by $14$: $x=\\frac{210}{14}$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{15}$.\nBoth methods reduce the equation to $x=\\frac{210}{14}$ and compute the same integer $x=15$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for a... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=14... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=15$ because $a=14\neq 0$. |
math-002275 | Algebra: Affine Functions — Injectivity | 2 | Solve and include a self-check: Solve for $x$ and verify your result:
(a) Solve $ 25x + (0) = -450 $.
(b) Substitute your value of $x$ back into the 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 $(0)$ from both sides: $25x=-450$.",
"Step 2: Since $25\\neq 0$, divide by $25$: $x=\\frac{-450}{25}$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-18}$.\nBoth methods reduce the equation to $x=\\frac{-450}{25}$ and compute the same integer $x=-18$; substitution confirms equality.",
"robustness_analysis": "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=25... | 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 $a=25\neq 0$. (Here the result is $\boxed{-18}$.) |
math-002276 | Algebra: Quadratics — Quadratic Formula | 2 | Solve (and briefly cross-validate): Find all real solutions and justify each step:
$$x^2 + (5)x + (0) = 0.$$
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 $b=5$ and $c=0$, s... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{-5,0\\}$.\nFactoring yields roots $r_1=0$ and $r_2=-5$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=25.",
"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... | 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 -5 and 0. (Here the result is $\boxed{\{-5,0\}$.) |
math-002277 | Precalculus: Polynomial Roots | 2 | Try to avoid pattern-matching; explain why: Find all real solutions and justify each step:
$$x^2 + (-5)x + (-126) = 0.$$
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 $x^2+(-5)x+(-126)=(x-(14))(x-(-9))$ by expansion.",
"Step 2: By the zero-product property, a p... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-9,14\\}$.\nFactoring yields roots $r_1=14$ and $r_2=-9$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=529.",
"robustness_analysis": "If the problem were perturbed: Factoring is ... | [
{
"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 -9 and 14. (Here the result is $\boxed{\{-9,14\}$.) |
math-002278 | Prealgebra: Solving for a Variable | 2 | Carefully track domains: Solve for $x$ and verify your result:
(a) Solve $ 22x + (-43) = 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... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=22x+(-43)$. Since the slope $22\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods reduce the equation to $x=\\frac{198}{22}$ and compute the same integer $x=9$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations w... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=22... | 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 $a=22\neq 0$. (Here the result is $\boxed{9}$.) |
math-002279 | Precalculus: Polynomial Roots | 2 | Make each step logically reversible (or explain if not): Find all real solutions and justify each step:
$$x^2 + (-1)x + (-90) = 0.$$
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": "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 $c=-90$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-9,10\\}$.\nFactoring yields roots $r_1=10$ and $r_2=-9$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=361.",
"robustness_analysis": "If the problem were p... | [
{
"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 -9 and 10. (Here the result is $\boxed{\{-9,10\}$.) |
math-002280 | Algebra: Quadratics — Vieta Cross-Check | 2 | Make each step logically reversible (or explain if not): Find all real solutions and justify each step:
$$x^2 + (15)x + (50) = 0.$$
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-che... | [
{
"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 $u+v=-(15)=-15$ and $uv=... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-10,-5\\}$.\nFactoring yields roots $r_1=-5$ and $r_2=-10$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=25.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 -5. |
math-002281 | Algebra: Affine Functions — Injectivity | 2 | Show all reasoning: Solve for $x$ and verify your result:
(a) Solve $ 20x + (-22) = 378 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-chec... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=20x+(-22)$. Since the slope $20\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{20}$.\nBoth methods reduce the equation to $x=\\frac{400}{20}$ and compute the same integer $x=20$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Inverse-operations works for any lin... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=20... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=20$ because $a=20\neq 0$. (Here the result is $\boxed{20}$.) |
math-002282 | Elementary Algebra: Linear Equations — Verification | 2 | Explain each transformation: Solve for $x$ and verify your result:
(a) Solve $ 8x + (-32) = 48 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cro... | [
{
"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: $8x=80$.",
"Step 2: Since $8\\neq 0$, divide by $8$: $x=\\frac{80}{8}$.",
... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{10}$.\nBoth methods reduce the equation to $x=\\frac{80}{8}$ and compute the same integer $x=10$; substitution confirms equality.",
"robustness_analysis": "Robustness note: Inverse-operations works for ... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=8\... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=10$ because $a=8\neq 0$. |
math-002283 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Work carefully and justify each inference: Solve for $x$ and verify your result:
(a) Solve $ 4x + (-28) = 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 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 $(-28)$ from both sides: $4x=88$.",
"Step 2: Since $4\\neq 0$, divide by $4$: $x=\\frac{88}{4}$.",
... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{22}$.\nBoth methods reduce the equation to $x=\\frac{88}{4}$ and compute the same integer $x=22$; substitution confirms equality.",
"robustness_analysis": "If the problem were perturbed: Invers... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=4\... | 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 $a=4\neq 0$. (Here the result is $\boxed{22}$.) |
math-002284 | Precalculus: Polynomial Roots | 2 | Make each step logically reversible (or explain if not): Find all real solutions and justify each step:
$$x^2 + (-7)x + (6) = 0.$$
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-chec... | [
{
"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 $x^2+(-7)x+(6)=(x-(1))(x-(6))$ by expansion.",
"Step 2: By the zero-product property, a produc... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{1,6\\}$.\nFactoring yields roots $r_1=1$ and $r_2=6$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=25.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 1 and 6. |
math-002285 | Elementary Algebra: Linear Equations — Verification | 2 | Solve with verification: Solve for $x$ and verify your result:
(a) Solve $ 20x + (78) = -2 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/cross-c... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=20x+(78)$. Since the slope $20\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-4}$.\nBoth methods reduce the equation to $x=\\frac{-80}{20}$ and compute the same integer $x=-4$; 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=20... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-4$ because $a=20\neq 0$. (Here the result is $\boxed{-4}$.) |
math-002286 | Prealgebra: Solving for a Variable | 2 | Solve and justify each step: Solve for $x$ and verify your result:
(a) Solve $ 14x + (43) = -307 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/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 $f(x)=14x+(43)$. Since the slope $14\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-25}$.\nBoth methods reduce the equation to $x=\\frac{-350}{14}$ and compute the same integer $x=-25$; 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... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-25$ because $a=14\neq 0$. (Here the result is $\boxed{-25}$.) |
math-002287 | Algebra: Quadratics — Vieta Cross-Check | 2 | Solve and sanity-check: Find all real solutions and justify each step:
$$x^2 + (-3)x + (-340) = 0.$$
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 $x^2+(-3)x+(-340)=(x-(20))(x-(-17))$ by expansion.",
"Step 2: By the zero-product property, a ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\{-17,20\\}$.\nFactoring yields roots $r_1=20$ and $r_2=-17$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=1369.",
"robustness_analysis": "Sensitivi... | [
{
"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 -17 and 20. |
math-002288 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Indicate where a theorem is used: Solve for $x$ and verify your result:
(a) Solve $ 28x + (6) = -442 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief 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 $(6)$ from both sides: $28x=-448$.",
"Step 2: Since $28\\neq 0$, divide by $28$: $x=\\frac{-448}{28}$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-16}$.\nBoth methods reduce the equation to $x=\\frac{-448}{28}$ and compute the same integer $x=-16$; substitution confirms equality.",
"robustness_analysis": "Sensitivity analysis: Inverse-op... | [
{
"error_description": "Divided by the coefficient without checking it is nonzero.",
"why_plausible": "In many classroom examples the coefficient is implicitly nonzero, so the condition is forgotten.",
"why_wrong": "If $a=0$ the equation becomes $b=c$ (either no solutions or infinitely many). Here $a=28... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-16$ because $a=28\neq 0$. |
math-002289 | Algebra: Quadratics — Factoring + Zero Product | 2 | Task: Find all real solutions and justify each step:
$$x^2 + (-16)x + (48) = 0.$$
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 $x^2+(-16)x+(48)=(x-(4))(x-(12))$ by expansion.",
"Step 2: By the zero-product property, a pro... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{4,12\\}$.\nFactoring yields roots $r_1=4$ and $r_2=12$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=64.",
"robustness_analysis": "If the problem were pert... | [
{
"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 4 and 12. (Here the result is $\boxed{\{4,12\}$.) |
math-002290 | Algebra: Quadratics — Vieta Cross-Check | 2 | Carefully track domains: Find all real solutions and justify each step:
$$x^2 + (4)x + (-117) = 0.$$
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 $x^2+(4)x+(-117)=(x-(9))(x-(-13))$ by expansion.",
"Step 2: By the zero-product property, a pr... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-13,9\\}$.\nFactoring yields roots $r_1=9$ and $r_2=-13$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=484.",
"robustness_analysis": "Generality 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... | 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 9. (Here the result is $\boxed{\{-13,9\}$.) |
math-002291 | Elementary Algebra: Linear Equations — Inverse Operations | 2 | Exercise: Solve for $x$ and verify your result:
(a) Solve $ 21x + (-79) = 278 $.
(b) Substitute your value of $x$ back into the 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 $f(x)=21x+(-79)$. Since the slope $21\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{17}$.\nBoth methods reduce the equation to $x=\\frac{357}{21}$ and compute the same integer $x=17$; substitution confirms equality.",
"robustness_analysis": "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=21... | 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 $a=21\neq 0$. (Here the result is $\boxed{17}$.) |
math-002292 | Elementary Algebra: Linear Equations — Verification | 2 | Do not skip justification steps: Solve for $x$ and verify your result:
(a) Solve $ 24x + (-32) = -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 verificat... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-32)$ from both sides: $24x=-168$.",
"Step 2: Since $24\\neq 0$, divide by $24$: $x=\\frac{-168}{24}$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-7}$.\nBoth methods reduce the equation to $x=\\frac{-168}{24}$ and compute the same integer $x=-7$; 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=24... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-7$ because $a=24\neq 0$. |
math-002293 | Algebra: Quadratics — Factoring + Zero Product | 2 | Answer using clear logical steps: Find all real solutions and justify each step:
$$x^2 + (16)x + (39) = 0.$$
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 $b=16$ and $c=39$,... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-13,-3\\}$.\nFactoring yields roots $r_1=-13$ and $r_2=-3$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=100.",
"robustness_analysis": "Robustness note: Fa... | [
{
"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 -3. (Here the result is $\boxed{\{-13,-3\}$.) |
math-002294 | Algebra: Quadratics — Vieta Cross-Check | 2 | Give an answer and a quick verification: Find all real solutions and justify each step:
$$x^2 + (-31)x + (228) = 0.$$
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 $u+v=-(-31)=31$ and $uv=... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{12,19\\}$.\nFactoring yields roots $r_1=19$ and $r_2=12$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=49.",
"robustness_analysis": "If the problem were perturbed: Factorin... | [
{
"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 19. (Here the result is $\boxed{\{12,19\}$.) |
math-002295 | Algebra: Affine Functions — Injectivity | 2 | Use two approaches if possible: Solve for $x$ and verify your result:
(a) Solve $ 22x + (37) = 15 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"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: $22x=-22$.",
"Step 2: Since $22\\neq 0$, divide by $22$: $x=\\frac{-22}{22}$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-1}$.\nBoth methods reduce the equation to $x=\\frac{-22}{22}$ and compute the same integer $x=-1$; substitution confirms equality.",
"robustness_analysis": "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=22... | Remember: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-1$ because $a=22\neq 0$. (Here the result is $\boxed{-1}$.) |
math-002296 | Algebra: Quadratics — Vieta Cross-Check | 2 | Question: Find all real solutions and justify each step:
$$x^2 + (3)x + (-4) = 0.$$
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 $x^2+(3)x+(-4)=(x-(1))(x-(-4))$ by expansion.",
"Step 2: By the zero-product property, a produ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\{-4,1\\}$.\nFactoring yields roots $r_1=1$ and $r_2=-4$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=25.",
"robustness_analysis": "If the problem were pert... | [
{
"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 -4 and 1. (Here the result is $\boxed{\{-4,1\}$.) |
math-002297 | Algebra: Quadratics — Vieta Cross-Check | 2 | Do not skip justification steps: Find all real solutions and justify each step:
$$x^2 + (-1)x + (-6) = 0.$$
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=-6... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\{-2,3\\}$.\nFactoring yields roots $r_1=3$ and $r_2=-2$. The quadratic formula gives the same two values because the discriminant is $(r_1-r_2)^2=25.",
"robustness_analysis": "Robustness note: Factoring is fast when integer... | [
{
"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 3. |
math-002298 | Precalculus: Polynomial Roots | 2 | Give reasoning, not just computation: Find all real solutions and justify each step:
$$x^2 + (-10)x + (9) = 0.$$
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 $x^2+(-10)x+(9)=(x-(1))(x-(9))$ by expansion.",
"Step 2: By the zero-product property, a produ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\{1,9\\}$.\nFactoring yields roots $r_1=1$ and $r_2=9$. 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 is... | [
{
"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 1 and 9. (Here the result is $\boxed{\{1,9\}$.) |
math-002299 | Elementary Algebra: Linear Equations — Verification | 2 | Derive the result step-by-step: Solve for $x$ and verify your result:
(a) Solve $ 28x + (62) = 90 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verification/... | [
{
"method_name": "Function + Injectivity",
"approach": "View the left side as an affine function with nonzero slope, hence injective; solving identifies the unique preimage of $c$.",
"steps": [
"Step 1: Let $f(x)=28x+(62)$. Since the slope $28\\neq 0$, $f$ is injective on $\\mathbb{R}$.",
"S... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\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-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... | Takeaway: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=1$ because $a=28\neq 0$. (Here the result is $\boxed{1}$.) |
math-002300 | Elementary Algebra: Linear Equations — Verification | 2 | Answer with a short justification: Solve for $x$ and verify your result:
(a) Solve $ 28x + (-19) = -187 $.
(b) Substitute your value of $x$ back into the original equation to confirm equality.
Your solution should explicitly justify any division step (i.e., explain why the divisor is nonzero).
Include a brief verific... | [
{
"method_name": "Inverse Operations",
"approach": "Undo the addition/subtraction first, then undo the multiplication by dividing by the nonzero coefficient.",
"steps": [
"Step 1: Subtract $(-19)$ 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... | Key idea: For $ax+b=c$, always isolate $x$ using valid inverse operations, and verify by substitution; here the unique solution is $x=-6$ because $a=28\neq 0$. (Here the result is $\boxed{-6}$.) |
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