用加减消元法解二元一次方程组 {3(x-1)=y+5 2(y+1)=x+1 {4(x+2)=3(5-y) 5(x+1)=2(4+y)
用加减消元法解二元一次方程组 {3(x-1)=y+5 2(y+1)=x+1 {4(x+2)=3(5-y) 5(x+1)=2(4+y)
数学人气:356 ℃时间:2019-10-23 08:59:43
优质解答
1、3(x-1)=y+5x+1=2(y+1)式1减去式2×3,得到-6=y+5-6(y+1)所以5y=5,解得y=1,x=2y+1=3故方程组的解为x=3,y=12、式1×2加上式2×3,得到8(x+2)+15(x+1)=6(5-y)+6(4+y)即23x+31=30+24所以23x=23解得x=1,那么y=5 -4*(1+2)...
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