1. **State the problem:** We have a farm with chickens and cows. Chickens have 2 legs each, cows have 4 legs each. There are 50 animals total and 140 legs total. We want to find how many chickens and cows there are. 2. **Define variables:** Let $x$ = number of chickens, $y$ = number of cows. 3. **Write the system of equations:** $$\begin{cases} x + y = 50 \\ 2x + 4y = 140 \end{cases}$$ 4. **Express the system in matrix form:** $$\begin{bmatrix} 1 & 1 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 50 \\ 140 \end{bmatrix}$$ 5. **Solve using matrix methods:** Let $A = \begin{bmatrix} 1 & 1 \\ 2 & 4 \end{bmatrix}$, $\mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix}$, and $\mathbf{b} = \begin{bmatrix} 50 \\ 140 \end{bmatrix}$. 6. **Find the inverse of $A$:** $$\det(A) = (1)(4) - (1)(2) = 4 - 2 = 2$$ $$A^{-1} = \frac{1}{2} \begin{bmatrix} 4 & -1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -0.5 \\ -1 & 0.5 \end{bmatrix}$$ 7. **Calculate $\mathbf{x} = A^{-1} \mathbf{b}$:** $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & -0.5 \\ -1 & 0.5 \end{bmatrix} \begin{bmatrix} 50 \\ 140 \end{bmatrix} = \begin{bmatrix} 2 \times 50 - 0.5 \times 140 \\ -1 \times 50 + 0.5 \times 140 \end{bmatrix} = \begin{bmatrix} 100 - 70 \\ -50 + 70 \end{bmatrix} = \begin{bmatrix} 30 \\ 20 \end{bmatrix}$$ 8. **Interpret the solution:** There are 30 chickens and 20 cows on the farm. **Final answer:** $x=30$ chickens, $y=20$ cows.