解二元一次方程组x/2+y/3=2和x/3+y/4=17/6,
2019-05-28
解二元一次方程组
x/2+y/3=2和x/3+y/4=17/6,
优质解答
x/2+y/3=2 两边乘以6得:
3x+2y=12 (1)
x/3+y/4=17/6 两边乘以12得:
4x+3y=34 (2)
(2)-(1)得:
x+y=22 (3)
(3)两边乘以3,
3x+3y=66 (4)
(4)-(1)得:
y=54
带入(3)得:
x=22-54
x=-32.
x/2+y/3=2 两边乘以6得:
3x+2y=12 (1)
x/3+y/4=17/6 两边乘以12得:
4x+3y=34 (2)
(2)-(1)得:
x+y=22 (3)
(3)两边乘以3,
3x+3y=66 (4)
(4)-(1)得:
y=54
带入(3)得:
x=22-54
x=-32.