{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261012-advance-20261016-r05-system-linear-equations-0005",
  "pilot_number": "R0014-07805",
  "number": 7805,
  "family_id": "qf-density16-system-candidate-two-residuals",
  "title": {
    "zh-Hant": "第 7805 題：候選同時滿足兩式的殘差",
    "en": "Question 7805: Two Rational Substitution Residuals for a System"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261012-advance-20261016-r05-system-linear-equations-0005",
    "batch_id": "batch-20261012-advance-20261016-r05-system-linear-equations-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-system-linear-equations"
    ],
    "title_zh": "候選同時滿足兩式的殘差",
    "stem_zh": "R上兩元兩式；rows每列[a,b,c]代表a*x+b*y=c。rows=[[1, 0, -8], [0, 1, 0]]，有序候選(x,y)=[(0)/(1), (0)/(1)]。求 (兩列a*x+b*y−c的有理殘差, 兩列同時零資格)。",
    "answer": {
      "kind": "expression",
      "value": "([8/1, 0/1], 否)",
      "explanation_zh": "有限結構核對=([8/1, 0/1], 否)。先規約合法候選分數，再分別代入兩列；只過一列不是系統的解。有限精確核對不是一般形式證明。候選須同時滿足兩式，零縮放不當可逆列操作。"
    },
    "difficulty": {
      "band": "upper_secondary",
      "level": 1,
      "skills": [
        "候選同時滿足兩式的殘差"
      ]
    },
    "variant": {
      "family_id": "qf-density16-system-candidate-two-residuals",
      "seed": 202610165005,
      "parameters": {
        "rows": [
          [
            1,
            0,
            -8
          ],
          [
            0,
            1,
            0
          ]
        ],
        "candidate": [
          [
            0,
            1
          ],
          [
            0,
            1
          ]
        ],
        "domain": "R2_linear_system_integer_slice",
        "operation": "system-candidate-two-residuals",
        "semantic_template": "qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,-8],[0,1,0]]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_integer_rational_arithmetic",
        "producer_finite_enumeration",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-12",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261012-advance-20261016-r05-system-linear-equations-0005",
    "batch_id": "batch-20261012-advance-20261016-r05-system-linear-equations-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "9e5cb775e353002132685ead5c22dc532e93cb35921a6d940f38f68a4160944a",
      "family_id": "qf-density16-system-candidate-two-residuals",
      "seed": 202610165005,
      "semantic_template": "qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,-8],[0,1,0]]}"
    },
    "fields": {
      "title": "Two Rational Substitution Residuals for a System",
      "stem": "Over R, rows=[[1, 0, -8], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).",
      "answer": {
        "kind": "expression",
        "value": "([8/1, 0/1], No)",
        "explanation": "Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([8/1, 0/1], No)."
      },
      "skills": [
        "Two Rational Substitution Residuals for a System"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "R上兩元兩式；rows每列[a,b,c]代表a*x+b*y=c。rows=[[1, 0, -8], [0, 1, 0]]，有序候選(x,y)=[(0)/(1), (0)/(1)]。求 (兩列a*x+b*y−c的有理殘差, 兩列同時零資格)。",
        "en": "Over R, rows=[[1, 0, -8], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "rows=[[1, 0, -8], [0, 1, 0]]",
      "after": "each row [a,b,c] means a*x+b*y=c over R",
      "relation": "implies",
      "rule_id": "two_equations_real_system_scope",
      "conditions": [],
      "explanation": {
        "zh-Hant": "兩式須同時成立，右側常數不是第三個未知數。退化零係數列也保留，不能先假設行列式非零。",
        "en": "Both equations must hold simultaneously.The right-hand constant is not a third unknown.Retain degenerate zero-coefficient rows rather than presuming a nonzero determinant."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "raw rational pairs=[[0, 1], [0, 1]]",
      "after": "candidate=[0/1, 0/1]",
      "relation": "implies",
      "rule_id": "candidate_rational_denominator_guard",
      "conditions": [],
      "explanation": {
        "zh-Hant": "兩個原候選分母都須非零，負分母合法但先規約符號與公因子，不拿唯一解覆寫候選。",
        "en": "Both original candidate denominators must be nonzero.Negative denominators are legal;normalize sign and common factors without replacing the candidate by a solved answer."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "candidate row0 ax := mul(1, 0)",
      "after": "candidate row0 ax = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "用本位置明示的純量與分量相乘，不改成內積或逐項矩陣乘法。",
        "en": "Multiply the specified scalar and component at this position;do not substitute an inner product or entrywise matrix product."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row0 ax，不以小數近似替代。",
          "en": "Recompute candidate row0 ax with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "0"
        ],
        "output": "0",
        "label": "candidate row0 ax"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "candidate row0 by := mul(0, 0)",
      "after": "candidate row0 by = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "用本位置明示的純量與分量相乘，不改成內積或逐項矩陣乘法。",
        "en": "Multiply the specified scalar and component at this position;do not substitute an inner product or entrywise matrix product."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row0 by，不以小數近似替代。",
          "en": "Recompute candidate row0 by with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "0",
          "0"
        ],
        "output": "0",
        "label": "candidate row0 by"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "candidate row0 lhs := sum(0, 0)",
      "after": "candidate row0 lhs = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "把已對齊的同一分量或同一位置相加，不跨分量配對。",
        "en": "Add the already aligned component or position,without mixing coordinates."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row0 lhs，不以小數近似替代。",
          "en": "Recompute candidate row0 lhs with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "0"
        ],
        "output": "0",
        "label": "candidate row0 lhs"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "candidate row0 residual := sub(0, -8)",
      "after": "candidate row0 residual = 8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按題目指定的前項減後項，保留殘差的正負。",
        "en": "Subtract the second operand from the first in the requested order,retaining the residual sign."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row0 residual，不以小數近似替代。",
          "en": "Recompute candidate row0 residual with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "0",
          "-8"
        ],
        "output": "8",
        "label": "candidate row0 residual"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "candidate row1 ax := mul(0, 0)",
      "after": "candidate row1 ax = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "用本位置明示的純量與分量相乘，不改成內積或逐項矩陣乘法。",
        "en": "Multiply the specified scalar and component at this position;do not substitute an inner product or entrywise matrix product."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row1 ax，不以小數近似替代。",
          "en": "Recompute candidate row1 ax with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "0",
          "0"
        ],
        "output": "0",
        "label": "candidate row1 ax"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "candidate row1 by := mul(1, 0)",
      "after": "candidate row1 by = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "用本位置明示的純量與分量相乘，不改成內積或逐項矩陣乘法。",
        "en": "Multiply the specified scalar and component at this position;do not substitute an inner product or entrywise matrix product."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row1 by，不以小數近似替代。",
          "en": "Recompute candidate row1 by with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "0"
        ],
        "output": "0",
        "label": "candidate row1 by"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "candidate row1 lhs := sum(0, 0)",
      "after": "candidate row1 lhs = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "把已對齊的同一分量或同一位置相加，不跨分量配對。",
        "en": "Add the already aligned component or position,without mixing coordinates."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row1 lhs，不以小數近似替代。",
          "en": "Recompute candidate row1 lhs with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "0"
        ],
        "output": "0",
        "label": "candidate row1 lhs"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "candidate row1 residual := sub(0, 0)",
      "after": "candidate row1 residual = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按題目指定的前項減後項，保留殘差的正負。",
        "en": "Subtract the second operand from the first in the requested order,retaining the residual sign."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 candidate row1 residual，不以小數近似替代。",
          "en": "Recompute candidate row1 residual with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "0",
          "0"
        ],
        "output": "0",
        "label": "candidate row1 residual"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "residuals=[8/1, 0/1]",
      "after": "simultaneous solution=否",
      "relation": "implies",
      "rule_id": "system_candidate_both_rows",
      "conditions": [],
      "explanation": {
        "zh-Hant": "兩個有理殘差必須同時為0；只通過一式，或把兩個相反殘差相加成0，都不夠。",
        "en": "Both rational residuals must individually be zero.Passing one equation,or summing two opposite residuals to zero,is insufficient."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      [
        "8/1",
        "0/1"
      ],
      false
    ]
  },
  "conclusion": {
    "canonical_value": "([8/1, 0/1], 否)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261012-advance-20261016-r05-system-linear-equations-foundations",
      "batch_path": "public/data/questions/batches/batch-20261012-advance-20261016-r05-system-linear-equations-foundations/questions.jsonl",
      "batch_sha256": "600d1fcb1634af8d3233900a81715f255bfc3a3d4fff09369711f40dea2bf385",
      "original_line": 5,
      "original_line_sha256": "e241f3e4fca83d5b9e6f60c55b58204974043c21540b5cc7dffe72a669a4bf47",
      "question_compact_sha256": "9e5cb775e353002132685ead5c22dc532e93cb35921a6d940f38f68a4160944a",
      "english_row_compact_sha256": "04785bf17343d0fdc87c2257b84bd3b6fffd86095bdd9781c28ca66c4ad6bde5",
      "english_manifest_path": "public/data/i18n/editions/batch-20261012-advance-20261016-r05-system-linear-equations-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261012-advance-20261016-r05-system-linear-equations-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-system-linear-equations": {
        "sha256": "62dc0d59fbb66d23a8ef47c7b2725a56713d925c7f2c8ef0a380c46552972403",
        "english_sha256": "7fc6d87498d9c2a7943b1715c26c7d3e2baf79d0f54d4beb9019b628da0010c4",
        "original_path": "public/data/mko/system-linear-equations.json",
        "english_path": "public/data/i18n/editions/mko-system-linear-equations.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "d7860245c70062705fa4858c13821db6ff286678dec63f22464364f5d24cfcc1",
    "original_candidate_sha256": "e3dc597a292d6af8b8054fdd15e51468b878066249b45c38e20515f1244c0b9a",
    "original_candidate_line_sha256": "e3dc597a292d6af8b8054fdd15e51468b878066249b45c38e20515f1244c0b9a"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-005-p15",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "23ba8196e85a09c03add0a87f96bec5d67a00dc648d1d68eef89101525b41a77",
      "question_first_witness_sha256": "17d59afc327b7cb8b6e588686d6dc1abbccd384d8ba32b8c046652f96821d705"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 11,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "e0353b0ce0b0c3541f979c6610c6cd50e71eabeb44fa96ab9976eb2aef192c0c",
      "arithmetic_steps": 8
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}