{x+y+z=6 2x+3y+6z=11 -3x+y+z=2
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/04 22:32:53
1.x+y+z2.-x-3y-1
(1)2x+3y-4z=-5(2)x+y+z=6(两边同时×33x+3y+3z=18(与(1)相减得(5)(3)x-y+3z=10(与(2)相加得(4))(4)2x+4z=16(5)x+7z=23(两
{3x+y-z=4①,2x-y+3z=12②,x+y+z=6③}①+②得5x+2z=16④,②+③得3x+4z=18⑤④×2—⑤得7x=14,x=2所以z=3、y=1所以方程组的解为x=2、y=1、z
有这样的公式:a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3(x+y
方程(1)+(2)得:3x-z=9④,方程(2)+(3)得:2x-z=7⑤,④-⑤得:x=2,把它代入⑤得:z=-3,把它代入(1)得:y=-3,∴原方程的解为x=2y=−3z=−3.
(1)x+y+z=6①3x-y+2z=12②x-y-3z=-4③①+②4x+3z=18④②-③2x+5z=16⑤⑤×24x+10z=32⑥⑥-④7z=14解得z=2代入⑤2x+5×2=16解得x=3将
2x-y+2z=-17①3x+y-3z=-4②x+y+z=6③③*2:2x+2y+2z=12④④-①:3y=29y=29/3带入③:x+z=-11/3⑤带入②:3x-3z=-41/3即x-z=-41/
x+y+z-6=02x+3y-z-12=02x-y-z=0组成方程组再解x=2y=3z=1
1.x+y=16①y+z=12②z+x=10③①-②x-z=4④③+④2x=14x=7⑤⑤代入①y=9⑥⑥代入②z=3x=7,y=9,z=3(2)3x-y+z=4①2x+3y-z=12②x+y+z=6
X+Y+Z=4,2*(X+Y+Z)+X=10,可以解出X=2.套入第二个和第一个.4+3Y-Z=66+2Y+2Z=10那么3Y=Z+2,2Y+2Z=4.Y=1,X=1X+Y+Z=4=2+1=12x+3
x-z=2.5y=7/2
x+2y-7z=a3x-4y+6z=b4x-2y-z=c则a+b=ca^3+b^-c^3=(a+b)^-3a^2b-3ab^2-c^3=-3ab(a+b)=-3abc=-3(x+2y-7z)(3x-4
x+y+z=6(1)2x-y+z=3(2)3x+9y+z=24(3)(1)-(2)得:2y-x=3(4)(3)-(1)得:2x+8y=18即x+4y=9(5)(4)+(5)得:6y=12y=2代入(4
9:12:10,我可是上课的时候给你回答的,
①x+2y-4z=0②3x+y-z=0①-2②x-6x-4z+2z=05x=2z代入①z=5x/2x+2y-10x=02y=9xy=9x/2x:y:z=1:9/2:5/2=2:9:5
x+2y+3z=6①2x+3y+z=6②3x+y+2z=6③,①+②+③得6x+6y+6z=18,所以x+y+z=3④,②-①得x+y-2z=0⑤,④-⑤得3z=3,解得z=1,③-①得2x-y-z=
2x+y+3z=38①3x+2y+4z=56②4x+y+5z=66③③-①得:2x+2z=28,即x+z=14④,①×2-②得:x+2z=20⑤,由④和⑤组成方程组:x+z=14x+2z=20,解得:
x+2y+3z=1 ①2x+3y+z=2 &nb
3元一次方程,好像是初一的问题哦.根据前面两个等式可以得出x=3zy=z(平方)/32x+3y+4z=2*(3z)+3*(z方/3)+4z现在变成了一元二次方程,你应该会解吧.
应该是3X=4Y,5Y=6Z吧?X+Y:Y+Z=[(4Y/3)+Y]:(Y+5Y/6)=14;11