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 =
years.

y2 = - 92.57x + 1052 ;

11.331 years

Step-by-step explanation:

Given that :

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 boater fish is :

Rate of change, slope = - 92.57 ; initial population, intercept = 1052

Representing the equation in the form ;

y = bx + c

b = slope or rate of change ; c = intercept

y2 = - 92.57x + 1052

To express equality :

y1 = y2

- 19.76x + 227 = - 92.57x + 1052

Collect like terms :

-19.76x + 92.57x = 1052 - 227

72.81x = 825

x = 825 / 72.81

x = 11.331 years

y2 =  –92.57x + 1,052

–19.76x + 227 = –92.57x + 1052

11.33

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.