問連比題目!?...

Question

這題該怎麼解?...
若(x+2):(y-4):(z-1)=8:5:4 , 且2x-y-z=26 , 試求x+y+z=?

Answer 1

x+2=8r => x=8r-2
y- 4=5r => y=5r+4 代入 2x-y-z=26
z -1=4r => z=4r+1

2(8r-2)-(5r+4)- (4r+1)=26
16r- 4- 5r- 4- 4r- 1= 26
7r = 35 r= 5

r=5代入
x=8r-2 => x= 38
y=5r+4 => y= 29
z=4r+1 => z=21

所以 x+y+z=88 #

Answer 2

設：x+2 = 8rx = 8r-2y-4 = 5ry = 5r+4z-1 = 4rz = 4r+12x-y-z = 26將上面那些式子代入 →16r-4-(5r+4)-(4r+1) = 267r = 35r = 5x+y+z= 8r-2+5r+4+4r+1= 17r+3= 17*5+3(r = 5)= 88#

Answer 3

(x+2):(y-4):(z-1)=8r:5r:4r
(x+2)=8r---x=8r-2
(y-4)=5r---y=5r+4
(z-1)=4r---z=4r+1
把ｘｙｚ帶入2x-y-z=26
2（8r-2）-（5r+4）-（4r+1）=26
(16r-5r-4r)-4-4-1=26
7r=35
r=5(x=8r-2---y=5r+4---z=4r+1)
所以...x=40-2=38
...y=25+4=29
....z=20+1=21
故:x+y+z=38+29+21=88
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