数学题``````````````````````````````````````~一列快车长306m,一列慢车长344m,两车相向而行,从相遇到离开需要13s;如果同向而行,快车从追及到超过慢车需要65s,求快车、慢车各自的速度列方程组13(x+y)=306

Question

数学题``````````````````````````````````````~
一列快车长306m,一列慢车长344m,两车相向而行,从相遇到离开需要13s;如果同向而行,快车从追及到超过慢车需要65s,求快车、慢车各自的速度
列方程组13(x+y)=306+344
65(x-y}=344+306
为什么列这样的方程组?
65(x-y}=344-306

Answer 1

13(x+y)=306+344
13秒走过的路程是两辆列车的车长
65(x-y}=344+306
快车65秒比慢车多走过的距离是慢车的车长