Question 155897

OK.
Let x=amount of time it takes Heidi to milk the cows alone
Then Heidi can milk the cows at the rate of 1/x cows per hour
Let y=amount of time it takes Sarah to milk the cows alone
Then Sara can milk the cows at the rate of 1/y cows per hour
Now we are basically told that Sarah and Heidi can milk the cows together at the rate of 1/2 of the cows per hour
So, 1/x+1/y=1/2 multiply each term by 2xy
2y+2x=xy---------------------------------------eq1

Alone Heidi can milk the cows in x hours (1/x)*x=1 (all the cows)
And alone, Sarah can milk the cows in y hours (1/y)*y=1 
But we are also told that alone, it takes Heidi three hours longer than it takes Sarah, So:

x=y+3---------------eq2

substitute x=y+3 into eq1

2x+2(y+3)=(y+3)*y  get rid of parens

2y+2y+6=y^2+3y simplify
4y+6=y^2+3y  sybtract 4y and also 6 from eahc side

4y-4y+6-6=y^2+3y-4y-6  collect like terms

0=y^2-y-6 ot
y^2-y-6=0 quadratic in standard form and it can be factored
(y-3)(y+2)=0
y=3 hrs------------------------amount of time it takes Sarah to milk the cows
(Note: we disregard the solution y=-2 for in this case time is positive)

Substitute y=3 into eq1
x=3+3 =6 hrs ---------------------amount of time it takes Heidi to milk the cows

CK
1/3 + 1/6 =1/2
2/6 + 1/6=1/2
1/2=1/2

and
6=3+3
6=6

Hope this helps-----ptaylor