Application of linear system

Applications of Linear System

The main objective for the applications of Linear System or Linear equation is to solve various problems using two variables where one is known and the other is unknown, also dependent on the first. Some of these applications of linear system are 

• Geometry problems by using two variables.
• Money problems by using two variables
• A mixture of problems by using two variables.
• Distance-Rate-Time problems by using two variables.
• Application of linear equation in business and economics

We know most of the student find it difficult to solve story/word problems where applications of Linear equation is required. 

The actual reason is that most students do not understand how to work with them. But trust us once you understand the process for working with Applications of Linear Equation, it will be fairly easy.

Read the Problem Again

You must read the problem carefully to understand what is given, what kind of application of linear system it is, and what is to be found. If required read it as many times as it takes. 
It is imperative for you to completely understand and identify all the given information and what you are being asked to find. Be mindful as at times one-word can completely change how the application of linear system is worked.

Assign Variables

Represent one of the unknown values/quantity with a variable and try to relate all the other unknown values (if any) to this variable, using diagrams or tables, etc. Pen down what each variable represents.

Write Equations

Form equations that relate unknown quantities to known quantities. We can use some of the known formulas and often the figure/equations sketched in the previous step can be used to determine the relevant equation leading to answer what we need to find.

Solve the System of Equations

Solve the equations formed in the previous step and answer all the questions asked as we may be asked several values but the equation will only give one of them.

Check Answers 

You can check the answer by plugging into the equation but also ensure that it makes sense. Sometimes mistakes can be identified by acknowledging that the answer does not make sense. 

Let’s understand this with a few examples of applications of linear system:

Application of linear equation in business and economics

Example 1

 

 

A dress is on 35 % sale and is for Rs. 15.00 How much was the dress being sold for before the sale?

Let’s start with defining x to be the price of the dress before the sale. It has been marked down by 35%. This means that 0.35x has been subtracted off from the original price. Therefore, the equation is,

x−0.35x=15

0.65x=15

p=15/ 0.65

  =23.0769

So, with rounding it looks like the shirt was originally sold for Rs. 23.08.

There is some standard formula that is used in some of the word problems. In our next set of examples, we will be using some of the standard formulae like 

• Distance=Rate × Time
• Work Done = Work Rate × Time Spent Working
• Amount of secondary liquid in the water = Percentage of Solution × Volume of Solution 

Application of system of linear equation

Example 2

 

 

There are two cars moving towards each other and are 600 miles apart. One car is moving at a speed of 100 mph and the other is moving at 80 mph. Assuming that the cars start moving at the same time how long does it take for the two cars to meet?
 
Here let t represent the amount of time that the cars are traveling before they meet. Now, we need to sketch a figure for this one. This figure will help us to write down the equation that we’ll need to solve.

From this figure, we can see that the Distance Car A travels plus the Distance Car B travels must equal the total distance separating the two cars, 600 miles.

Here is the linear equation in two separate forms.

The distance of Car A + Distance of Car B =600

Rate of Car A * Time of Car A + Rate of Car B * Time of Car B = 600

We used the standard formula here twice, once for each car. We know that the distance a car travels is the rate of the car times the time traveled by car. In this case, we know that Car A travels at 100 mph for t hours and that

Car B travels at 80 mph for t hours as well. Plugging these into the linear equation and solving it gives us,

100t+80t=600

180t=600

t=600/180

=3.33 hrs

So, the two cars travel 3.33 hours to meet.

Application of linear system

Example 3

 

 

Shyam can clean a 4 BHK residential flat in 6 hours. If Ram and Shyam work together then they can clean it in 4 hours. How long would it take Ram to clean the 4 BHK residential flat by himself?

Let t be the amount of time it would take Ram to clean the 4 BHK residential flat by himself. The linear equation for this problem is,

(Portion of the job done by Shyam) +(Portion of the job done by Ram) =1 Job ……. (1)

(Work Rate of Shyam) *(Time Spent Working) +(Work Rate of Ram) * (Time Spent Working) = 1 …………. (2)

Here we know that the time is taken working together = 3.5 hrs.

So, we need to find out the rate of work for each person.

Rate of work for Shyam = no. of Job done/ Hours = 1/6 …………. (a)

As we want to know how much time it will take Ram to complete the same job of cleaning the flat by himself. Let take t as the time is taken for him to complete the task himself.

Rate of work for Ram = no. of Job done/ Hours = 1/t ……………. (b)

Now plugging in (a) and (b) in the 2^nd  linear equation we get 

1/6(4) +1/t (4) =1……Multiplying both sides by 6t

4t+4*6 = 6t

2t = 24

t = 12

 

so, it seems Ram would take 12 hrs to clean the 4 BHK Residential flat by himself

---

---