{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-system-linear-equations"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-system-linear-equations-01","title":{"zh-Hant":"二元聯立一次方程組：唯一解公式、兩列核對與 D=0 的一致性判定","en":"Linear systems in two variables: the unique-solution formula, checking both rows, and consistency when D=0"},"summary":{"zh-Hant":"(I) 解 2x+3y=7、x−y=1：證明 D=ae−bd≠0 時的唯一解公式，求解並代入兩列核對；(II) 判定 x+2y=3、2x+4y=7；(III) 判定 x+2y=3、2x+4y=6；並以 (IV) 0x+0y=1、0x+0y=2 說明 D 與兩個 Cramer 分子都為零時不能斷定有無限多解。","en":"(I) Solve 2x+3y=7, x−y=1: prove the unique-solution formula when D=ae−bd≠0, solve, and check by substituting into both rows; (II) decide x+2y=3, 2x+4y=7; (III) decide x+2y=3, 2x+4y=6; and use (IV) 0x+0y=1, 0x+0y=2 to show that D and both Cramer numerators being zero does not imply infinitely many solutions."},"mko_ids":["mko-system-linear-equations"]}],"title":"概念推導導讀"}