Basic algebra word problem: Number of sheep?

Question

A farmer raises sheep and chickens at his farm. The number of sheep is $\frac{1}{3}$ the number of chickens. There are $96$ fewer sheep legs than chicken legs. How many sheep are there at the farm?

My work:

Number of sheep = $S$

Number of Chickens = $C$

We know that there is one third of sheep than chickens. so, $\frac{1}{3}S = C$

We know that there are 96 fewer sheep legs than chicken leg. so, $2C = 4S - 96$

I got stuck here.

Answer 1

You need to change your first equation. If the number of sheep is $\frac{1}{3}$ the number of chicken then $3S = C$.

Also in your second equation, the $-$ needs to be a $+$ because if there are fewer sheep legs than chickens then you'll need to add 96 on the sheep value.

Then combine the equations and you'll get the answer.

Answer 2

We know that there is one-third of sheep than chickens. so, $S=\frac{1}{3}C$

We know that there are 96 fewer sheep legs than the chicken leg. so,$ 2C-96=4S$

$$C=3S$$ $$2C-96=4S$$

Two equations two variables can you solve from here?

Answer 3

The number of sheep $S$ is $\frac{1}{3}$ the number of chickens $C$ so $S =\frac{1}{3}C $ and also $4S=2C - 96$ $$\begin{cases} S=\frac{1}{3}C \\ 4S=2C - 96 \end{cases} \Leftrightarrow \begin{cases} C=3S \\ 4S=6S - 96 \end{cases} \Leftrightarrow \begin{cases} C=3S \\ 2S= 96 \end{cases} \Leftrightarrow \begin{cases} C=3S \\ S= 48 \end{cases} \Leftrightarrow \begin{cases} C=144 \\ S= 48 \end{cases}$$