Let's solve the given problem step by step.
### Step 1: Define the Linear Functions
1. Population of Rainbow Smelt (y₁):
The population of rainbow smelt is modeled by the linear function:
[tex]\[
y_1 = -19.76x + 227
\][/tex]
where [tex]\( x \)[/tex] represents the number of years since 1990.
2. Population of Bloater Fish (y₂):
The population of bloater fish is modeled by the linear function:
[tex]\[
y_2 = -92.57x + 1052
\][/tex]
where [tex]\( x \)[/tex] represents the number of years since 1990.
### Step 2: Set the Two Functions Equal to Each Other
Next, we need to determine when the populations of the two species are equal, i.e., when [tex]\( y_1 = y_2 \)[/tex].
[tex]\[
-19.76x + 227 = -92.57x + 1052
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
1. Move all terms involving [tex]\( x \)[/tex] to one side and constants to the other:
[tex]\[
-19.76x + 92.57x = 1052 - 227
\][/tex]
2. Combine like terms:
[tex]\[
72.81x = 825
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{825}{72.81}
\][/tex]
By performing the division, we find:
[tex]\[
x = 11.330861145447054
\][/tex]
Therefore, the linear function that models the population of bloater fish is:
[tex]\[
y_2 = -92.57x + 1052
\][/tex]
The linear equation that determines when the two populations were equal is:
[tex]\[
-19.76x + 227 = -92.57x + 1052
\][/tex]
The solution is:
[tex]\[
x = 11.33 \text{ years}
\][/tex]
Thus, approximately 11.33 years after 1990, the populations of rainbow smelt and bloater fish were equal.
