•In many problems, generating an algebraic equation is the most efficient way of solving the problem. In this case, assign a variable to a base quantity and write one or more equations that represent the relationships given in the question.
–Example: Three cars were sold at a dealership on Saturday. Car B sold for twice the price of Car A. Car C sold for $5000 less than Car A. If the sum of all three cars was $40,000, what was the price of Car A?
In this problem, it is clear that Car A is the base quantity that the other car’s prices are being compared to, so let:
a = price of car A
b = price of car B
c = price of car C
Then, translate the question from English to algebra:
b = 2a (Equation 1)
c = a – 5000 (Equation 2)
a + b + c = 40,000 (Equation 3)
Substitute Equations 1 & 2 into Equation 3 for the variables b & c.
a + (2a) + (a – 5000) = 40,000
4a – 5000 = 40,000
4a = 35,000
a = 8,750
Hence, the price of Car A was $8,750.