# Algebraic Model: Definition & Examples

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• 0:02 What Are Algebraic Models?
• 2:25 Using Models to Solve Problems
• 4:57 Another Example
• 6:21 Lesson Summary

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Lesson Transcript
Instructor: Laura Pennington

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

Algebraic models are used frequently in mathematics. This lesson will offer a definition of algebraic models and use multiple examples to familiarize you with the concept. We'll learn what algebraic models are, how to create them, and how to use them.

## What Are Algebraic Models?

Have you ever gone to the bank to deposit some money into an interest bearing account, and the banker was able to tell you how much interest you will have after a certain number of years? If so, you may have wondered how they were able to calculate this amount. One way of finding this is by using what's called an algebraic model. An algebraic model takes a real-world situation described in words and describes that situation using algebra.

Consider our bank example. There are algebraic models to describe different types of investments. For example, when a bank is offering r% simple interest on an investment of P dollars, then the amount of interest earned, I, at the end of t years can be found by multiplying the amount of money you're investing times the interest rate in decimal form times the number of years the money will be in the account. We can describe this using algebra as I = Prt, where P is your original investment, r is your interest rate in decimal form, and t is the number of years the money is in the account. The expression I = Prt is an algebraic model.

Thus, we see that if you were to go to the bank in the initial scenario we described with \$1,000 to invest, and the bank is offering 5% simple interest, then the banker can tell you how much interest you will have after a certain number of years using the algebraic model:

I = Prt

Suppose you ask how much money you will have after two years. Then I is our unknown quantity that we are trying to find, P = 1,000, r = 0.05 (5% in decimal form), and t = 2. We plug this into our algebraic model to get:

I = 1,000 * 0.05 * 2 = 100

So we see that after two years, we would have \$100 in interest.

## Using Models to Solve Problems

The banking scenario makes use of a well-known interest formula to create our algebraic model representing the situation. It is useful when there is a formula we can use, but this is not always the case. We can create an algebraic model just from the information given - without a formula.

For example, assume Bob goes to the store to buy some oranges, and the oranges cost \$0.50 each. Bob has \$7.00 to spend, so he wants to know how many oranges he can buy. In this case there is not a well-known formula for my situation. However, Bob can still create an algebraic model of the situation. He wants to know how many oranges he can buy, so let's call that number x, and let's call his total cost C. Bob knows that his total cost will be the number of oranges he buy times the cost for each orange, so his total cost C will be 0.50x. Putting this in equation form, we have:

C = 0.50x

This is our algebraic model representing the situation.

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