Algebra in Every Day Life

Algebra is used very frequently in our daily lives. Its use is so common that most of the times we use Algebra without even realizing it.

For Example:

1. When you go for a gas refill in your car with $15 in your pocket, and let us say the gas price is $3 per kilogram, how do you calculate the number of kilograms of gas that you can get filled into your car? You simply do the following calculation:

`($15) /( $3//kg ) = 5 kgs`

The figure below illustrates this problem graphically. Here each block represents $1. Each bunch of 3 dollars is used to purchase 1kg of gas.

Algebra in Daily Life example gas refill

2. Let us say that you are in a shopping mall, and you have purchased 10 grocery items. You need to get shopping bags that can accommodate all the 10 items, where each bag has a room for 3 items. How many bags do you need? You will quickly perform the following calculation in your mind to know the number of bags needed:

`(text(10 items))/(text (3 items/bag)) = 3.33 bags ≈ 4 bags `

The figure below illustrates this problem: The different shapes inside the bags illustrate the purchased items and the numbers inside these shapes depict the item numbers.

Algebra in Daily usage - Shopping Bag

In both these examples, you are using a simple Algebraic formula:

`a/( b )`

In the first example:

a = Money in your pocket = $15

b = Price of 1kg of gas in Dollars = $3/kg

In the second example:

a = Number of items purchased = 10 items

b = Capacity of a single shopping bag = 3 items/bag

In simple words, we can say that a and b simply denote the values involved in the calculation. That is, a and b simply generalize our calculation.

Algebra plays this very role in our everyday life. It supplies us with simple generic formulas with the help of which we can solve our daily life problems with extraordinary ease. It uses symbols such as a, b, c, d, x, y, z etc (instead of actual values) to represent our daily life problems and to find their solutions.

3.Suppose you have to buy one dozen eggs worth $10, three breads where price of one bread is $5 and five bottles of juice each worth $8. How much money do you need to take with you to the grocery store?

This is a simple calculation but simple arithmetic may not work very well here. Using algebra, you can solve this problem easily:


  • a = price of one dozen eggs = $10
  • b = price of one bread = $5
  • c = price of one juice bottle = $8

=> Money needed = a + 3b + 5c

=> Money needed = $10 + 3($5) + 5($8) = $10 + $15 + $40 = $65

Algebra in Daily Life Another Exmaple with shopping

The figure shown above illustrates the way you compute the amount needed. You perform “Algebra” quickly in your mind without knowing that you do!

Presented above are a few simple examples of the use of Algebra in our daily life. Algebra is not just limited to schools and various courses as part of the syllabus or curriculum. It is used by almost every person on the face of the earth on a daily basis.

There are thousand such examples where Algebra helps us perform calculations of rather complex nature. Our mind is so much habitual of using it that most of the times we apply it unconsciously.

Our everyday life is full of such examples where we apply simple algebraic techniques for simple arithmetic calculations without "consciously" knowing that we are using algebra. Algebraic techniques are also useful in more advanced calculations, and one again these techniques are in common practice by almost all of us. We present an example here.

Advanced Example: An Optimization Problem

Very often we come across a situation where we have a couple of different choices. But we want to choose the best option that brings us maximum benefit while staying under some constraints. There is a branch of Algebra called “Linear Programming” that deals with finding the best possible solution to an optimization problem while staying within prescribed limitations / constraints. The real world problem can be represented mathematically through algebraic expressions called linear inequalities. A solution to these linear inequalities gives the optimum solution to the problem.