Linear Equations: Definition, Examples, Graphs, Solutions

Linear Equations:  Linear equations are the equations, which are in the first order. In general, linear equations describe the equations of straight lines. The graph of the linear equation is always a straight line.

We have linear equations in one variable and linear equations in two variables, and so on. The linear equations in one variable are generally denoted by \(a x+b=0\), and the general form of linear equation in two variables is \(a x+b y+c=0\). In this article, we will discuss about linear equations, their types, graphs and solved examples.

Linear Equations: Definition

If any linear equation is plotted on graph paper, it will look like a straight line. Hence, the name is given for those equations as linear equations. Linear equations are the equations, which are in the first order. In general, linear equations describe the equations of straight lines. The highest degree of the linear equations is one.

A linear equation should contain the variables and constants (zero-degree terms). Any linear equation in \(x\) and \(y\) can follow a certain set of rules. The highest degree of both the variables \(x\) and \(y\) should be \(1\). Apart from that, constants (zero-degree variables) can also be present.

Learn About Linear Equations in Two Variables

Linear equations in one variable are generally represented in the form of \(a x+b=0\).

Here, \(x\) – variable and \(a, b\) – constants of the given equation

Linear equations in two variables are generally represented in the form of \(a x+b y+c=0\).

Here, \(x, y\) – variables and \(a, b\) – constants of the given equation

Linear Equation in One Variable

A linear equation is an equation with the first order. Any equation is said to be a linear equation in one variable if it has one variable (\(x\) or \(y\)) of the highest degree one, and also, they contain the constant (zero-degree variable) term.
The general form of the linear equation in one variable is given by \(A x+B=0\).
Here,

\(x\) – Variable with degree one

\(A\) – Coefficient of the variable \(x\)

\(B\) – Constant term of the given equation

Example:
In the equation, \(2 x+4=0\) with the variable \(x\) and \(2, 4\) are the constants.

Linear Equation in Two Variables

Any equation is said to be a linear equation in two variables if it has two variables (\(x\) and \(y\)) of the highest degree one, and they contain the constant (zero-degree variable) term.

The general form of the linear equation in two variables is given by \(A x+B y=C\).

Here,

\(x, y\) – Variables with degree one

\(A\) – Coefficient of the variable \(x\)

\(B\) – Co-efficient of the variable \(y\)

\(C\) – Constant term of the given equation

Example:
The equation \(2x+3y=7\) is the linear equation with two variables \(x\) and \(y\).

Pair of Linear Equations in Two Variables

The pair of linear equations in two variables is the system of the concurrent lines.  The graphical representation of the pair of linear equations in two variables is straight lines. 

The general way of representing the pair of linear equations with two variables is \(ax+by+c=0\).

Thus, the standard form of the pair of linear equations in two variables is given by

\(a_{1} x+b_{1} y+c_{1}=0\)
\(a_{2} x+b_{2} y+c_{2}=0\)

where, \(a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}\) are real numbers and \(a_{1}^{2}+b_{1}^{2} \neq 0\) and \(a_{2}^{2}+b_{2}^{2} \neq 0 .\) Here, \(a_{1}, a_{2}, b_{1}, b_{2}\) are the coefficients of the variables \(x, y\) respectively.

The pair of linear equations with two variables are classified into three types based on their solution. They are

1. Intersecting lines
2. Parallel lines
3. Coincident lines

Example:
The pair of equations in two variables (\(x\) and \(y\)) are \(7x-15y=2, x+2y=3\) are intersecting lines.

Simultaneous Linear Equations

Two linear equations with two variables are called simultaneous linear equations.

The pair of linear equations with two variables is also known as simultaneous linear equations, as they can be solved to find the solution of the linear equations.

Example:
The equations \(2 x+3 y=11\) and \(5 x+7 y=13\) are known as simultaneous linear equations in two variables \(x, y\).

Graph of Linear Equations

In graph paper, linear equations are represented as straight lines. The reason behind the degree of the linear equation is one is the graphical representation of the linear equation is a straight line.

The standard form of the linear equation in two variables is \(a x+b y+c=0\) and in the slope-intercept form is \(y=m x+c\), and the graph of the slop-intercept form is given below:

Solution of Linear Equations in One Variable

The value substituted for the variable of the given linear equation is called the root or solution of the equation.  A linear equation is like a weighing balance with equal weights on both sides.  The equation’s value does not change by adding or subtracting the same number from both sides of an equation. If we divide or multiply the equations by the same value on both sides of an equation, the equation remains unchanged.

The left side of the equal sign is called L.H.S, and the right side of the equal sign is R.H.S of the equation.

Example:
Solve the equation \(3 x+1=4\)
Subtracting \(1\) from both sides of the equation
\(\Rightarrow(3 x+1)-1=4-1\)
\(\Rightarrow 3 x=3\)
Dividing both sides of the given equation by \(3\)
\(\Rightarrow \frac{3 x}{3}=\frac{3}{3}\)
\(\Rightarrow x=1\)

Solving Linear Equations

The pair of linear equations with two variables or system of linear equations are solved by using algebraical or graphical methods:

Graph Method

The method utilises graph paper for solving the given equations by taking a suitable scale.

1. Write the equations in the form of \(y=m x+c\)
2. Plot the points on the graph paper which satisfy the above equations.
3. Observe the solution of the equations from the graph.

Example: The graph of the pair of lines x+2y=3, 4x+5y=6 given below:

The intersection point (-1,2) is the solution of the given equations.

Elimination Method

In this method, the solution of the equations is found by eliminating any one of the variables of the given equations.

Example: Solve the equations \(x+y=5\) and \(2 x-3 y=4\)

Solution: Let, \(x+y=5…..(i)\) and \(2 x-3 y=4…..(ii)\)
Equate the co-efficient of variable \(x\), multiply the equation \((i)\) with \(2\).
\((i) \times 2 \Rightarrow 2 x+2 y=10 \ldots \ldots(i i i)\)
Subtract equations \((ii)\) and \((iii)\),
\(\begin{array}{l} \,\,\,\,\,\,\,\,2x – 3y = 4\\ \,\,\,\,\,\,\,\,2x + 2y = 10\\ \left( – \right)\,\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \,\,\,\,\,\,\,\,\, – 5y = \, – 6 \end{array}\)
\(\Rightarrow y=\frac{6}{5}\)

Then, from the equation \((i)\),
\(\Rightarrow x+\frac{6}{5}=5\)
\(\Rightarrow x=5-\frac{6}{5}=\frac{25-6}{5}\)
\(\Rightarrow x=\frac{19}{5}\)

Therefore, the values of \(x, y\) are \(\frac{19}{5}, \frac{6}{5}\).

Substitution Method

In this method, the solution of the equations is found by substituting the value of the variable of any one equation in another equation and solve.

Example: Let \(7 x-15 y=2 \cdots(i)\) and \(x+2 y=3-\cdots(i i)\)

From equation \((ii)\),
\(\Rightarrow x=3-2 y-\cdots(i i i)\)
Then, equation \((i)\),
\(\Rightarrow 7(3-2 y)-15 y=2\)
\(\Rightarrow 21-14 y-15 y=2\)
\(\Rightarrow-29 y=2-21=-19\)
\(\Rightarrow y=\frac{19}{29}\)

Then, from the equation \((iii)\),
\(\Rightarrow x=3-2\left(\frac{19}{29}\right)\)
\(\Rightarrow x=\frac{87-38}{29}$\)
\(\Rightarrow x=\frac{49}{29}\)

Therefore, the solution of given equations is \(x=\frac{49}{29}, y=\frac{19}{29}\).

Cross-Multiplication Method

Let two linear equations \(a_{1} x+b_{1} y+c_{1}=0\) and \(a_{2} x+b_{2} y+c_{2}=0\).

1. Write the coefficients of the variables \((x, y)\) along with the constant values as shown below:

2. Write the products by cross multiplying as shown below:

\(\frac{x}{b_{1} c_{2}-b_{2} c_{1}}=\frac{y}{c_{1} a_{2}-c_{2} a_{1}}=\frac{1}{a_{1} b_{2}-b_{1} a_{2}}\)

3. Solve the obtained equations for variables.

\(\frac{x}{b_{1} c_{2}-b_{2} c_{1}}=\frac{1}{a_{1} b_{2}-b_{1} a_{2}} ; \frac{y}{c_{1} a_{2}-c_{2} a_{1}}=\frac{1}{a_{1} b_{2}-b_{1} a_{2}}\)

4. Find the values of the variables \((x, y)\)

\(x=\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-b_{1} a_{2}} ; y=\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-b_{1} a_{2}}\)

Example: Let \(2 x+3 y-46=0\) and \(3 x+5-74=0\)

Write the values of the coefficients as mentioned below:

\(\Rightarrow \frac{x}{3(-74)-(5)(-46)}=\frac{y}{(-46)(3)-2(-74)}=\frac{1}{2(5)-3(3)}\)
\(\Rightarrow \frac{x}{-222+230}=\frac{y}{-138+148}=\frac{1}{10-9}\)
\(\Rightarrow \frac{x}{8}=\frac{y}{10}=1\)
\(\Rightarrow \frac{x}{8}=1 ; \frac{y}{10}=1\)
\(\Longrightarrow x=8\) and \(y=10\)

Hence, the solution of the given equations is \(x=8, y=10\).

Solved Examples – Linear Equations

Q.1. \(T – 20\) cricket match was organised in Hyderabad. India and Australia were the two participating teams. Two Indian batsmen together scored \(183\) runs. Write the above information in the form of a linear equation?
Ans: Let the individual runs scored by two Indian batsmen are \(x\) and \(y\).
Given that two Indian batsmen scored together \(183\) runs.
So, the sum of individual runs scored by two Indian batsmen is equals to \(183\) runs.
\(x+y=183\)

Q.2. Asit trying to balance the two expressions by using weight balance, as shown in the figure. Help the Asit finding the unknown value.

Ans: Given weights are \(4 x+1\) and \(5\).
They are balanced, which means they can be written as \(4 x+1=5\)
Subtract one from both sides of the above equation,
\(\Rightarrow 4 x+1-1=5-1\)
\(\Rightarrow 4 x=4\)
Divide both sides of the above equation by \(4\)
\(\Rightarrow \frac{4 x}{4}=\frac{4}{4}\)
\(\Rightarrow x=1\)
So, the value of the unknown given in the above problem is \(1\).

Q.3. The difference between two numbers is \(12\), and one number is four times the other. Find them.
Ans: Let the number be \(x\) and \(y\).
Given the difference of the two numbers is \(12\).
\(x-y=12 \cdots(i)\)
According to the question,
\(x=4 y\) – – – – \((ii)\)
Substitute equation \((ii)\) in equation \((i)\), thus
\(\Rightarrow 4 y-y=12\)
\(\Rightarrow 3 y=12\)
\(\Rightarrow y=\frac{12}{3}\)
\(\Rightarrow y=4\)
Substitute the above value of \(y\) in the equation \((ii)\),
\(\Rightarrow x=4(4)\)
\(\Rightarrow x=16\)
Therefore, the numbers are \(16\) and \(4\).

Q.4. The cost of the book is twice the cost of a pen. Write the linear equation for the given condition.
Ans: Let the cost of the book is \(R s . x\)
And, the cost of the pen is \(R s . y\)
According to the question, the cost of the book is twice the cost of a pen.
The above condition can be written mathematically, as \(x=2 \times y\)
\(\Rightarrow x=2 y\)
\(\Rightarrow x-2 y=0\)
The above equation \(x-2 y=0\) represents the given condition.

Q.5. Five times the number added to half of the number gives \(55\). Find the number.
Ans: Let the number be \(x\).
Five times the number is \(5 x\), and half of the number is \(\frac{x}{2}\)
Given, five times the number added to half of the number gives \(55\).
\(\Rightarrow 5 x+\frac{x}{2}=55\)
\(\Rightarrow \frac{10 x+x}{2}=55\)
\(\Rightarrow \frac{11 x}{2}=55\)
\(\Rightarrow 11 x=110\)
\(\Rightarrow x=\frac{110}{11}\)
\(\Rightarrow x=10\)
Therefore, the required number is \(10\).

Summary

In this article, we studied the linear equations definition and understood how to plot the graph of linear equations. We looked at the linear equations in one variable, linear equations in two variables, pair of linear equations in two variables and simultaneous linear equations with examples.

We also studied the solution of linear equations and the methods of solving them by using the balancing method, elimination method, graphical method, substitution method and cross-multiplication method with the help of solved examples.

Learn About Solution of a Linear Equation

FAQs About Linear Equations

Let’s look at some of the commonly asked questions about linear equations:

Q.1. How to solve linear equations?
Ans: Linear equations are solved graphically and algebraically by using the elimination, substitution and cross-multiplication method.

Q.2. What is the formula of linear equations?
Ans: The linear equation in one variable is \(a x+b=0\). Here, \(x-\) variable and \(a, b\) – constants of the given equation. The linear equations in two variables are in the form of \(a x+b y+c=0\). Here, \(x, y\) -variables and \(a, b\) – constants of the given equation.

Q.3. What is a linear equation? Explain with an example.
Ans: Linear equations are the equations, which are in the first order. The linear equations in one variable are represented in the form of \(a x+b=0\).
Example:
\(2 x+3=0\)

Q.4. What are the three forms of linear equations?
Ans: The three forms of linear equations are
1. Point-slope form
2. Intercept form
3. Standard form

Q.5. What is the degree of the linear equations?
Ans: The degree of the linear equations is one.

Q.6. How to solve linear equations in three variables?
Ans: Follow the below steps for solving linear equations in three variables:
1. Consider any two equations and solve them for one variable.
2. Again consider the other two equations and solve for the same variable.
3. Now, we have a system of two linear equations with two unknown variables. 
4. Solve the two equations
5. Again substitute the obtained values of the variables into any of the given equations and solve for unknown variables.

We hope this detailed article on linear equations helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!