Three pasture fields have areas of $\frac{10}{3}$, $10$ and $24$ acres, respectively. The fields initially are covered with grass of the same thickness and new grass grows on each at the same rate per acre. If $12$ cows eat the first field bare in $4$ weeks and $21$ cows eat the second field bare in $9$ weeks, how many cows will eat the third field bare in $18$ weeks? Assume that all cows eat at the same rate.
I set up a system of equations where $T =$ thickness, $g =$ rate at which grass grows back per week, $x_1 =$ rate at which cows eat per week, $x =$number of cows that will eat the third field in 18 weeks.
We then have $\begin{cases} \frac{10}3T+4g = 48x_1 \\10T + 9g = 21 \cdot 9x_1 \\ 24T+18g = 18x \cdot x_1 \end{cases}$.
Is this correct? And if so how should I solve it or is there a simpler way?
Here's a non-algebraic way,
using the unit cow-weeks (c-w) like man-days, and realizing that
c-w will be directly proportional to acreage
Initially, I shall transform figures to compute for $\frac{10}3$ acres
Fodder needed for $4$ weeks $=12\cdot4 = 48\;$ c-w
Fodder needed for $9$ weeks $=\frac13\cdot21\cdot9 = 63\;$ c-w
Fodder growth in $5$ weeks $=63-48 = 15\;$ c-w
$\therefore$ Fodder growth in $9$ weeks $= 27\;$ c-w
Fodder needed for $18$ weeks $= 63+27 =90\;$ c-w
For $24$ acre field, enough fodder for $90\cdot24\div\frac{10}3= \color{red}{36}\cdot18\;$ c-w
Thus $\color{red}{36}$ cows can be fed