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. Assumin; 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?

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Define Variables: Let's define the variables: Let P_CF be the population of Canyon Falls. Let P_SC be the population of Swift Creek. Let t be the number of years from today.  We are given: P_CF = 22,500 P_SC = 15,200 The rate of decrease for Canyon Falls is 740 people per year. The rate of increase for Swift Creek is 1,500 people per year.  We want to find t such that P_SC = 2 * P_CF.  We can write two equations that describe the populations of Canyon Falls and Swift Creek after t years: P_CF(t) = 22,500 - 740t (since the population is decreasing) P_SC(t) = 15,200 + 1,500t (since the population is increasing)  We want to find t when P_SC(t) = 2 * P_CF(t).Set Up Equation: Let's set up the equation based on the condition that the population of Swift Creek equals twice the population of Canyon Falls: 15,200 + 1,500t = 2 * (22,500 - 740t)  Now we will solve for t.Solve Equation: First, distribute the 2 on the right side of the equation: 15,200 + 1,500t = 45,000 - 1,480t  Now, let's combine like terms by moving all terms involving t to one side and constants to the other side: 1,500t + 1,480t = 45,000 - 15,200  This simplifies to: 2,980t = 29,800  Now, divide both sides by 2,980 to solve for t: t = 29,800 / 2,980Find Value of t: Perform the division to find the value of t: t = 10  So, it will take 10 years for the population of Swift Creek to be twice the population of Canyon Falls, assuming the rates of increase and decrease remain constant.

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