# Renting apartments

## Situation (costs)

Annie is renting an apartment. The cost is a deposit fee of \$530 plus \$320 per month. John is renting an apartment of the same size, however, his cost is a deposit fee of \$200 plus \$480 per month.

## Variable identification

Let C represent the cost
Let M represent the number of months

## Equations

• Annie: C = 320M+530 (-320M+C=530)
• John: C = 480M+200 (-480M+C=200)

## Algebraically solving for the point of intersection

480M+200 = 320M+530
160M+200 = 530
160M = 330
M = 2.0625 (33/16)

## The significance of this solution

This solution is significant because it displays the rate of change for each scenario, and also which is cheaper at what specific number of months. Deciphering the rate of change is important in a situation like this one, because it allows the buyer to know how much they will be paying over the course of multiple months. Knowing which scenario costs less at what specific number of months, will help the buyer save money in the sense that it would allow them to use the cheaper method based on the number of months they plan on renting the apartment.

## Summary

Annie and John will both be paying the same amount of money (\$1190) at exactly 2.0625 months. This means that the better option in terms of finance would be John's apartment for less than 2.0625 months, and Annie's apartment for more than 2.0625 months.