There are 33 animals in the barn. Some are chickens and some are pigs. There are 106 legs in all. How many of each animal are there? 25 chickens and 13 pigs 22 chickens and 15 pigs 20 chickens and 18 pigs 10 chickens and 28 pigs

Let's denote the number of chickens as \( c \) and the number of pigs as \( p \). We are given the following information: 1. The total number of animals is 33: \( c + p = 33 \) 2. The total number of legs is 106: \( 2c + 4p = 106 \) We can solve this system of equations to find the values of \( c \) and \( p \). Let's solve it using the substitution method. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}c+p=33\\2c+4p=106\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}c=33-p\\2c+4p=106\end{array}\right.\) - step2: Substitute the value of \(c:\) \(2\left(33-p\right)+4p=106\) - step3: Simplify: \(66+2p=106\) - step4: Move the constant to the right side: \(2p=106-66\) - step5: Subtract the numbers: \(2p=40\) - step6: Divide both sides: \(\frac{2p}{2}=\frac{40}{2}\) - step7: Divide the numbers: \(p=20\) - step8: Substitute the value of \(p:\) \(c=33-20\) - step9: Simplify: \(c=13\) - step10: Calculate: \(\left\{ \begin{array}{l}c=13\\p=20\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}c=13\\p=20\end{array}\right.\) - step12: Rewrite: \(\left(c,p\right) = \left(13,20\right)\) The solution to the system of equations is \( c = 13 \) and \( p = 20 \). Therefore, there are 13 chickens and 20 pigs in the barn. So, the correct answer is: 13 chickens and 20 pigs.