3110014: System of Linear Equations: In daily life for domestic budget management.

            Let us consider a problem of monthly home budget. The quantities required and budget of the family are given in the following table.                        

Requirements         (in Kg)		January	February	March
	Tea	1	1.25	0.5
	Rice	2	1	2
	Wheat	6	4	5
	Total Monthly Expenditure limit in (Rs.)	750	775	475

       The Problem here is to find the costs of the items which the family can afford? 

         To address this case first of all we have to identify the unknown quantities which are to be found.

Here we have the unknowns (costs of Tea, Rice and Wheat ).

Next we define the unknowns as x,y,z as the cost/unit of Tea, Rice and Wheat respectively.

For the month January the requirement is 1 kg of Tea, 2 kg of Rice and 6 kg of Wheat. 

Hence the total expenditure of this month is given by the equation

                                                            

                                 x+2y+6z=750

Similarly for the next months the expenditure equation will be

                                

                          1.25 x+y+4z=775

                                                                                                                                                         0.5 x+2y+5z=475

We can Solve this using the Wolfram-alpha computational engine. 

Click for the Solution

The wolfram alpha gives the solution as 

x=500,  




y=50,





 z=25

How we interpret the solution?

The costs of the Tea is Rs. 500/kg, cost of the Rice is 

.

Rs.50/kg and the cost of the wheat is Rs.25/kg.


Any Rise in the costs of any of these three items will affect the family's monthly budget.

More example on the following links :

http://www.shelovesmath.com/algebra/intermediate-algebra/systems-of-linear-equations/#InvestmentWordProblem


http://study.com/academy/lesson/writing-and-evaluating-real-life-linear-models.html


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