3X=5Y=7Z=11

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已知5x=3y,4z=7y,求x:y:z

x:y:z=(3y/5):y:(7y/4)=(3/5):1:(7/4)=12:20:35再问:已知x+2y-z=02x+3y+z=0求x:y

3x+7y+z=5 ,4x+10y+z=6 ,x+y+z=

x+y+z=A(3x+7y+z)+B(4x+10y+z)易知3A+4B=1,7A+10B=1,A+B=1,解得A=3,B=-2,所以原式=3*5-2*6=3,以后也是这样做就好了!给分吧!

5x+3y+2z=2011 4x+6y+7z=2012 求x+y+z

(5x+3y+2z)+(4x+6y+7z)=2011+20129(x+y+z)=4023x+y+z=447

x+2y+4z=17 2x+y+z=7 3x+y+2z=11

X+2Y+4Z=17.①2X+Y+Z=7.②3X+Y+2Z=11.③③-②,得:x+z=4.④②+③-①,得:4x-z=1...⑤④+⑤,得:5x=5x=1代入④,得:1+z=4z=3再代入②,得:2

解下列方程组(1)3x-y+z=3 2x+y-3z=11 x+y+z=12(2)5x-4y+4z=13 2x+7y-3z

第一题3x-y+z=3①;2x+y-3z=11②;x+y+z=12③;①减③2X-2Y=-9④;3倍的③即3X+3Y+3Z=36⑤;⑤加②5X+4Y=47⑥;2倍的④4X-4Y=-18⑦;⑥式加⑦9X

matlab解方程怎么出错了[x,y,z]=('x^2-5*y^2+7*z^2+12=0 ','3*x*y+x*z-11

[x,y,z]=solve('x^2-5*y^2+7*z^2+12=0','3*x*y+x*z-11*x=0','2*y*z+40*x=0')

已知x,y,z满足3x+7y+z=5 4x+10y+z=3 求x+y+z的值

93x+7y+z=5所以6x+14y+2z=10又因为4x+10y+z=3所以2x+4y+z=7原题中两式相减得x+3y=-2所以x+y+z=9

x+2y+3z=14 2x+y+z=7 3x+y+2z=11 求x y z

x=1,y=2,z=3联立任意两个,消除z,x+2y+3z=142x+y+z=75x+y=7再联立其他的的两个.消除z2x+y+z=73x+y+2z=11x+y=3然后联立这两个x+y=35x+y=7

2x+5y+4z=0,3x+y-7z=0,则x+y-z=?

解法2:2x+5y+4z=0式①3x+y-7z=0式②x+y-z=?式①×3-式②×23(2x+5y+4z)-2(3x+y-7z)=015y+12z-2y+14z=013y+26z=0式③式①-式②×

2x+3y-z=11 2x+y-5z=8 -2+7y+z=19

顺序将三个算式设为①②③①-②=2y+4z=3……④②+③=8y-4z=27……⑤④+⑤=10y=30y=3带入⑤z=-3/4带入①x=17/8多思考一下吧2x+3y-z=11……(1)2x+y-5z

x+2y+3z=14 2x+y+z=7 3x+y+2z=11

2x+y+z=7,3x+y+2z=11可得x+z=42(2x+y+z)=14,x+2y+3z=14可得3x-z=0x+z=4,3x-z=0可得x=1,所以z=3,y=2

x-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 分别求出x=?y=?z=?

x-y-z=-1(1)3x+5y+7z=11(2)4x-y+2z=-1(3)(1)*2+(3)得6x-3y=-32x-y=-1(4)所以2x-y=4x-y+2z=-1x+z=0代入(2)有5y+4z=

已知x+4y+3z=3x-2y-5z=0,求x+2y-z/2x-3y+7z的值

x+4y+3z=3x-2y-5z=0则x+4y+3z=0①3x-2y-5z=0,则6x-4y-10z=0②①②两式相加,得7x-7z=0,所以x=z代入①,得z+4y+3z=0,所以y=-z所以x+2

已知x、y、z满足{3x+7y+z=5 ,求x+y+z的值 {4x+10y+z=3

3x+7y+z=5.(1)4x+10y+z=3.(2)(1)*3-(2)*2有9x+21y+3z-(8x+20y+2z)=5*3-3*2x+y+z=15-6x+y+z=9

{5x-3y+z=2{5x+2y-4z=3{-5x+y-z=2 {x-y-z=-1{3x+5y+7z{4x-y+2x=-

是三元一次方程组吗?是的话过程很多……再问:是三元一次方程再答:5x-3y+z=2(1)5x+2y-4z=3(2)-5x+y-z=2(3)(1)+(3),得:-2y=4y=-2(4)(2)+(3),得

2x+5y+4z=6,3x+y-7z=-4,x+y-z=?

已知,2x+5y+4z=6,3x+y-7z=-4,可得:2(2x+5y+4z)+3(3x+y-7z)=2*6+3*(-4)=0;即有:13(x+y-z)=0,所以,x+y-z=0.

已知2x+5y+4z=6 3x+y-7z=-4求x+y-z

解法1:2x+5y+4z=0式①3x+y-7z=0式②x+y-z=?式③式①=0,式②=0,所以式①-式③=式②-式③即:2x+5y+4z-x-y+z=3x+y-7z-x-y+zx+4y+5z=2x+

若3x+7y+z=5,4x+10y+z=3,则x+y+z=?

3x+7y+z=5.(1)4x+10y+z=3.(2)(1)*3-(2)*2得x+y+z=15-6=9所以x+y+z=9

{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2

这个题目没有问题么,我是说最后一个式子确定是z+5y+8z=-2?如果没有问题的话:x+y+z=1;①x+3y+7z=-1;②z+5y+8z=-2③①-②2Y+6Z=-2Y=(-2-6Z)/2=-1-

x/2=y/3=z/5 x+3y-z/x-3y+z

设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是:(x+3y-z)/(x-3y+z)?若是,如下:(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3