小学算术题：X+Y=60 Y+z=51 z+x=39,x,y,z分别是多少？

小学算术题：X+Y=60 Y+z=51 z+x=39,x,y,z分别是多少？

解：
x+y = 60 (1)
y+z = 51 (2)
z+x = 39 (3)
(1) - (2) 得
x - z = 9 (4)
(3) + (4) 得
2x = 48
x = 24
代入(1) 得
y = 60 -x = 36

再由(3) 得
z = 39 - x = 15

所以，
x = 24
y = 36
z = 15

三式相加 2X+2Y+2Z=150 X+Y+Z=75……4
4-1 Z=15
4-2 X=24
4-3 Y=36

x+y+t+z+z+x=150
2(x+y+z)=150
x+y+z=75
X+Y=60
z=75-60=15
y=51-15=36
x=39-15=60-36=24

x+y+y+z+z+x=150
2(x+y+z)=150
x+y+z=75
x+y=60
z=15
z+x=39
x=24
y+z=51
y=36

x+y+t+z+z+x=150
2(x+y+z)=150
x+y+z=150
x+y=50
y=10
x=12

由题意可得｛X+Y=60 ①
｛Y+z=51 ②
｛z+x=39 ③
①+②+③ 2x+2y+2z=150
x+y+z=75 ④
④-① z=15
将z=15带入② 得 y=36
将z=15带入③ 得 x=24 ｛x=24
｛y=36 是原方程组的解
｛z=15