Question 149794
let x="speed in still water" __ d=rt __ t=d/r __ time upstream plus time downstream equals 9 hr


upstream __ t=60/(r-3)


downstream __ t=60/(r+3)


[60/(r-3)]+[60/(r+3)]=9 __ multiplying by (r+3)(r-3) __ 60(r+3)+60(r-3)=9(r+3)(r-3)


120r=9r^2-81 __ 0=9r^2-120r-81 __ dividing by 3 __ 0=3r^2-40r-27