Solving a Simultaneous Equation to Determine the Cost of Onions

Imagine a scenario where you have to solve a real-life problem using algebra. Given two equations, we can determine the exact cost of an onion based on the cost of potatoes and onions together. This method is not only useful in mathematics but also in practical situations like grocery shopping or budgeting.

Introduction to the Problem

Suppose you are faced with the following scenario: If 5 potatoes and 6 onions cost 1.22 and 6 potatoes and 5 onions cost 1.31, what is the cost of one onion?

Setting the Equations

To solve this problem, let's assume the cost of one potato is x and the cost of one onion is y. We can then write the following equations:

5x 6y 1.22 —— eq. 1

6x 5y 1.31 —— eq. 2

Solving the Equations

First, we need to add both equations to eliminate the variable y:

5x 6y 6x 5y 1.22 1.31

11x 11y 2.53

x y 0.23 —— eq. 3

Next, subtract equation 1 from equation 2 to eliminate x:

(6x 5y) - (5x 6y) 1.31 - 1.22

x - y 0.09 —— eq. 4

Now, subtract equation 3 from equation 4:

(x - y) - (x y) 0.09 - 0.23

-2y -0.14

y 0.07

Conclusion

Therefore, the cost of one onion is 0.07, which is 7 paisa. This method of solving simultaneous equations is a fundamental algebraic technique and can be applied in various real-life scenarios to make informed decisions and solve practical problems.

Keywords: simultaneous equations, cost calculation, algebraic problem