Cocks and Cows Confuse too ?

Let number of hens= $x$ Let number of cows= $y$ Each animal obviously had $1$ head so the first equation in mathematical terms is : $x+y=48$ A hen has $2$ feet (the number of hen,s feet will be twice the number of hens) and a cow has $4$ feet (the number of cow,s feet will be $4$ times the number of cows) so the second equation in mathematical terms is: $2x+4y=140$ Taking out the common factor $2$ on both sides we get : $x+2y=70$ So the system of equations we have to solve is : $x+y=48\rightarrow \;(1)$ $x+2y=70\rightarrow \;(2)$ With a little rearranging in $(2)$ we get: $x+2y=(x+y)+y=48+y=70$ Solving this, we get: $y=70-48=\boxed{22}$ So number of cow= $22$ .Plugging this value in the second equation,we get: $x+22=48\longrightarrow \;x=48-22=\boxed{26}$

\[\begin{cases} h + c = 48 \\ 2h + 4c = 140 \end{cases}

\implies

\begin{cases} \boxed{h = 26} \\ c = 22 \end{cases}

\]

A hen has 1 head and 2 feet while a cow has 1 head and 4 feet. Let $h$ be the number of hens and $c$ be the number of cows. Then,

$h+c=48$ $\color{#D61F06}(1)$

$2h+4c=140$ $\color{#D61F06}(2)$

Solving the system of equations gives $h=26$ and $c=22$ .

Hello,

let h = hens, c= cows,

For those heads,

h + c = 48(1st equation),

For those feets,

2h + 4c = 140

h + 2c = 70(2nd equation),

solve them by substitution, or elimination, then you will get hens = 26

thanks.....