Step 1: Rewrite the Year 3 Population Using the Distributive Property
The population of bears in Year 3 is given as \( 2(x + 10) \).
Using the Distributive Property: \[ 2(x + 10) = 2x + 20. \]
So, the bear population in Year 3 without parentheses is \( 2x + 20 \).
Step 2: Determine the Number of Bears in Year 1 Based on Year 2 Population
We know that the total number of bears in Year 2 is 42: \[ x + 10 = 42. \]
To solve for \( x \): \[ x = 42 - 10, \] \[ x = 32. \]
So, the number of bears in Year 1 is 32.
Step 3: Determine the Number of Bears in Year 1 Based on Year 3 Population
We know that the total number of bears in Year 3 is 50: \[ 2(x + 10) = 50. \]
First, we can simplify the equation: \[ 2x + 20 = 50. \]
Now, solving for \( x \): \[ 2x = 50 - 20, \] \[ 2x = 30, \] \[ x = 15. \]
So, the number of bears in Year 1 is 15.
Step 4: Solve the Inequality for Year 4
We know that \( y + 5 > 38 \).
To solve for \( y \): \[ y > 38 - 5, \] \[ y > 33. \]
Possible values for \( y \) can be any numbers greater than 33. Therefore, three possible values for \( y \) could be:
- 34
- 35
- 36
Thus, values for \( y \) could be \( 34, 35, \) or \( 36 \).