SOLUTION: THE POPULATION OF TWO TOWNS ARE CHANGING AT STEADY RATES. ONE TOWN HAS A POPULATION OF 25,500. ITS POPULATION IS INCREASING BY 200 PEOPLE EACH YEAR. THE OTHER TOWN HAS A POPULATION

Algebra ->  Rectangles -> SOLUTION: THE POPULATION OF TWO TOWNS ARE CHANGING AT STEADY RATES. ONE TOWN HAS A POPULATION OF 25,500. ITS POPULATION IS INCREASING BY 200 PEOPLE EACH YEAR. THE OTHER TOWN HAS A POPULATION      Log On


   

Question 510709: THE POPULATION OF TWO TOWNS ARE CHANGING AT STEADY RATES. ONE TOWN HAS A POPULATION OF 25,500. ITS POPULATION IS INCREASING BY 200 PEOPLE EACH YEAR. THE OTHER TOWN HAS A POPULATION OF 47,900. ITS POPULATION IS DECREASING BY 800 PEOPLE EACH YEAR. IF THE RATE FOR EACH TOWN REMAINS THE SAME,IN HOW MANY YEARS WILL THE POPULATION BE THE SAME ?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
THE POPULATION OF TWO TOWNS ARE CHANGING AT STEADY RATES.
ONE TOWN HAS A POPULATION OF 25,500. ITS POPULATION IS INCREASING BY 200 PEOPLE EACH YEAR.
THE OTHER TOWN HAS A POPULATION OF 47,900. ITS POPULATION IS DECREASING BY 800 PEOPLE EACH YEAR.
IF THE RATE FOR EACH TOWN REMAINS THE SAME,IN HOW MANY YEARS WILL THE POPULATION BE THE SAME ?
:
let t = no. of years for this to happen
An equation for the 1st town
f(t) = 200t + 25500
An equation for the 2nd town
f(t) = -800t + 47900
:
Find when the populations are equal
1st town = 2nd town
200t + 25500 = -800t + 47900
200t + 800t = 47900 - 25500
1000t = 22400
t = 22400%2F1000
t = 22.4 yrs the will be the same
:
:
Check this by finding the population of each after 22.4 yrs
200(22.4) + 25500 = 29980
-800(22.4)+ 47900 = 29980