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Applications of Linear Equations
Applications of linear equations are used by people on a daily basis even without using a line graph because the situations faced by them might have an unknown quantity that can be represented as a linear equation such as calculating mileage rates, income over time, etc. The important aspect here is the four main arithmetic operations – addition, subtraction, multiplication, and division. Let us learn more about the applications of linear equations in this article and solve some reallife examples.
1.  What is the Application of Linear Equations? 
2.  RealLife Applications of Linear Equations 
3.  Application of Linear Equations in Word Problems 
4.  FAQs on Application of Linear Equations 
What is the Application of Linear Equations?
The main objective for the applications of linear equations or linear systems 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 equations are:
 Geometry problems by using two variables.
 Money problems by using two variables
 A mixture of problems by using two variables.
 DistanceRateTime problems by using two variables.
 Application of linear equation in business and economics
Definition of Linear Equations
An equation with degree 1 is considered a linear equation. In other words, a linear equation is an equation that is written for two different variables. This equation will be a linear combination of these two variables, and a constant can be present. The standard form is ax + by = c, where a and b are coefficients, x and y are variables, and c is the constant. A linear equation can also be plotted on a graph where it is a straight line. The image below shows a few examples of linear equations.
RealLife Applications of Linear Equations
Applications of linear equations are required almost frequently in our life even if that means doing calculations mentally at times. And this is where the first application of linear equation comes. Some of the reallife applications of linear equations could be calculating the cost of hiring a taxi on vacation, it could be a useful tool to compare the better rates of payment for work or budgeting or making any sort of predictions. Let us look at an example to understand this better. Let us find electricity consumed on days 1,2,3 and so on. Every day, the average amount of electricity consumed would be 23 units. So, electricity consumed at the end of the nth day is X = 23 × n + 0.
Let us look at another example. Suppose we rent a car with a charge of $200 plus $25 for every hour. Here you don’t know how many hours you will travel so by using “t” to represent the number of hours to your destination and “x” to represent the cost of that taxi ride, this can be framed in an equation as x = 25 × t + 200.
Application of Linear Equations in Word Problems
Application of linear equations is used in solving word or story problems which can be difficult in the beginning but once the process is understood, solving word problems will become simple. The process is as follows:
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 a 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 it into the equation but also ensure that it makes sense. Sometimes mistakes can be identified by acknowledging that the answer does not make sense.
There is some standard formula that is used in some of the word problems, they are:
 Distance=Rate × Time
 Work Done = Work Rate × Time Spent Working
 Amount of secondary liquid in the water = Percentage of Solution × Volume of Solution
Related Topics
Listed below are a few topics related to applications of linear equations, take a look.
Examples on Applications of Linear Equations

Example 1: Five years ago, the age of Niraj was seven times that of his son Girish. Five years from, his age will be three times that of his son. What is the ratio of their ages today?
Solution: Let the ages of Niraj and Girish be x and y respectively. Then, their ages five years ago would have been:
(x – 5), (y – 5)
And their ages five years from today will be:
(x + 5) , (y + 5)
From the information given in the problem, we have:
\[\begin{array}{l}\left\{ \begin{array}{l}x – 5 = 7\left( {y – 5} \right)\\x + 5 = 3\left( {y + 5} \right)\end{array} \right.\\ \Rightarrow \qquad x – 7y + 30 = 0\\ \qquad\quad x – 3y – 10 = 0\end{array}\]
Subtracting the two equations, we have:
\[\begin{array}{l} – 4y + 40 = 0\\ \Rightarrow y = 10\end{array}\]
Plugging this value of y back into the first equation, we have:
\[\begin{array}{l}x – 5 = 7\left( {y – 5} \right)\\ \Rightarrow \qquad x – 5 = 7\left( {10 – 5} \right) = 35\\ \Rightarrow \qquad x = 40\end{array}\]
Thus, their present ages are 40 years and 10 years.

Example 2: Solve the following pair of equations:
\[\begin{align}&\frac{2}{x} + \frac{3}{y} = 18\\&\frac{5}{x} – \frac{2}{y} = 7\end{align}\]
Solution: Clearly, this is not a pair of linear equations, but it can easily be reduced to one, using the following substitutions:
\[\frac{1}{x} \to p,\;\frac{1}{y} \to q\]
Thus, our pair of equations becomes:
\[\begin{array}{l}\left\{ \begin{array}{l}2p + 3q = 18\\5p – 2q = 7\end{array} \right.\\ \Rightarrow \;\;\;2p + 3q – 18 = 0\\\;\;\;\;\;\;5p – 2q – 7 = 0\end{array}\]
Let’s use crossmultiplication to solve this pair:
\[\begin{align}&\frac{p}{{\left( {3 \times – 7} \right) – \left( { – 2 \times – 18} \right)}} = \frac{{ – q}}{{\left( {2 \times – 7} \right) – \left( {5 \times – 18} \right)}}\\ &\qquad\qquad\qquad\qquad\qquad\;\;= \frac{1}{{\left( {2 \times – 2} \right) – \left( {5 \times 3} \right)}}\\ &\Rightarrow \;\;\;\frac{p}{{ – 21 – 36}} = \frac{{ – q}}{{ – 14 + 90}} = \frac{1}{{ – 4 – 15}}\\ &\Rightarrow \;\;\;\frac{p}{{ – 57}} = \frac{{ – q}}{{76}} = \frac{1}{{ – 19}}\\ &\Rightarrow \;\;\;p = 3,\;q = 4\end{align}\]
Note that these values correspond to p and q, whereas the original unknowns were x and y, so we still have one more step left:
\[\begin{align}{l}x = \frac{1}{p} = \frac{1}{3}\\y = \frac{1}{q} = \frac{1}{4}\end{align}\]
This completes our solution.

Example 3: Solve the following pair of equations:
\[\begin{align}{l}\frac{5}{{x – 1}} + \frac{1}{{y – 2}} = 2\\\frac{6}{{x – 1}} – \frac{3}{{y – 2}} = 1\end{align}\]
Solution: Once again, these equations are not linear, but we can easily reduce them to linear form:
\[\frac{1}{{x – 1}} \to p,\;\frac{1}{{y – 2}} \to q\]
Thus, our equations become:
\[\begin{array}{l}\left\{ \begin{array}{l}5p + q = 2\\6p – 3q = 1\end{array} \right.\\ \Rightarrow \;\;\;5p + q – 2 = 0\\\;\;\;\;\;\;6p – 3q – 1 = 0\end{array}\]
By crossmultiplication, the solution is:
\[\begin{align}&\frac{p}{{\left( {1 \times – 1} \right) – \left( { – 3 \times – 2} \right)}} = \frac{{ – q}}{{\left( {5 \times – 1} \right) – \left( {6 \times – 2} \right)}}\\ &\qquad\qquad\qquad\qquad\qquad= \frac{1}{{\left( {5 \times – 3} \right) – \left( {6 \times 1} \right)}}\\ &\Rightarrow \;\;\;\frac{p}{{ – 7}} = \frac{{ – q}}{7} = \frac{1}{{ – 21}}\\ &\Rightarrow \;\;\;p = \frac{1}{3},\;q = \frac{1}{3}\end{align}\]
Now, we evaluate the original unknowns:
\[\begin{align}&\frac{1}{{x – 1}} = p = \frac{1}{3}\\ &\Rightarrow \;\;\;x – 1 = 3\;\; \Rightarrow \;\;x = 4\\&\frac{1}{{y – 2}} = q = \frac{1}{3}\\ &\Rightarrow \;\;\;y – 2 = 3\;\; \Rightarrow \;\;y = 5\end{align}\]
Thus, the final solution is:
\[x = 4,\;y = 5\]
Practice Questions on Applications of Linear Equations
FAQs on Applications of Linear Equations
What are the Applications of Linear Equations in Real Life?
In reallife situations where there is an unknown quantity or identity, the use of linear equations comes into play, for example, figuring out income over time, calculating mileage rates, or predicting profit. Most of the time mental calculations are used in some reallife situations without drawing a line graph.
What is a Linear Equation? Explain with an Example.
An equation of the form ax + by = c is called a linear equation. Here, x and y are variables, and a, b and c are constants. Examples of the linear equation are:
 y = 4x – 3
 7y – 5x = 1
 y = 2
What are the 4 Methods of Solving Linear Equations?
The methods for solving linear equations are given below:
 Substitution method
 Elimination method
 Cross multiplication method
 Graphical method
What are the Applications of Linear Equations in Two Variables?
The application of linear equations in two variables is solving equations involving two different variables.
How Do you Graph a Linear Equation?
The basic methods of the graphing linear equation:
 The first method is by plotting all points on the graph and then drawing a line through the points.
 The second is by using the yintercept of the equation and the slope of the equation.
What is the General Form of a linear equation in One Variable?
The general form of the linear equations in one variable is Ax + B = 0 where x is the variable and A is the coefficient of x and b is the constant term.
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