Algebra as a decision-making tool

Suppose you are in the market for a new car, and the particular one you like is offered in

both gas-only and hybrid models. You are wondering for the hybrid at what specific

number on the odometer does the savings in gasoline purchased offset the additional

amount in the purchase price? I enjoy teaching algebra because this question can be

answered with the knowledge of writing and solving systems of algebraic equations. 

Algebra allows one to simplistically model the cost of driving based on the purchase

price (P), the rate of the cost of driving per mile (m), and the variable of miles driven (x).

(Cost = P + mx) Some online research gives you the values for the purchase price,

gas price, and fuel economy (mpg). Graph your two equations, and where they

intersect is the exact mileage point at which operating the two types of car is equal. 

Any miles driven beyond this point and the higher-priced hybrid starts saving you money.

Students in my Algebra 1 class conducted this research and solved the systemsof equations written for operating the two models in order to answer the original

question for themselves. Some of them were surprised by the answer.