若x+2y+3z=11,3x+5y+7z=27,求x+y+z的值RT
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若x+2y+3z=11,3x+5y+7z=27,求x+y+z的值RT
若x+2y+3z=11,3x+5y+7z=27,求x+y+z的值
RT
若x+2y+3z=11,3x+5y+7z=27,求x+y+z的值RT
3x+5y+7z=27减x+2y+3z=11
得出2x+3y+4z=16再减x+2y+3z=11 得:
x+y+z=5
x+2y+3z=11得到z=(11-x-2y)/3;
3x+5y+7z=27得到y=8-2x.
带入即可得到了,x全部消去了。
一式乘以2,得2x+4y+6z=22 三式
二式减去三式 得x+y+z=5
x+2y+3z=11<1> 3x+5y+7z=27<2> <2>-<1>得2x+3y+4z=16<3> <3>-<1>得:x+y+z=5
1式*3与2式相减 得Y=6-2Z
1式*5与2式*2后两式相减 得X=Z-1
故 X+Y+Z=z-1+6-2z+z=5
先用后面的一个式子减去前面的式子得到:2x+3y+4z=16
再用此式减去题中的x+2y+3z=11,得到:x+y+z=5
5
若X:Y:Z=5:6:7那么(Y-X):(Y+Z)和(X+2Y+3Z):(3X+2Y+Z)
1.{x:y:z=3:4:5 2x+3y-5z=-142.{x-y-z=-1 3x+5y+7z=11 4x-y+2z=-13.{x+y=1 x+z=0 2x+y+2z=44.{x-y+z=-3 2x-y-z=-2 -3x+y+z=-15.{x-y+z=-3 3x+y-4z=6 2x+y-z=36.{x-2y+5z=16 2x+y-5z=7 2x-y+z=38
x-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 分别求出x=?y=?z=?
三元一次方程 X-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 X=?Y=?Z=?
x+y-z=3 y+z-x=5 z+x-y=7 求x,y,z
{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2
2x+3y-z=11 2x+y-5z=8 -2+7y+z=19
x/2=y/3=z/5 x+3y-z/x-3y+z
若x+y+z=3y=2z,则x/x+y+z=?
..1.x+y+z=26,2.4y+3z=7,y-z=1,3x+2z=17,x+2y-z=18 9x-7y=173.x-7y+3z=17 4.2x+4y+3z=95x-6y-z=24 3x-2y+5z=113x+7y-2z=1 5x-6y+7z=131.x+y+z=26,y-z=1x+2y-z=18 2.4y+3z=7,3x+2z=11,9x-7y=17 3.x-7y+3z=17 5x-6y-z=24 3x+7y-2z=14.2x+4y+3z=93x-2y+5z=11 5x-6y+7z=13
(1)2X+4Y+3Z=93X-2Y+5Z=115X-6Y+7Z=13(2)X:Y=3:2Y:Z=5:4X+Y+Z=66(3)3X-Y+Z=32X+Y-3Z=11X+Y+Z=13以上3个题目为三元一次方程~
已知x-y/x+y=y+z/2(y-z)=z+x/3(z-x),求证8x+9y+5z=0THX..
六年级三元一次方程组第一题:{5x-3y+z=2 5x+2y-4z=3 -5x+y-z=2第二题:{x+y=-2 x+z=13 y+z=1第三题:{x-y-z=-1 3x+5y+7z=11 4x-y+2z=-1
三元一次方程组题目x+y-z=11 y+z-x=5 x-y+z=1x+y=4 y+z=3 x+z=5若已知x-3y+2z=0 2x+3y-8z=0, 且xyz≠0,则x:y:z=
x+Y+Z=1 X+3Y+7Z=-1 X+5Y+8Z=2x+Y+Z=1X+3Y+7Z=-1X+5Y+8Z=2
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z
1.已知x.y.z均为整数,若11∣(7x+2y-5z),求证:11∣(3x-7y=12z)