解下列三元一次方程组:x/2=y/3=z/5,3x-y+2z=26

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解下列三元一次方程组:x/2=y/3=z/5,3x-y+2z=26
解下列三元一次方程组:x/2=y/3=z/5,3x-y+2z=26

解下列三元一次方程组:x/2=y/3=z/5,3x-y+2z=26
x/2=y/3 => x=2/3y -------(1)
y/3=z/5 => z=5/3y ------(2)
将(1)(2)代入3x-y+2z=26中
3*2/3y-y+2*5/3y=y+10/3y=13/3y=26 => y=6 => x=4 z=10

x/2=y/3=z/5=t
x=2t, y=3t , z=5 t

3x-y+2z=26
6t- 3t +10t=26
13t=26
t=2

x=4
y=6
z=10

解设一份为A
那么x/2=y/3=z/5=t
解得 {X=2A
{Y=3A
{Z=5A
{X=2A
把 {Y=3A 带入3x-y+2z=26中
{Z=5A
解得 X=4 Y=6 Z=10