第 32971 題：精確有理候選的雙列驗證

原題與已知條件

實數論域中兩個限制 (2)x+(3)y=11 與 (4)x+(-3)y=13 必須同時成立。最簡正分母候選 x=4/1、y=1/1 是否滿足整個系統？

• G001 — 實數論域中兩個限制 (2)x+(3)y=11 與 (4)x+(-3)y=13 必須同時成立。最簡正分母候選 x=4/1、y=1/1 是否滿足整個系統？

定義與適用範圍

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。

• mko-system-linear-equations

逐步推導

source roles = [qf-linear-system-solution-verification, verify_solution, reals, []]

derived = [[ALL_rows_YesNo_STRING]]

source roles = [[[2, 3, 11], [4, -3, 13]]]

derived = [[coefficient_x, coefficient_y, right_side]]

System ORIGINAL a times e main product := mul(2, -3)

System ORIGINAL a times e main product = -6

System ORIGINAL b times d other product := mul(3, 4)

System ORIGINAL b times d other product = 12

System ORIGINAL signed determinant ae minus bd := sub(-6, 12)

System ORIGINAL signed determinant ae minus bd = -18

source roles = [[[2, 3, 11], [4, -3, 13]]]

derived = [-6, 12, -18]

source roles = [x candidate, [4, 1]]

derived = [4/1]

source roles = [y candidate, [1, 1]]

derived = [1/1]

System candidate ORIGINAL two positive denominators product := mul(1, 1)

System candidate ORIGINAL two positive denominators product = 1

System cleared row0 ORIGINAL coefficient x times numerator := mul(2, 4)

System cleared row0 ORIGINAL coefficient x times numerator = 8

System cleared row0 ORIGINAL x term times y denominator := mul(8, 1)

System cleared row0 ORIGINAL x term times y denominator = 8

System cleared row0 ORIGINAL coefficient y times numerator := mul(3, 1)

System cleared row0 ORIGINAL coefficient y times numerator = 3

System cleared row0 ORIGINAL y term times x denominator := mul(3, 1)

System cleared row0 ORIGINAL y term times x denominator = 3

System cleared row0 ORIGINAL rhs times both denominators := mul(11, 1)

System cleared row0 ORIGINAL rhs times both denominators = 11

System cleared row0 ORIGINAL two left terms sum := sum(8, 3)

System cleared row0 ORIGINAL two left terms sum = 11

System cleared row0 ORIGINAL residual := sub(11, 11)

System cleared row0 ORIGINAL residual = 0

source roles = [0, 2, 3, 11, 4, 1, 1, 1]

derived = [8, 3, 11, 0, 是]

System cleared row1 ORIGINAL coefficient x times numerator := mul(4, 4)

System cleared row1 ORIGINAL coefficient x times numerator = 16

System cleared row1 ORIGINAL x term times y denominator := mul(16, 1)

System cleared row1 ORIGINAL x term times y denominator = 16

System cleared row1 ORIGINAL coefficient y times numerator := mul(-3, 1)

System cleared row1 ORIGINAL coefficient y times numerator = -3

System cleared row1 ORIGINAL y term times x denominator := mul(-3, 1)

System cleared row1 ORIGINAL y term times x denominator = -3

System cleared row1 ORIGINAL rhs times both denominators := mul(13, 1)

System cleared row1 ORIGINAL rhs times both denominators = 13

System cleared row1 ORIGINAL two left terms sum := sum(16, -3)

System cleared row1 ORIGINAL two left terms sum = 13

System cleared row1 ORIGINAL residual := sub(13, 13)

System cleared row1 ORIGINAL residual = 0

source roles = [1, 4, -3, 13, 4, 1, 1, 1]

derived = [16, -3, 13, 0, 是]

Candidate direct reverse ORIGINAL row0 x term := mul(2, 4)

Candidate direct reverse ORIGINAL row0 x term = 8

Candidate direct reverse ORIGINAL row0 y term := mul(3, 1)

Candidate direct reverse ORIGINAL row0 y term = 3

Candidate direct reverse ORIGINAL row0 terms sum := sum(8, 3)

Candidate direct reverse ORIGINAL row0 terms sum = 11

Candidate direct reverse ORIGINAL row0 residual minus rhs := sub(11, 11)

Candidate direct reverse ORIGINAL row0 residual minus rhs = 0

source roles = [Candidate direct reverse, 0, 2, 3, 11, 4/1, 1/1]

derived = [8, 3, 11, 0/1]

Candidate direct reverse ORIGINAL row1 x term := mul(4, 4)

Candidate direct reverse ORIGINAL row1 x term = 16

Candidate direct reverse ORIGINAL row1 y term := mul(-3, 1)

Candidate direct reverse ORIGINAL row1 y term = -3

Candidate direct reverse ORIGINAL row1 terms sum := sum(16, -3)

Candidate direct reverse ORIGINAL row1 terms sum = 13

Candidate direct reverse ORIGINAL row1 residual minus rhs := sub(13, 13)

Candidate direct reverse ORIGINAL row1 residual minus rhs = 0

source roles = [Candidate direct reverse, 1, 4, -3, 13, 4/1, 1/1]

derived = [16, -3, 13, 0/1]

逐步反向核對

• S035 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S034 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row1 residual minus rhs，不以小數近似替代。
• S033 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row1 terms sum，不以小數近似替代。
• S032 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row1 y term，不以小數近似替代。
• S031 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row1 x term，不以小數近似替代。
• S030 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S029 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row0 residual minus rhs，不以小數近似替代。
• S028 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row0 terms sum，不以小數近似替代。
• S027 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row0 y term，不以小數近似替代。
• S026 · premise_and_result_audit — 以精確有理數重算 Candidate direct reverse ORIGINAL row0 x term，不以小數近似替代。
• S025 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S024 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL residual，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL two left terms sum，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL rhs times both denominators，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL y term times x denominator，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL coefficient y times numerator，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL x term times y denominator，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 System cleared row1 ORIGINAL coefficient x times numerator，不以小數近似替代。
• S017 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S016 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL residual，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL two left terms sum，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL rhs times both denominators，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL y term times x denominator，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL coefficient y times numerator，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL x term times y denominator，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 System cleared row0 ORIGINAL coefficient x times numerator，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 System candidate ORIGINAL two positive denominators product，不以小數近似替代。
• S008 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S007 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S006 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S005 · premise_and_result_audit — 以精確有理數重算 System ORIGINAL signed determinant ae minus bd，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 System ORIGINAL b times d other product，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 System ORIGINAL a times e main product，不以小數近似替代。
• S002 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。
• S001 · premise_and_result_audit — 以原始資料、欄位角色、型別、論域及本步前提反向代回核對；不表示結論能唯一還原前提。來源綁定候選、獨立審查、形式證明與發布分開。

結論

是

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。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。