一道初一数学题！！！

y=(5-2t)/(7-4t) 等同于7y-4ty=5-2t
即7y-5=4ty-2t=t(4y-2)
得t=(7y-5)/(4y-2) (#)
将（#）代入 x=1-2t得x=1-(7y-5)/(2y-1)=(4-5y)/(2y-1)

你题目里的(5-2t)/7-4t是(5-2t)/(7-4t) 还是(5-2t)/7 -4t呢?我按照第一种解的

y=(5-2t)/7-4t ------>首先用y来表示t ---> t=(5-7y)/30
把 t=(5-7y)/30 带入x=1-2t
即x=1-(5-7y)/15

t=(1-x)/2 y=(5-2t)/(7-4t) 带入 t与x的关系式 所以y=(4+x)/(5+2x)

y=(5-2t)/7-4t
=5/7-2t/7-4t
=5/7-30t/7
所以t=(5/7-y)*(7/30)
=1/6-7y/30
所以x=1-2*(1/6-7y/30)
=1-1/3+7y/15
=2/3+7y/15

y = (5 - 2t)/(7 - 4t)
7y - 4yt = 5 - 2t
2t(2y - 1) = 7y - 5
t = (7y - 5)/[2(2y - 1)]

x = 1 - 2t = 1 - (7y - 5)/(2y - 1) = (-5y + 4)/(2y - 1)