# Cows And Grass: A 3 By 3 System Problem

2 cows in 4 weeks can eat all the grass on 2 acres and all the grass that grows on these 2 acres during the 4 weeks. 3 cows in 2 weeks can eat all the grass on 2 acres and all the grass that grows on these 2 acres during the two weeks. How many cows can eat in 6 weeks all the grass on 6 acres and all the grass that grows on these 6 acres during the 6 weeks? Assume the grass has a constant growth rate and is distributed evenly.

I feel like I should use a 3 by 3 system to solve this problem but I only got to:

$x$= amount of grass 1 cow eats in a week

$y$= amount of grass that grows on an acre per week.

What should I do next?

• You also need z = initial amount of grass per acre – WW1 Dec 13 '16 at 22:50
• @WW1 That's what I'm missing! Thanks! :) – Yuna Kun Dec 13 '16 at 22:56

Case 1:the amount of grass consumed by the cows is $$2+4\times 2\times a$$ Then the rate at which one cow can eat is simply $$\frac{(2+4\times 2\times a)}{2\times 4}$$ Case 2: the amount of grass consumed by the cows is $$2+2\times 2\times a$$ And so the rate at which one cow can eat is $$\frac{(2+2\times 2\times a)}{3\times 2}$$ Now we set the expressions from case 1 and case 2 equal: $$\frac{(2+4\times 2\times a)}{2\times 4}=\frac{(2+2\times 2\times a)}{3\times 2}$$ We get a = 0.25 acres/week
Case 3: the amount of grass consumed by the cows is $$6+6\times 6\times a=6+6\times 6\times0.25=15$$ So the rate at which one cow can consume grass is $$\frac{15}{6\times (number\;of\;cows)}$$ Since we know one cow can eat at a rate of 0.5 acres/week $$\frac{15}{6\times (number\;of\;cows)}=0.5$$ $$number\;of\;cows=5$$