Question

The number of rainbow smelt in Lake Michigan had an average rate of change of −19.76 per year between 1990 and 2000. The bloater fish population had an average rate of change of −92.57 per year during the same time. If the initial population of rainbow smelt was 227 and the initial population of bloater fish was 1,052, after how many years were the two populations equal? The linear function that models the population of rainbow smelt is y1 = −19.76x + 227, where x = the years since 1990 and y1 = the number of rainbow smelt. The linear function that models the population of bloater fish is y2 = . The linear equation that determines when the two populations were equal is . The solution is x =

Answer 1

Let x =  number of years between 1990 and  2000.

Rainbow smelt:
Initial population = 227
Average yearly rate of change = -19.76
The linear function that models the population is
y₁ = 227 - 19.76x

Bloater fish
Initial population = 1052
Average yearly rate of change = -92.57
The linear function that models the population is
y₂ = 1052 - 92.57x

When the two populations are equal, then y₁ = y₂.
That is
227 - 19.76x = 1052 - 92.57x
(92.57 - 19.76)x = 1052 - 227
72.81x = 825
x = 11.33 years
   = 11 years, 0.33*12 months
   = 11 years, 4 months
Add x to 1990 to obtain the year April 2011.

Answer:
x = 11.33 years.
This occurs in April 2001.

Answer 2

X= 11.33

That's hopefully gonna help you a bit