{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5

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{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5

{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
二元一次方程 和 三元
{y-1/2x+1=0 {2y+3(x-2/2)-10=10
{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5

{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
你这个是分两题的吗.
∵y-1/2x+1=0 ∴y=1且x≠-1/2
∵ 2y+3(x-2/2)-10=10 ∴x=14
同理下一题利用11x+3z=9将z用x带掉,即z=(9-11x)/3
再带入{3x+2y+z=8 {2x-6y+4z=5将他们看成二元一次方程 进行计算