解:(1)
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,
把①代入②得:2y-z=16…④,
把①代入③得:4y+z=164…⑤,
④+⑤得:6y=180,解得:y=30,
把y=30代入①得:x=66,
把x=66,y=24代入③得:z=50,
則方程組的解是:
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;
(2)
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,
①+②得:5x-y=7…④,
②×2+③得:8x+5y=-2…⑤,
解方程組:
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,解得:
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,
把
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代入②得:2-2-z=4,則z=-4.
故方程組的解是:
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;
(3)
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,
①+②+③得:2x+2y+2z=2,即x+y+z=1…④,
④-①得:z=-4,
④-②得:x=2,
④-③得:y=3.
故方程的解是:
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;
(4)
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,
③-①得:x-2y=-8…④,
②-④得:y=26,
把y=26代入②得:x=27,
把x=27,y=26代入①得:z=-27.
故方程組的解是:
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.
分析:(1)把第一個方程分別代入另外兩個方程,即可消去x,得到關(guān)于y、z的兩個方程,求解y、z的值,進而求得x的值;
(2)通過①+②和②×2+③即可消去z,得到關(guān)于x、y的方程組求得x、y的值,進而代入求得z的值;
(3)三個式子相加求得x+y+z的值,然后分別減去每個方程,即可求解;
(4)③-①即可消去z,然后利用加減法即可求得y的值,進而代入求得x、z的值.
點評:本題考查二元一次方程組和三元一次方程組的解法,有加減法和代入法兩種,一般選用加減法解二元一次方程組較簡單.