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This problem is about solving a system of equations based on the number of animals and the number of legs they have.

### Part 1:
- There are 18 animals (cows and chickens) in total.
- Cows have 4 legs, and chickens have 2 legs.
- There are 50 legs in total.

Let:
- \( x \) be the number of cows.
- \( y \) be the number of chickens.

From the problem, we know the following:
1. \( x + y = 18 \) (the total number of animals).
2. \( 4x + 2y = 50 \) (the total number of legs).

#### Solving the system:
- From the first equation, solve for \( y \): 
  \[
  y = 18 - x
  \]

- Substitute this into the second equation:
  \[
  4x + 2(18 - x) = 50
  \]

- Simplify:
  \[
  4x + 36 - 2x = 50
  \]
  \[
  2x + 36 = 50
  \]
  \[
  2x = 14
  \]
  \[
  x = 7
  \]

So, there are 7 cows.

- Substitute \( x = 7 \) into the first equation to find \( y \):
  \[
  7 + y = 18
  \]
  \[
  y = 11
  \]

So, there are 11 chickens.

### Part 2 (Extension):
- Now there are 55 animals, and there are 150 legs.

Let:
- \( x \) be the number of cows.
- \( y \) be the number of chickens.

From the problem, we know:
1. \( x + y = 55 \)
2. \( 4x + 2y = 150 \)

#### Solving the system:
- Solve for \( y \) from the first equation:
  \[
  y = 55 - x
  \]

- Substitute this into the second equation:
  \[
  4x + 2(55 - x) = 150
  \]

- Simplify:
  \[
  4x + 110 - 2x = 150
  \]
  \[
  2x + 110 = 150
  \]
  \[
  2x = 40
  \]
  \[
  x = 20
  \]

So, there are 20 cows.

- Substitute \( x = 20 \) into the first equation:
  \[
  20 + y = 55
  \]
  \[
  y = 35
  \]

So, there are 35 chickens.

### Final Answers:
- For the first problem, there are 7 cows and 11 chickens.
- For the extension, there are 20 cows and 35 chickens.

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. If there are 24 animals with 74 legs in total, how many cows and how many chickens are there?
2. What if there were 40 legs in total but still 18 animals, how would the numbers change?
3. If we know there are 10 cows, how many chickens are there among 50 animals?
4. Can you create a new problem with 30 animals and 100 legs? Solve it!
5. If the number of cows doubled, how would that affect the total number of legs?

**Tip:** For systems of equations with word problems, identify variables for unknowns and carefully write down relationships before solving.

There are 18 animals loose in the barnyard. Some are cows, and some are chickens. There are 50 legs in total. How many cows and chickens are in the barnyard?

Learn how to solve a system of equations to determine the number of cows and chickens based on the total number of animals and legs. Suitable for students in Grades 6-8.

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