Let’s talk about this one thing I’m pretty sure no one likes doing – going to a vegetable shop on a Saturday morning. Apart from being a boring chore, it also means having to deal with your parents’ disapproval if you get vegetables that aren’t fresh enough or are too pricey.

Along with quadratic equation and graphs, a linear equation can also help you in your day to day life. Choosing vegetables that are fresh enough is something you tend to learn with practice, but figuring out if you’re buying a pumpkin at a reasonable price, is something you can work out with a simple linear equation. You obviously can compare the prices per kilo, at the different shops near your house, and if that’s the case, you’re lucky. However, suppose the one shop charges you 15 rupees per pumpkin, and another has an offer wherein you buy 3 kilos of cut pumpkins for 50 rupees, then all you need is a little math to sort it out.

A simple linear equation wherein you find the price of the pumpkin per kilo, at both shops, would help solve this conundrum. Let the price of the pumpkin per kilo at shop 1 be x, and the weight of the pumpkin be 1.5 kilograms. X would then be equal to 15*1.5, which lands you with the value of Rs 22.5 per kilogram, at shop 1. This was a simple linear equation with one variable.

At shop 2, assuming the price of a pumpkin per kilo to be another variable, y, we come up with the equation of y=50/3, which would mean Rs 16.67 per kilogram. Moreover, since this is cut, it saves the weight occupied by the seeds in the pumpkin at shop 1, meaning a much more reasonable price.

Pretty sure you would impress any parent with this one

Linear equations with one variable are easier and seem exceedingly more reasonable than those with 2 variables. There are many situations similar to the vegetable shopping chore that would need one variable equation-solving. But when are two variable equations used? They seem ridiculous, with so many unknown parameters you feel it’s like being blindsided.

Let me attempt to change your mind on that.

Let’s assume you’re going to college soon, and are looking for the right mobile phone plan to choose. There are the obvious options of Vodafone and Airtel, the best in the field.

Say Vodafone charges you Rs 300 a month, with 0.5p per minute, for STD calls, while Airtel charges Rs 400 a month, with 0.2p per minute.

The number of minutes of the calls you will make per month is not constant, so let us assign that the variable x.

The cost of each plan per month depends on x, and so we assign it the variable y.

Thus, we end up with the equations –

Y=0.5x+300, for Vodafone (equation1), and Y=0.2x+400, for Airtel (equation 2). Equating these to each other, to solve them, we get

0.5x+300=0.2x+400

0.5x-0.2x=400-300

0.3x=200

X=200/0.3 = 666.67 minutes.

Y would then become 633 rupees for Vodafone, and 533 rupees for Airtel.

Airtel thus becomes the better choice.

And there you go, you just solved two real-time problems with linear equations.

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