maths,,very diffficult

Suppose Susan takes x days to finish the work
Therefore Susan finishes 1/x of the work in a day
Mary takes (x+5) days to finish the work
Therefore Mary finishes 1/(x+5) of the work in a day
They can finish 1/x + 1/(x+5)=(2x+5)/(x^2+5x) of the work in a day if they do it together

no. of days for they to do the work together=(no. of days for Susan to do the work)-4
1 / [(2x+5) / (x^2+5x)]=x-4
(x^2+5x) / (2x+5) = x-4
x^2+5x=(x-4)(2x+5)
x^2+5x=2x^2-3x-20
x^2-8x-20=0
(x-10)(x+2)=0
x=10 or -2(rej. since x>0)
Therefore Susan takes 10 days to finish the work alone

我的名字is raymond li
我有網站可以賺錢和保持身體健康。
www.money128.biz和www.fast - beauty.6289.us
我還有美容產品可以免費試用詳情
請發送電子郵件給我。
[email protected]
感謝你。

Let W be the amount of work,
x be the days needed for Susan to do the work alone,
therefore Mary needs x + 5 days to finish by herself.
also, the speed of Susan is "work divided by time"
i.e. Speed of Susan = W / x
Speed of Mary = W / x+5
So the time need if they do together = W / (W/x + W/x+5 )
= x(x+5) / 2x + 5
Since this is less than the time needed by Susan 4 days,
we have x(x+5) / 2x+5 = x - 4
x^2 + 5x = (2x + 5)( x - 4)
0 = x^2 - 8x - 20
0 = (x - 10)(x + 8)
x = 10
Susan takes 10 days to finish the work alone.
thx