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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

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NCERT Solutions for Maths Class 9 Linear Equations in Two Variables Chapter 4 - FREE PDF Download

NCERT Maths Chapter 4 Linear Equations in Two Variables Class 9 is an important chapter to master in this curriculum. This chapter is intended to help students understand the ideas involved in producing graphs of linear equations on a Cartesian plane. Students will learn how to represent these equations graphically, understand their solutions, and interpret the significance of these solutions on a coordinate plane.

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Table of Content
1. NCERT Solutions for Maths Class 9 Linear Equations in Two Variables Chapter 4 - FREE PDF Download
2. Glance on Maths Chapter 4 Class 9 - Linear Equations in Two Variables
3. Access Exercise wise NCERT Solutions for Chapter 4 Maths Class 9
4. Exercises Under NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
5. Access NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables
6. Overview of Deleted Syllabus for CBSE Class 9 Maths Linear Equations in Two Variables
7. Class 9 Maths Chapter 4: Exercises Breakdown
8. Other Study Material for CBSE Class 9 Maths Chapter 4
9. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs


Follow the expert-designed class 9 linear equations in two variables chapter answers to grasp the context and how to solve the questions in this chapter. Discover the correct answers to all the questions included in the chapter's exercise with ease.


Glance on Maths Chapter 4 Class 9 - Linear Equations in Two Variables

  • Chapter 4 linear equation in two variables class 9 Maths deals with solving Linear Equations In Two Variables

  • A linear equation in two variables is an equation where the highest power of each variable is 1.

  • The x-axis represents the variable x and the y-axis represents the variable y.

  • The y-intercept is the point where the line crosses the y-axis (when x = 0).

  • The slope tells you the steepness and direction of the line:

  • Positive slope: Line goes up from left to right.

  • Negative slope: Line goes down from left to right.

  • Zero slope: Line is horizontal (parallel to x-axis).

  • Solve for one variable in terms of the other to get the equation in slope-intercept form.

  • This article contains chapter notes, important questions, exemplar solutions and exercises links for Chapter 4 - Linear Equations In Two Variables, which you can download as PDFs.

  • There are 2 exercises (6 fully solved questions) in class 9 chapter 4 maths Linear Equation In Two Variables.


Access Exercise wise NCERT Solutions for Chapter 4 Maths Class 9

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Exercises Under NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Exercise 4.1

The first exercise of this chapter consists of 2 questions, with question number two having eight sub questions in exercise 4.2 of NCERT Solutions for Maths Class 9 Chapter 4. Most of the questions of this exercise are based on the standard form of linear equation with two variables which is a potent technique to compute the value of a,b and c of any given equation. There are basically three types of questions found from this section:


  • Type 1: Representing a statement in a linear equation with two variables.

  • Type 2: Expressing the linear equation in its standard form.

  • Type 3: Identification of a, b and c in a linear equation with two variables.


Exercise 4.2

The second exercise in chapter 4 class 9 maths consists of 4 questions and is mostly based on the solution of linear equations with two variables. Once you grasp the concept of finding the solution, you will be able to identify equations having unique solutions, two solutions or many solutions. Given below are the types of questions found related to the topic:


  • Type 1: Identification of the number of solutions of the given equations.

  • Type 2: Finding the solution of the given equation.

  • Type 3: Cross-checking the solutions of the equation.

  • Type 4: Finding the value of an unknown term if the solution of the equation is given.


Access NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables

Exercise (4.1)

1: Construct a linear equation in two variables to express the following statement.

The cost of a textbook is twice the cost of an exercise book.

Ans. Let the cost of a textbook be $\text{x}$ rupees and the cost of an exercise book be $\text{y}$ rupees.

The given statement:  The cost of a textbook is twice the cost of an exercise book

So, in order to form a linear equation, 

the cost of the textbook $\text{=}\,\text{2 }\!\!\times\!\!\text{ }$ the cost of an exercise book.  

$\Rightarrow \text{x=2y}$

$\Rightarrow \text{x-2y=0}$.


2: Determine the values of $\text{a}$, $\text{b}$, $\text{c}$ from the following linear equations by expressing each of them in the standard form \[\text{ax+by+c=0}\].

(i) $\text{2x+3y=9}\text{.}\overline{\text{35}}$ 

Ans. The given linear equation is

$\text{2x+3y=9}\text{.}\overline{\text{35}}$

Subtracting $9.\overline{35}$ from both sides of the equation gives

$\text{2x+3y}-\text{9}\text{.}\overline{\text{35}}\text{=0}$ 

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as  

$\text{a=2}$, 

$\text{b=3}$, and

$\text{c}=-\text{9}\text{.}\overline{\text{35}}$


(ii) $\text{x-}\frac{\text{y}}{\text{5}}\text{-10=0}$

Ans. The given linear equation is

$\text{x-}\frac{\text{y}}{\text{5}}\text{-10}=\text{0}$ 

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{1}$,

$\text{b}=-\frac{\text{1}}{\text{5}}$, and

$\text{c}=-\text{10}$.

(iii) $\text{-2x+3y=6}$

Ans. The given linear equation is

$\text{-2x+3y=6}$

Subtracting $6$ from both sides of the equation gives 

$-\text{2x+3y}-\text{6}=\text{0}$

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=-\text{2}$,

$\text{b}=\text{3}$, and

$\text{c}=-\text{6}$.


(iv) $\text{x=3y}$

Ans. The given linear equation can be written as

$\text{1x}=\text{3y}$

Subtracting $3y$ from both sides of the equation gives 

$\text{1x-3y+0=0}$

Now, by comparing the above equation with the standard form of the linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=\text{1}$,

$\text{b}=-\text{3}$, and

$\text{c}=\text{0}$.


(v) \[\text{2x}\mathbf{=-}\,\text{5y}\]

Ans. The given linear equation is

\[\text{2x}=-\text{5y}\].

Adding $5y$ on both sides of the equation gives

\[\text{2x+5y+0=0}\].

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=\text{2}$,

$\text{b}=\text{5}$, and

$\text{c}=\text{0}$.


(vi) $\text{3x+2=0}$

Ans. The given linear equation is

$\text{3x+2=0}$. 

Rewriting the equation gives

$\text{3x+0y+2=0}$

Now, by comparing the above equation with the standard form of linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{3}$,

$\text{b}=\text{0}$, and

$\text{c}=\text{2}$.


(vii) $\text{y-2=0}$

Ans. The given linear equation is

$\text{y-2=0}$ 

The equation can be expressed as

$\text{0x+1y-2}=\text{0}$

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{0}$,

$\text{b}=\text{1}$, and

$\text{c}=-\text{2}$.


(viii) $\text{5=2x}$

Ans: The given linear equation is

$\text{5=2x}$.

The equation can be written as 

$\text{-2x+0y+5=0}$.

Now, by comparing the above equation with the standard form of the linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=-\text{2}$,

$\text{b}=\text{0}$, and

$\text{c}=\text{5}$.


Exercise (4.2) 

1: Complete the following statement by choosing the appropriate answer and explain why it should be chosen?

$\text{y=3x+5}$ has ___________. 

(a) A unique solution, 

(b) Only two solutions, 

(c) Infinitely many solutions. 

Ans: Observe that, $\text{y}=\text{3x+5}$ is a linear equation. 

Now, note that, for $\text{x}=\text{0}$, $\text{y}=\text{0+5=5}$.

So, $\left( \text{0,5} \right)$ is a solution of the given equation.

If $\text{x=1}$, then $\text{y}=\text{3 }\!\!\times\!\!\text{ 1+5}=\text{8}$.

That is, $\left( \text{1,8} \right)$ is another solution of the equation. 

Again, when $\text{y}=\text{0}$, $\text{x}=-\frac{5}{3}$ .

Therefore, $\left( -\frac{5}{3},0 \right)$ is another solution of the equation.

Thus, it is noticed that for different values of $\text{x}$ and $\text{y}$, different solutions are obtained for the given equation.

So, there are countless different solutions exist for the given linear equation in two variables. Therefore, a linear equation in two variables has infinitely many solutions. 

Hence, option (c) is the correct answer.


2: Determine any four solutions for each of equations given below. 

(i) $\mathbf{2x}+\mathbf{y}=\mathbf{7}$. 

Ans: The given equation 

$\text{2x+y}=\text{7}$ is a linear equation.

Solving the equation for $y$ gives

$\text{y=7-2x}$. 

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For  $\text{x=0}$, 

$\text{2}\left( \text{0} \right)\text{+y=7}$

$\Rightarrow \text{y=7}$

So, one of the solutions obtained is $\left( \text{x,y} \right)\text{=}\left( \text{0,7} \right)$. 

For  $\text{x=1}$, 

$\text{2}\left( \text{1} \right)\text{+y=7}$

$\Rightarrow \text{y=5}$

Therefore, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{1,5} \right)$.

For $\text{x=2}$, 

$\text{2}\left( \text{2} \right)\text{+y=7}$

$\Rightarrow \text{y=3}$

That is, a solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{3,1} \right)$.

Also, for $\text{x=3}$, 

\[\text{2}\left( \text{3} \right)\text{+y=7}\]

$\Rightarrow \text{y=1}$ 

So, another one solution is $\left( \text{x,y} \right)\text{=}\left( \text{3,1} \right)$.

Thus, four solutions obtained for the given equations are $\left( \text{0,7} \right)$ , $\left( 1,5 \right)$, $\left( 2,3 \right)$, $\left( 3,1 \right)$.


(ii) $\mathbf{\pi x}+\mathbf{y}=\mathbf{9}$.

Ans.The given equation 

$\pi x+y=9$                                                                          …… (a)

is a linear equation in two variables.

By transposing, the above equation (a) can be written as

$\text{y=9- }\!\!\pi\!\!\text{ x}$. 

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For $\text{x=0}$, 

$\text{y=9- }\!\!\pi\!\!\text{ }\left( \text{0} \right)$

$\Rightarrow \text{y=9}$

Therefore, one of the solutions obtained is $\left( \text{x,y} \right)\text{=}\left( \text{0,9} \right)$.

For $\text{x=1}$,

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{1} \right)$

$\Rightarrow \text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }$.

So, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{1,9- }\!\!\pi\!\!\text{ } \right)$.

For  $\text{x=2}$, 

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{2} \right)$

$\Rightarrow \text{y}=\text{9}-\text{2 }\!\!\pi\!\!\text{ }$

That is, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{2,}\,\text{9-2 }\!\!\pi\!\!\text{ } \right)$.

Also, for $\text{x=3}$, 

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{3} \right)$

$\Rightarrow \text{y}=9-\text{3 }\!\!\pi\!\!\text{ }$.

Therefore, another one solution is $\left( \text{x,y} \right)\text{=}\left( \text{3,}\,\text{9-3 }\!\!\pi\!\!\text{ } \right)$.

Thus, four solutions obtained for the given equations are $\left( 0,9 \right)$, $\left( \text{1,9,- }\!\!\pi\!\!\text{ } \right)$, $\left( \text{2,9-2 }\!\!\pi\!\!\text{ } \right)$, $\left( \text{3,9-3 }\!\!\pi\!\!\text{ } \right)$.


(iii) $\mathbf{x}=\mathbf{4y}$.

Ans. The given equation 

$\text{x=4y}$ is a linear equation in two variables.

By transposing, the above equation can be written as

$\text{y=}\frac{\text{x}}{4}$ .

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For $\text{x=0}$, 

$\text{y}=\frac{0}{4}=0$.

Therefore, one of the solutions is $\left( \text{x,y} \right)\text{=}\left( \text{0,0} \right)$.

For $x=1$,

$\text{y}=\frac{1}{4}$.

So, another solution of the given equation is $\left( \text{x,y} \right)\text{=}\left( \text{1,}\frac{1}{4} \right)$.

For $\text{x=2}$,

$\text{y}=\frac{2}{4}=\frac{1}{2}$.

That is, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{2,}\frac{1}{2} \right)$.

Also, for $\text{x=3}$.,

$\text{y}=\frac{3}{4}$.

Therefore, another one solution is $\left( \text{x,y} \right)=\left( 3,\frac{3}{4} \right)$.

Thus, four solutions obtained for the given equations are $\left( 0,0 \right)$, $\left( \text{1,}\frac{1}{4} \right)$, $\left( \text{2,}\frac{1}{2} \right)$, $\left( 3,\frac{3}{4} \right)$.


3: Identify the actual solutions of the linear equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] from each of the following solutions.

(i)  $\left( \mathbf{0},\mathbf{2} \right)$

Ans: Substituting $\text{x=0}$ and $\text{y=2}$ in the Left-hand-side of the equation \[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=0-2\left( 2 \right) \\ & =-4 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 0,2 \right)$.

Hence, $\left( 0,2 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 


(ii) $\left( \mathbf{2},\mathbf{0} \right)$

Ans. Substituting $\text{x=2}$ and $\text{y=0}$ in the Left-hand-side of the equation \[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=2-2\left( 0 \right) \\ & =2 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 2,0 \right)$.

Hence, $\left( 2,0 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 


(iii) $\left( \mathbf{4},\mathbf{0} \right)$ 

Ans. Substituting $\text{x=4}$ and $\text{y=0}$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=4-2\left( 0 \right) \\ & =4. \end{align}$

Therefore, Left-hand-side is equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 4,0 \right)$.

Hence, $\left( 4,0 \right)$ is a solution of the equation \[\text{x-2y=4}\]. 


(iv) $\left( \sqrt{\mathbf{2}}\mathbf{,4}\sqrt{\mathbf{2}} \right)$

Ans. Substituting $\text{x=}\sqrt{2}$ and $\text{y=4}\sqrt{2}$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=\sqrt{2}-2\left( 4\sqrt{2} \right) \\ & =\sqrt{2}-8\sqrt{2} \\ & =-7\sqrt{2} \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( \sqrt{2},4\sqrt{2} \right)$.

Hence, $\left( \sqrt{2},4\sqrt{2} \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 


(v) $\left( \mathbf{1},\mathbf{1} \right)$

Ans. Substituting $\text{x}=1$ and $\text{y}=1$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=1-2\left( 1 \right) \\ & =1-2 \\ & =-1 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 1,1 \right)$.

Hence, $\left( 1,1 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 


4: If $\left( \mathbf{x},\mathbf{y} \right)=\left( \mathbf{2},\mathbf{1} \right)$ is a solution of the equation \[\text{2x+3y=k}\], then what is the value of $\mathbf{k}$?

Ans: By substituting $\text{x}=2$, $\text{y}=1$ and into the equation

\[\text{2x+3y=k}\] gives

$\text{2}\left( \text{2} \right)\text{+3}\left( \text{1} \right)\text{=k}$

$\Rightarrow \text{4+3=k}$

$\Rightarrow \text{k=7}$.

Hence, the value of $\text{k}$ is $7$.


Summary of Linear Equations

An equation includes equal sign (=) which indicates that the terms on the left-hand side are equal to the terms on the right-hand side. A Linear equation is an equation for a straight line containing variables and constants in the form given below:


\[a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}\]..... + b = 0


Where \[a_{1}, a_{2}, a_{3}\]... are coefficients, b is a constant and \[x_{1}, x_{2}, x_{3}\].... are the variables. If the value of any coefficient or variable is zero then the term containing that coefficient or variable becomes zero. This is because anything multiplied to zero is equal to zero.


A linear equation is a simple equation containing coefficients, constants and one or more variables, but a linear equation can never have exponents and roots.


One, Two, Three What Variables Would Be?- Types of Linear equations

One variable linear equation: A linear equation which has only one variable (unknown term) represented by alphabets or symbols is known as one variable linear equation. It is represented as ax+b = 0, where a is a coefficient of variable x and b is a constant. The coefficient can never be zero.


Examples: 7x + 6 = 13


Two variable linear equation: A linear equation is an equation which has two variables (unknown terms) represented by alphabets or symbols known as a two-variable linear equation. It is represented as ax+by+c = 0, where a and b are coefficients,  x and y are variables and c is a constant. If any coefficient becomes zero then the two-variable linear equation changes to one variable linear equation.


Examples: 2x +3y = 24


Three or more variable linear equations: A linear equation containing more than two variables is called a linear equation of three or more variables. It can be represented as: 

\[a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}\]..... + b = 0.


Examples: 5x+ 21y - 3z = -2


Linear Equation with Two Variables

Till now in the journey of algebraic equations, we have learned solving single equations with only one variable (unknown). For example something like 9x + 4 = 22. Simple, Isn’t it?


But what happens if there is more than one unknown in an equation something like 5x + 3y = 15. We solve it differently. So before looking into the solution of a linear equation with two variables let us understand the Linear equation with two variables mathematically.


Definition

An equation of the type ax+by+c = 0, where a,b,c are real numbers such that a and b are non-zero, is called a linear equation in two variables x and y.


Example: x+y-5 = 0 is a linear equation in the two variables(unknowns) x and y. Note that x=2 and y=3 satisfy this linear equation.


A Single Linear Equation with Two Variables Cannot be Solved.

There’s no way anyone could legitimately ask you to solve a single equation with two variables because that would give you infinite solutions. But two equations having two variables each can be solved to find the value of x and y simultaneously. A group of two or more equations is called a system of equations.


Each equation represents a straight line. If two lines are taken then there are high chances that those two lines intersect at a unique point which satisfies both the equations. In order to find the intersecting point, pick two random lines and solve.


Solution of a Linear Equation

We know that every linear equation in one variable has a unique solution. What about the solution of a linear equation with  two variables? There will always be a pair of values one for x and the other for y which satisfy the given equation. Also, note that there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.


There are many ways to solve a system of linear equations with two variables. Given below are the two basic methods to solve a linear equation with two variables.

 

1. Graphical Method of Solving Linear Equation

Instead of finding the solution of two the linear equations separately we find the solution of the system instead. If we graph both the lines in the same coordinate system then the point of intersection of two lines will be the solution of the system.


For Example: To solve the system of equations having two equations -


2x+2 = y and x-1 = y, we need to consider a value of x and find its corresponding value of y for each equation. For equations 2x+2 = y and x-1 = y a random value of x is taken and its corresponding value of y is to be calculated. The points will be (1,4), (2,6), (3,8) for equation 2x+2 = y. And the points (1,0), (2,1), (3,2) for equation x-1 = y. The points are to be plotted on a graph. The point of intersection of these two lines will be the solution of the system.


Geometric Representations


2. Substitution Method of Solving Linear Equation

The other way of solving a system of linear equations is by substitution method. This system shows how to solve linear equations easily by finding the value of one variable in terms of another variable by using one equation and then replacing this value in another equation.


Let's find the solution of the same system of linear equations.

2x+2 = y

x-1 = y

From equation (2) we can say that y = x-1.

Substituting the value of y in equation (1).

2x+2 = x-1

2x-x = -2-1

x = -3

Thus, we can find the actual value of y by substituting the value of x as -3 in equation (1).

2x + 2 = y

2(-3) + 2 = y

-6 + 2 = y

-4 = y

or, y= -4.

Therefore, the solution of the system of linear equations is (-3,-4).


Summary

  • An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables. 

  • A linear equation in two variables has infinitely many solutions. 

  • The graph of every linear equation in two variables is a straight line. 

  • x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis. 

  • The graph of x = a is a straight line parallel to the y-axis. 

  • The graph of y = a is a straight line parallel to the x-axis. 

  • An equation of the type y = mx represents a line passing through the origin. 

  • Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.


Overview of Deleted Syllabus for CBSE Class 9 Maths Linear Equations in Two Variables

Chapter

Dropped Topics


Linear Equations in Two Variables

Graph of a Linear Equation in Two Variables

Equations of Lines Parallel to the x-axis and y-axis


Class 9 Maths Chapter 4: Exercises Breakdown

Exercise

Number of Questions

Exercise 4.1

2 Questions & Solutions

Exercise 4.2

4 Questions & Solutions


Conclusion 

The NCERT Solutions for Maths Chapter 4 on Linear Equations in Two Variables for Class 9, provided by Vedantu, offer comprehensive guidance for mastering this topic. Key points to focus on include understanding the graphical representation of linear equations, methods of solving equations, and interpreting solutions. These concepts are crucial for building a strong foundation in algebra. Typically, previous year question papers have included around 3-4 questions from this chapter, emphasizing its importance in the curriculum. Students should practice the solved examples and exercise questions thoroughly to excel in exams.


Other Study Material for CBSE Class 9 Maths Chapter 4


Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Study Materials for CBSE Class 9 Maths

FAQs on NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

1. How do Vedantu’s NCERT Solutions for Maths Class 9 Chapter 4 help all the students to score good marks?

In this chapter of Maths, Class 9 students will study about Linear Equations In Two Variables. And below are the list of benefits of how these NCERT Solutions will help the students. Take a glance through the list. 

  • The solutions are written in simple and easy-to-understand language keeping in mind every kind of the students.

  • If there is any complicated solution, that is broken down into simple steps to make it understandable for every student to grasp the concept in less time.

  • The solutions are updated as per the latest NCERT curriculum and guidelines.

  • It covers the entire syllabus and concept in the form of solutions.

  • The answers are treated systematically and in an interesting manner.

  • The content is designed in a concise manner, which is brief and self-explanatory.

  • Some answers include necessary infographics and images to facilitate the understanding of the concept.

  • It is handy and serves as a note during exam revision.

  • The solutions are kept easy for you to solve maximum questions and get a command of the chapter. Also, it Improves the problem solving speed

2. What can I learn from Maths Class 9 Chapter 4?

In this chapter, the knowledge of linear equations in one variable is recalled and extended to that of two variables. Any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is defined as a linear equation in two variables. The NCERT solutions for Class 9 Maths provide chapter-wise answers of the questions asked in the exercises in the textbook and students can get the command over this Linear Equation concept of Algebra with the help of easy examples provided.

3. Give me an overview of the chapter.

Linear Equations In Two Variables” is chapter 4 of the NCERT Textbook, and it falls under the unit 2 Algebra. Interesting topics such as Graph of a Linear Equation in Two Variables and Equations of Lines Parallel to x-axis and y-axis will be learnt in this chapter with the help of plotting the two variables of a linear equation on a graph sheet. The solutions provided for all these topics are step by step so that the learner can imbibe the concept step by step.

  • 4.1 – Introduction

  • 4.2 – Linear Equations

  • 4.3 – Solution of a Linear Equation

  • 4.4 – Graph of a Linear Equation In Two Variables

  • 4.5 – Equations of Lines Parallel to the x-axis and y-axis

4. How many exercises are there in this chapter?

NCERT Grade 9 CBSE Chapter 4 Linear Equations in Two Variables belongs to the Algebra segment. This chapter has five topics and sub-topics including the introduction part. The Introduction part depicts solving a linear equation in two variables and how does the solution look like on the Cartesian plane. The topic Linear Equations explains the crucial points that should be memorised while solving a linear problem. In this chapter, you will learn about the Solution of a Linear Equation. This topic deals with a solution of a linear equation with two variables with a pair of values, one for x and one for y which satisfies the given equation. 


The Chapter 4 named Linear Equation in Two Variables Class 9 is divided into five sections and four exercises. Below is the list of exercises along with the number and types of questions.

  • Exercise 4.1: 2 Questions (1 Short Answer, 1 Main Question with 8 short answer questions)

  • Exercise 4.2: 4 Questions (2 MCQs, 1 Main Questions with 3 equations, 1 Short Answer Questions)

  • Exercise 4.3: 8 Questions (4 Long Answer Questions, 2 Short Answer Questions, 1 MCQ, 1 Main Question with 5 Part Questions)

  • Exercise 4.4: 2 Questions (1 main question with 2 sub-section, 1 Main question with 2 sub-section)

5. What are all the important topics that need to be covered to score in Class 9 Maths Chapter 4?

The topics that you must not miss in order to fully complete the Class 9 Maths Chapter 4 Linear Equations in Two Variables are - expressing linear equations in Ax + By + C = 0; different solutions of a linear equation with two variables; graph of a linear equation in two variables; and equations of the line parallel to x-axis and y-axis. To get clarity of these concepts, you need to solve all examples and exercise questions. 

6. Where can I get the best solutions for Class 9 Maths Chapter 4? 

Vedantu is the best place to get free solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables. Vedantu’s website and Vedantu app offers a comprehensive and detailed NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables that has explanations to all the different sub-topics. These solutions have complete step by step solutions to the examples and exercises that will give clarity to all the concepts. These are prepared by experts who have decades of experience, so these solutions are credible and up to date.

7. Is Chapter 4 of Linear Equation in Two Variables of Class 9 Maths tough?

No, the NCERT Class 9 Mathematics Linear Equation in Two Variables is not difficult for those who thoroughly practise the chapter. Your main aim should be to finish the chapter thoroughly, including all examples, exercise questions, and other questions. NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables by Vedantu is the best resource for completing Class 9 Maths Chapter 4 without problem.

8. How can I solve Class 9 Maths Chapter 4?

You can easily solve Class 9 Maths Chapter 4 Linear Equations in Two Variables with the help of Vedantu’s NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables. These solutions have full solved numerical problems step by step that will help you in clearing your concepts. You can also get access to many miscellaneous questions that will enhance your base even further.  

9. Can the graph of a linear equation with two variables be a curve?

No, the graph of a two-variable linear equation cannot be a curve. The path is always straight. Get the greatest companion available only at Vedantu for additional NCERT Class 9 Mathematics Chapter 3 questions that will assist you in completing comprehensive exercises along with examples and random questions. You may obtain entire solutions to the exercises by downloading the NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry.

10. How do you know if an equation is linear in two variables?

A linear equation in two variables is of the form 𝑎𝑥+𝑏𝑦+𝑐=0 where a, b, and 𝑐 are real numbers, 𝑥 and 𝑦 are the variables. The equation is considered linear because the highest power of the variables x and y is 1.

11. Can linear equations have 2 answers?

A linear equation in two variables has infinitely many solutions, represented by the points on a line in a Cartesian plane. Each point on the line is a solution to the equation. However, a single linear equation does not have just two specific answers but a continuum of solutions.

12. What are the steps in solving linear equations in two variables?

The steps to solve a system of linear equations in two variables typically involve:

  • Graphical Method: Plotting both equations on a graph and finding the point of intersection.

  • Substitution Method: Solving one equation for one variable and substituting this into the other equation.

  • Elimination Method: Adding or subtracting the equations to eliminate one variable, making it possible to solve for the other variable.

13. What is a linear equation cannot have?

A linear equation cannot have:

  • Variables with exponents other than 1 (e.g., 𝑥2 or 𝑦3)

  • Products of variables (e.g., 𝑥𝑦).

  • Variables in the denominator. (e.g., 1/𝑥).

Linear equations must maintain the variables to the first degree and should not involve any form of non-linear terms.