Question

unit 4 linear equations and linear systems test review
a farm has chickens and cows. all the cows have 4 legs and all the chickens have 2 legs. all together, there are 82 cow and chicken legs on the farm. complete the table to show some possible combinations of chickens and cows to get 82 total legs. use this information to complete the chart and part a.

number of chickens (x)	number of cows (y)
7
	10
19
	5

a. if the farm has 30 chickens and cows, and there are 82 chicken and cow legs all together, then how many chickens and how many cows could the farm have?

b. if the farm has 24 chickens and cows, and there are 62 chicken and cow legs all together, the how many chickens and how many cows could the farm have?

c. if the farm has 32 chickens and cows, and there are 96 chicken and cow legs all together, then how many chickens and how many cows could the farm have?

Explanation:

Step1: Define Variables

Let \( x \) be the number of chickens and \( y \) be the number of cows. Chickens have 2 legs, cows have 4 legs, so the leg equation is \( 2x + 4y = 82 \), and the total number of animals equation (for part a, b, c) is \( x + y = n \) (where \( n = 30, 24, 32 \) for parts a, b, c respectively). Simplify the leg equation: \( x + 2y = 41 \).

Part a: \( x + y = 30 \)

Step2: Solve System of Equations

We have \(

$$\begin{cases} x + y = 30 \\ x + 2y = 41 \end{cases}$$

\). Subtract the first equation from the second: \( (x + 2y) - (x + y) = 41 - 30 \) → \( y = 11 \). Then substitute \( y = 11 \) into \( x + y = 30 \): \( x = 30 - 11 = 19 \).

Part b: \( x + y = 24 \)

Step3: Solve for b

System: \(

$$\begin{cases} x + y = 24 \\ x + 2y = 41 \end{cases}$$

\). Subtract: \( y = 41 - 24 = 17 \). Then \( x = 24 - 17 = 7 \).

Part c: \( x + y = 32 \)

Step4: Solve for c

System: \(

$$\begin{cases} x + y = 32 \\ x + 2y = 41 \end{cases}$$

\). Subtract: \( y = 41 - 32 = 9 \). Then \( x = 32 - 9 = 23 \).

Completing the Table (Original Table)

For \( x = 35 \): Substitute into \( x + 2y = 41 \) → \( 35 + 2y = 41 \) → \( 2y = 6 \) → \( y = 3 \).
For \( x = 7 \): \( 7 + 2y = 41 \) → \( 2y = 34 \) → \( y = 17 \).
For \( y = 10 \): \( x + 2(10) = 41 \) → \( x = 41 - 20 = 21 \).
For \( x = 19 \): \( 19 + 2y = 41 \) → \( 2y = 22 \) → \( y = 11 \).
For \( y = 5 \): \( x + 2(5) = 41 \) → \( x = 41 - 10 = 31 \).

Answer:

(Table Completion and Parts a-c):

Table:

number of chickens (\( x \))	number of cows (\( y \))
7	17
21	10
19	11
31	5

Part a:

Number of chickens: 19, Number of cows: 11

Part b:

Number of chickens: 7, Number of cows: 17

Part c:

Number of chickens: 23, Number of cows: 9