A simple arithmetic problem In a field grass grows at uniform rate ; If the field can feed 36 cows for 4 days , 21 cows for 9 days , how many cows can be feed for 18 days ? 
My solution :  Let $x$ be the initial amount  of grass before the cows start eating ; let $y$ be amount a cow can eat in one day and $z$ be the uniform rate of growth ; then $x- 4 \times 36y+4z=0$ and $x- 21\times 9y+9z=0$ ; we get $z=9y$ , so $x-144y+36y=0$ i.e. $x=108y$ ; now if $m$ cows can be feed for 18 days then $x- 18my+18z=108y-18my+162y=0$ , so $18m=108+162=270$ , so $m=15$ . I don't know whether I am correct or not . And also I would like a more Arithmetical solution ; Please help .  
 A: Your answer is correct.
For an arithmetic solution, in the first case, assume the grass is let grow for $4$ days, then the initial grass and the grass accumulated over $4$ days is equal to $36*4$ or $144$ cow-days. Similarly, in the second case, the initial grass and the grass accumulated over $9$ days is equal to $189$ cow-days. The difference between these two cases is the grass accumulated over $5$ days which is equal to $(189-144)$ or $45$ cow-days. That means, the grass accumulation per day is equal to $9$ cow-days. 
Take the first case. The grass accumulated for $4$ days is equal to $36$ cow-days. Since the initial grass and the grass accumulated over $4$ days is equal to $144$ cow-days, the initial grass is equal to $(144-36)$ or $108$ cow-days. 
Now, to find the answer to the question: in $18$ days, the grass accumulation is equal to $18*9$ or $162$ cow-days. Adding the initial amount of $108$ cow-days, the amount at the end of $18$ days is equal to $270$ cow-days. Divide $18$ into $270$ to find the number of cows, which is $15$. 
By the way, even algebraically, a simpler solution is:
Assume $I$ is the initial grass and the growth for day is $d$. 
$I+4d= 144$
$I+9d= 189$
From these two, 
$d=9, I=108$
If $m$ cows can be fed for $18$ days, 
$108+18*9= 18m$
which gives $m=15$
