有一條數學題想問

50[(2y/5)+(y/10)=[(y/2)-(y/5)]y-5y
20y+5y=(y²/2)-(y²/5)-5y
250y=5y²-2y²-5y
3y²-255y=0
3y(y-85)=0
y=0 or y-85=0
y=0 or y=85

50[(2y/5)+(y/10)=[(y/2)-(y/5)]y-5y
20y+5y=(y²/2)-(y²/5)-5y
250y=5y²-2y²-5y
3y²-255y=0
3y(y-85)=0
y=0 or y-85=0
y=0 or y=85

50(2y/5+y/10)=(y/2-y/5)y-5y
20y+5y=(y/2-y/5)y-5y
20y+5y+5y=(y/2-y/5)y-5y+5y
30y/y=(y/2-y/5)y/y
30=(y/2-y/5)x10
30=5y-2y
30=3y
y=10

50(2y/5+y/10)=(y/2-y/5)y-5y

20y + 5y=y^/2 - y^/5 - 5y

25y=5y^-2y^/10 - 5y

25y + 5y=(5y^ - 2y^)/10

30y=3y^/10

300y=3y^

y^=100

y=10