Question

解下列三元一次方程组

(1)｛2x-5y=-11 ① 7y+4z=5 ② 5x+2z=2 ③｝
(2)｛3x-y+z=4 ① 2x+3y-2z=12 ② 3x+2y=66 ③｝
(3)｛3x+2y+z=14 ① x+y+z=10 ② 2x+3y-z=1 ③｝
(4)｛2x+y=42 ① 2y+z=48 ② x+2z=48 ③｝

(求求各位内,,我要步骤额,,)
各位行行好帮帮忙啦,,不然我今晚不要睡觉了

Answer 1

解： (1)｛2x-5y=-11 ① 7y+4z=5 ② 5x+2z=2 ③｝
由③×2→10x+4z=4④-②→10x-7y=-1-①×5→18y=54→y=3
代入①→x=2再后求得z=4
(2) ｛3x-y+z=4 ① 2x+3y-2z=12 ② 3x+2y=66 ③
①×2-②→8x+y=20→16+2y=40-③→13x=-26→x=-2代入③→y=36
再后求得z=46
(3) ｛3x+2y+z=14 ① x+y+z=10 ② 2x+3y-z=1 ③｝
②+③→3x+4y=11④
①-②→2x+y=4→y=4-2x⑤代入④→x=1代入⑤→y=2再后求得z=7
(4) ｛2x+y=42 ① 2y+z=48 ② x+2z=48 ③｝
②×2-③→4y-x=48④×2+①→9y=138→y=46/3代入①→x=40/3
再后求得z=52/3