已知x-y=5,y-z=3
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由x+3y-5z=0得x=5z-3y代入2x-y-3z=0中,得10z-6y-y-3z=07z=7y∴y=z代入x=5z-3y中得x=5y-3y=2y∴x=2y于是x∶y∶z=2y∶y∶y=2∶1∶1
5x+3y=3z--------a-x-3y=-z--------ba式+b式4x=2z得z=2x代入a中得x=3y,y=x/3x:y:z=x:x/3:2x=1:1/3:2=3:1:6
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
x=4y/3y=yz=2y/5所以,x:y:z=4/3:1:2/5=20:15:6
由x+y-z=0,2x-y+2z=0可得:z=-3xy=-4x则3x-2y+5z/5x-3y+2z=3x+8x-15x/5x+12x-6x=-4x/11x=-4/11
x+2y-3z=0⑴2x+3y+5z=0⑵⑵-⑴得x+y-2z=0⑶⑶-⑴得y=z代入⑶得x=y=z所以(x+y+z)除以(x-y+z)=3
将Z当成已知数,将X、y用Z来表示2x+3y-4z=03x+4y+5z=0整理得:2x+3y=4z3x+4y=-5z变成二元二次方程解之得:x=-31zy=22z代人x+y+z/x-y+z=2/13
哇,这个很复杂啊再问:你会吗?
设:x/4=y/5=z/6=k则有:x=4k,y=5k,z=6k(x+y+z)/(3x-2y+z)=(4k+5k+6k)/(12k-10k+6k)=15k/8k=15/8
两式相加,得到4x-4z=0x=z代入①即可求出y=2z∴(x+y+z)/(x+2y-3z)=(z+2z+z)/(z+4z-3z)=(4z)/(2z)=2二十年教学经验,专业值得信赖!如果你认可我的回
由题意得:x+y=3①y+z=4②x+z=5③①+②+③得:2x+2y+2z=12,即x+y+z=5.故选A.
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
4x-5y+2z=0(1)x+2y=3z(2)(2)×4-(1)得:13y=14zy=14/13z(1)×2+(2)×5得:13x=11zx=11/13z所以:x:y:z=11/13:14/13:1=
【解】视z为常数,由已知两方程,可解得x=3zy=2z将其代入待求值式中,得3x*x+2y*y+5z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+5z^2]/[5(3z)^2+(
设a(2x+5y+4z)+b(7x+y+3z)=x+y+z比较系数得2a+7b=5a+b=4a+3b=1a=1/11,b=2/11因此x+y+z=a(2x+5y+4z)+b(7x+y+3z)=1/11
/>x/4=y/5=z/6=t分别用t表示x,y,z然后带入到要求的式子x+y+z/3x-2y+z中最终解得结果
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+
X=3K,Y=4K,Z=5K3X+2Y-4Z=189K+8K-20K=18K=-6X=-18,Y=-24,Z=-30X+Y+Z=-72
应该是3X=4Y,5Y=6Z吧?X+Y:Y+Z=[(4Y/3)+Y]:(Y+5Y/6)=14;11