Question

解三元一次方程组（x-1）/3=(y-3)/4=(z+2)/5 2x-3y+2z=1

用加减消元法或者代入消元法，答案一定要准确，检验过。（不要设k法）

Answer 1

（x-1）/3=(y-3)/4

4x-4=3y-9
3y=4x+5
y=(4x+5)/3

（x-1）/3=(z+2)/5

5x-5=3z+6
z=(5x-11)/3
代入2x-3y+2z=1
2x-(4x+5)+2(5x-11)/3=1
-2x-5+10x/3-22/3=1
4x/3=40/3
所以
x=10
y=(4x+5)/3=15

z=(5x-11)/3=13

Answer 2

(X-1)/3=(Y-3)/4=(Z+2)/5;2X-3Y+2Z=1
令(X-1)/3=(Y-3)/4=(Z+2)/5=K
则(X-1)/3=K,X=3K+1
同理,Y=4K+3,Z=5K-2
代入2X-3Y+2Z=1中,
2*(3K+1)-3*(4K+3)+2*(5K-2)=1
6K+2-12K-9+10K-4=1
4K=12
K=3
所以
X=3K+1=3*3+1=10
Y=4K+3=4*3+3=15
Z=5K-2=3*5-2=13

Answer 3

x=3/5z+11/5 y=4/5z+23/5 代入2x-3y+2z=1
x=10
y=15
z=13

Answer 4

解：由题意，（x-1)/3=(z+2)/5 ①
(y-3)/4=(z+2)/5 ②
2x-3y+2z=1 ③
解①得 x=3/5z+11/5
解②得 y=4z/5+23/5
将①②代入③解得z= 13
所以x= 10
y= 15

Answer 5

5(x-1)/3-2=1,化简得x=10,
y=15; z=13