与えられた4つの連立方程式を解く。代数学連立方程式方程式代入法2025/7/61. 問題の内容与えられた4つの連立方程式を解く。2. 解き方の手順(1)y=3xy = 3xy=3x を 9x−2y=129x - 2y = 129x−2y=12 に代入します。9x−2(3x)=129x - 2(3x) = 129x−2(3x)=129x−6x=129x - 6x = 129x−6x=123x=123x = 123x=12x=4x = 4x=4y=3x=3(4)=12y = 3x = 3(4) = 12y=3x=3(4)=12(2)x=−3y+2x = -3y + 2x=−3y+2 を 2x+y=−12x + y = -12x+y=−1 に代入します。2(−3y+2)+y=−12(-3y + 2) + y = -12(−3y+2)+y=−1−6y+4+y=−1-6y + 4 + y = -1−6y+4+y=−1−5y=−5-5y = -5−5y=−5y=1y = 1y=1x=−3y+2=−3(1)+2=−1x = -3y + 2 = -3(1) + 2 = -1x=−3y+2=−3(1)+2=−1(3)y−3x=−1y - 3x = -1y−3x=−1 を y=3x−1y = 3x - 1y=3x−1 と変形し、5y+2x=125y + 2x = 125y+2x=12 に代入します。5(3x−1)+2x=125(3x - 1) + 2x = 125(3x−1)+2x=1215x−5+2x=1215x - 5 + 2x = 1215x−5+2x=1217x=1717x = 1717x=17x=1x = 1x=1y=3x−1=3(1)−1=2y = 3x - 1 = 3(1) - 1 = 2y=3x−1=3(1)−1=2(4)2x+y=22x + y = 22x+y=2 から y=2−2xy = 2 - 2xy=2−2x を得て、x+2y=1x + 2y = 1x+2y=1 に代入します。x+2(2−2x)=1x + 2(2 - 2x) = 1x+2(2−2x)=1x+4−4x=1x + 4 - 4x = 1x+4−4x=1−3x=−3-3x = -3−3x=−3x=1x = 1x=1y=2−2x=2−2(1)=0y = 2 - 2x = 2 - 2(1) = 0y=2−2x=2−2(1)=03. 最終的な答え(1) (x,y)=(4,12)(x, y) = (4, 12)(x,y)=(4,12)(2) (x,y)=(−1,1)(x, y) = (-1, 1)(x,y)=(−1,1)(3) (x,y)=(1,2)(x, y) = (1, 2)(x,y)=(1,2)(4) (x,y)=(1,0)(x, y) = (1, 0)(x,y)=(1,0)