【答案】
分析:觀察方程的特點,可發(fā)現(xiàn)方程(2)可分解為(2x+y)(x-3y)=0,則2x+y=0,x-3y=0,有y=-2x,x=3y,分別代入第一個方程,求值即可.
解答:解:
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,
由方程(2)可分解為(2x+y)(x-3y)=0,
則2x+y=0,x-3y=0,
∴y=-2x,x=3y,
分別代入(1)得:
x
2=1,2y
2=1,
解得:x
1=1,x
2=-1,y
3=
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,y
4=-
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,
再代入y=-2x,x=3y,可得
y
1=-2,y
2=2,x
3=
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,x
4=-
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.
點評:此題的關鍵是把方程(2)可分解為(2x+y)(x-3y)=0,再求解.