数学题{x+2Y+3z=14 2x+y+z=7 3x+y+2z=11

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数学题{x+2Y+3z=142x+y+z=73x+y+2z=11数学题{x+2Y+3z=142x+y+z=73x+y+2z=11数学题{x+2Y+3z=142x+y+z=73x+y+2z=11x+2Y

数学题{x+2Y+3z=14 2x+y+z=7 3x+y+2z=11
数学题{x+2Y+3z=14 2x+y+z=7 3x+y+2z=11

数学题{x+2Y+3z=14 2x+y+z=7 3x+y+2z=11
x+2Y+3z=14 (1)
2x+y+z=7 (2)
3x+y+2z=11 (3)
(2)*2-(1)得3x-z=0
(3)*2-(1)得5x+z=8
两式相加,8x=8
x=1
z=3
所以y=2

x=14-2y-3z对吧。把x分别带入第二个和第三个式子里,可得关于yz的二元一次方程组。可以解出来。带入原来的式子求x,过程一定自己解。这样更有成就感。

x+2y+3z=14 2x+y+z=7 3x+y+2z=11
(3x+y+2z)- (2x+y+z)=4
x+z=4 3x*2+y*2+2z*2=22 6x+2y+4z=22 (6x+2y+4z)-(x+2y+3z)=8
5x+z=8 ( 5x+z)-(x+z)=4 4x=4 x=1 z=3 y=2

将x+2Y+3z=14乘2得2x+4y+6z=28,用2x+4y+6z=28减2x+y+z=7得3y+5z=21
将2x+y+z=7乘3得6x+3Y+3z=21,,将3x+y+2z=11 乘2得6x+2Y+4z=22,用6x+2Y+4z=22减6x+3Y+3z=21得-y+z=1
-y+z=1和3y+5z=21解得y=2,z=3
将y=2,z=3代入x+2Y+3z=14 解得x=1
则方程的解为x=1,y=2,z=3

不知道你是几年级。这道题其实不难。我们把X+2Y+3Z=14方程一.2X+Y+Z=7看作方程二.3X+3Y+4Z=21看作方程三.方程三减去方程一和二的和。得Y+Z=5.代在方程二得X=1.代入方程一得Y=2.Z=3