Question

用加减消元法解决二元一次方程组 2x-5y=7 { x-y/3=x+y/4 2x-5y=7 { x-y/3=x+y/4

Answer 1

2x-5y=7 (1)
x-y/3=x+y/4 (2)
请确认一下（2）有没错,如果没错,直接由（2）得出y=0,代入（1）x=3.5
由题目及解题要求看,（2）应该有括号的,应该(x-y)/3=(x+y)/4吧?
2x-5y=7 (1)
(x-y)/3=(x+y)/4 (2)
(2)*12：(x-y)/3*12=(x+y)/4*12
4x-4y=3x+3y
x-7y=0 (3)
(1)-(2)*2：(2x-5y)-(x-7y)*2)=7
2x-5y-2x+14y=7
9y=7
y=7/9 (4)
(4)代入(3)：x-7*7/9=0
x=5/4/9
x=5/4/9(5又4/9)、y=7/9
经验算,x=5/4/9(5又4/9)、y=7/9 是方程组的解.