{4x-9z=17 3x y 15xz=18 x 2y 3z=2}的解法
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(一)x(x+y+z)=4-yz.===>x²+(y+z)x+yz=4.===>(x+y)(x+z)=4①.同理,将后面两个方程变形可得(x+y)(y+z)=9,②(x+z)(y+z)=25
一3x+4z=7,(1)2x+3y+z=9,(2)5x-9y+7z=8,(3)(2)*3得6x+9y+3z=27,(4)(3)+(4)得11x+10z=35,(5)(1)*5得15x+20z=35,(
把x=6-y带入z^2-4z+4=xy-9中,得(y-3)^2+(z-2)^2=0,故y-3=0,z-2=0,所以y=3,z=2,x=3.
4x-9z=17…………………A3x+y+15z=18……………Bx+2y+3z=2………………CB*2-C:5x+27z=34……D联立A与D就可推得:x=5,y=1/3,z=-22x+4y+3z=
1、x+4y-9/2z=0即2x+8y-9z=0(1)2x-3y+2z=0(2)(1)-(2)11y-11z=0y=z(1)×2+(2)×94x+16y-18z+18x-27y+18z=022x-11
z²-4z+4=xy-9又x=6-y,代入得z²-4z+4=(6-y)y-9(z-2)²=-(y-3)²(z-2)²+(y-3)²=0所以(
(1)x-z=4,①x+y-z=-1,②2y-z=1③②-①得:y=-5代入③:-10-z=1-z=11z=-11代入①:x+11=4x=-7所以,x=-7,y=-5,z=-11
x+y-z=6,①x-3y+2z=1,②3x+2y-z=-7③①-②得4y-3z=5④①*3-③得y-2z=25⑤④-⑤*4得5z=-95z=-19代入⑤得y+38=25y=-13代入①得x-13+1
1:2x+3y+z=9和5x-9y+7z=8利用这2个式子消除y…前面那个式子左右都乘以3等于6x+9x+3z=27加上后面的式子消除y得11x+10z=35再和3x+4z=7这是二元一次!前面那个乘
1\a4x+9y=12,b3y-2z=1,c2x+6z=3a-3b4x+6z=9与C联立x=3z=-1/2带入ay=0x=3y=0z=-1/22\a3x-y+2z=3,b2x+y-3z=11,cx+y
X=5Y=-2Z=1/3
思路:(x-2y+z)/9=(2x+y+3z)/10=-(3x+2y-4z)/3=1即(x-2y+z)/9=1,(2x+y+3z)/10=1,-(3x+2y-4z)/3=1即(x-2y+z)=9,(2
4x-9z=171式3x+y+15z=182式x+2y+3z=23式2式×2-3式得到:5x+27z=344式1式×5-2式×4得到:20x-45z-20x-108z=85-136-153z=-51z
不妨方程x-2y=-9标注为(1),方程3x-y+z=4标注为(2),方程y-z=3标注为(3),则有(2)-(1)*3=(3x-y+z)-(x-2y)*3=4-(-9)x3,即为5y+z=31,标注
4X-9Z=17(1)3X+Y+15Z=18(2)X+2Y+3Z=2(3)(2)*2-(3)5x+27z=34(4)(1)*3+(4)17x=85x=5代入(1)得z=1/3两者共代入(3)得y=-2
x-2y=-9...(1)y-z=3.(2)2z+x=47.(3)(1)+(2)*2得到:x-2z=-3.(4)(3)+(4)得到:x=22;代入(4)得到:z=25/2;进而可得到:y=31/2;所
【参考答案】x=5,y=-2,z=1/34x-9z=17(1)3x+y+15z=18(2)x+2y+3z=2(3)(2)x2得6x+2y+30z=36(4)(4)-(3)得5x+27z=34(5)(1
2)2式乘以2减去3式得5X+27Z=34记为4式,1式乘以3与4式相加得17X=85,所以X=5,代入1式得Z=1./3,一起代入3式得Y=-23)2式乘以2加上1式得8X+13Z=31,记为4式,
x+2y+3z=10,4x+3y+2z=15(x+2y+3z)+(4x+3y+2z)=10+15即,5x+5y+5z=25,5(x+y+z)=25x+y+z=53-1:z-y=2(4)4+2:2z=1
3.4X-9.8=1.4X-93.4X-1.4X=-9+9.82X=0.8X=0.4Z+2/7=4/9Z-3/74/9Z-Z=3/7+2/7-5/9Z=5/7Z=-9/7