{x-y=3 2x+3(x-y)=11 用加减消元法解方程组
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{x-y=32x+3(x-y)=11用加减消元法解方程组{x-y=32x+3(x-y)=11用加减消元法解方程组{x-y=32x+3(x-y)=11用加减消元法解方程组第二个方程,展开先得到.5x-3
{x-y=3 2x+3(x-y)=11 用加减消元法解方程组
{x-y=3 2x+3(x-y)=11 用加减消元法解方程组
{x-y=3 2x+3(x-y)=11 用加减消元法解方程组
第二个方程,展开先
得到.5x-3y=11,
第一个方程 x-y=3.变成,5x-5y=15.
,两式相减.
5x-5y-(5x-3y)=15-11.
得到y=-2.
所以,x=1
x=1,y=-2
由
x-y=3
2x+3(x-y)=11
有
x-y=3
5x-3y=11
有
3x-3y=9
5x-3y=11
两式相减,得:
2x=2
x=1
所以
y=-2
x-y=3 。。。。。。.。。。①
2x+3(x-y)=11。。。。。②
由②-3×①的x=1.。。。。。③
由③-①的 y=-2 。。。。④
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