几道关于六年级一次方程组的问题(1) {(x-1)/3-(y+2)/4=7{(x-1)/3+(y+2)/4=3(2) {(2x+3y)/4-1=y{(x/3)+(y/2)=4(3) {x+y+z=26{x-y=1{2x-y+z=18(4) {x+y=3{y+z=5{z+x=4(5) {5x+4y+z=0{3x+y-4z=11{x+y+z=-2(6) {(x/2)=(y/
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几道关于六年级一次方程组的问题(1) {(x-1)/3-(y+2)/4=7{(x-1)/3+(y+2)/4=3(2) {(2x+3y)/4-1=y{(x/3)+(y/2)=4(3) {x+y+z=26{x-y=1{2x-y+z=18(4) {x+y=3{y+z=5{z+x=4(5) {5x+4y+z=0{3x+y-4z=11{x+y+z=-2(6) {(x/2)=(y/
几道关于六年级一次方程组的问题
(1) {(x-1)/3-(y+2)/4=7
{(x-1)/3+(y+2)/4=3
(2) {(2x+3y)/4-1=y
{(x/3)+(y/2)=4
(3) {x+y+z=26
{x-y=1
{2x-y+z=18
(4) {x+y=3
{y+z=5
{z+x=4
(5) {5x+4y+z=0
{3x+y-4z=11
{x+y+z=-2
(6) {(x/2)=(y/3)=(z/5)
{x+y+z=20
几道关于六年级一次方程组的问题(1) {(x-1)/3-(y+2)/4=7{(x-1)/3+(y+2)/4=3(2) {(2x+3y)/4-1=y{(x/3)+(y/2)=4(3) {x+y+z=26{x-y=1{2x-y+z=18(4) {x+y=3{y+z=5{z+x=4(5) {5x+4y+z=0{3x+y-4z=11{x+y+z=-2(6) {(x/2)=(y/
1.两式相加可得X=16 代入一式可得Y=-10
2.(x/3)+(y/2)=4→x=12-3/2y→(2*(12-3/2y)+3y)/4-1=y→y=5 x=11/2=5
3.x=1+y→x+y+z=(1+y)+y+z=1+z+2y=26
2x-y+z=2(1+y)-y+z=2+y+z=18 两式相减可得:y=9
x-y=1→x=10 z=7
4.x=3-y→z+x=z+3-y=4
y+z=5 两式相加得:z=3 →x=1 y=2
5.z=-(5x+4y)代入其他两式得:23x+17y=11
4x+3y=2 两式相减得:y=2 x=-1 z=-3
6.x=2/5z y=3/5z 代入第三式得:z=10 x=4 y=6
楼上的第一题错了吧 。。。 应该是X=16 Y=-10吧