Today, the population of Canyon Falls is 22,500 and the population of Swift Creek is 15,200 . The population of Canyon Falls is decreasing at the rate of 740 people each year while the population of Swift Creek is increasing at the rate of 1,500 people each year. Assuming these rates continue into the future, in how many years from today will the population of Swift Creek equal twice the population of Canyon Falls?
Q. Today, the population of Canyon Falls is 22,500 and the population of Swift Creek is 15,200 . The population of Canyon Falls is decreasing at the rate of 740 people each year while the population of Swift Creek is increasing at the rate of 1,500 people each year. Assuming these rates continue into the future, in how many years from today will the population of Swift Creek equal twice the population of Canyon Falls?
Set up equation: Let's denote the number of years from today as y. We need to set up an equation that represents the situation where the population of Swift Creek equals twice the population of Canyon Falls after y years.Canyon Falls population after y years: 22500−740ySwift Creek population after y years: 15200+1500yThe equation we need to solve is: 15200+1500y=2(22500−740y)
Distribute and combine terms: Now, let's distribute the 2 on the right side of the equation:15200+1500y=45000−1480y
Solve for y: Next, we'll combine like terms by adding 1480y to both sides and subtracting 15200 from both sides:1500y+1480y=45000−152002980y=29800
Final population after 10 years: Now, we'll solve for y by dividing both sides of the equation by 2980:y=298029800y=10
Final population after 10 years: Now, we'll solve for y by dividing both sides of the equation by 2980:y=298029800y=10We have found that after 10 years, the population of Swift Creek will be twice the population of Canyon Falls.
More problems from Exponential growth and decay: word problems