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

Linear Equations in Two Variables Class 9 Questions And Answers PDF Free Download

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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables. Students can also access these solutions in PDF form.

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

Question 1: The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs x and that of a pen to be Rs y ).

Answer:

Let the cost of a notebook be Rs x and that of a pen be Rs y.

According to the given condition: The cost of a notebook is twice the cost of a pen.

Thus, $x=2y$

$\Rightarrow x-2y=0$

Question 2: (i) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $2x + 3y = 9.3\bar{5}$

Answer:

Given : $2x + 3y = 9.3\bar{5}$

$\Rightarrow 2x + 3y - 9.3\bar{5}=0$

Here , $a=2, b=3$ and $c = - 9.3\bar{5}$

Question 2: (ii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $x - \frac{y}{5} - 10 = 0$

Answer:

Given:

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

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

Here,

$a=1$,

$b=\frac{-1}{5}$

$c = -10$

Question 2: (iii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $-2x + 3y = 6$

Answer:

Given :

$-2x + 3y = 6$

$\Rightarrow -2x + 3y - 6=0$

Here , $a= -2, b=3$ and $c = -6$

Question 2: (iv) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $x =3y$

Answer:

Given : $x =3y$

$\Rightarrow x -3y=0$

Here , $a= 1, b= -3$ and $c =0$

Question 2: (v) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $2x = - 5y$

Answer:

Given : $2x = - 5y$

$\Rightarrow 2x + 5y=0$

Here , $a=2, b= 5$ and $c =0$

Question 2: (vi) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $3x + 2 = 0$

Answer:

Given : $3x + 2 = 0$

$\Rightarrow 3x + 2 = 0$

Here , $a= 3, b=0$ and $c =2$

Question 2: (vii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $y -2 = 0$

Answer:

Given : $y -2 = 0$

$\Rightarrow 0.x+y -2 = 0$

Here, $a=0, b= 1$ and $c = -2$

Question 2: (viii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $5 = 2x$

Answer:

Given: $5 = 2x$

$\Rightarrow 2x+0.y-5=0$

Here, $a=2, b= 0$ and $c = -5$

Question 1: Which one of the following options is true, and why? $y = 3x + 5$ has

(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

Answer:

Given : $y = 3x + 5$

This equation is of a line and a line has infinite points on it and each point is a solution Thus, (iii) infinitely many solutions is the correct option.

Question 2: Write four solutions for each of the following equations:

(i) $2x + y = 7$ (ii) $\pi x +y = 9$ (iii) $x = 4y$

Answer:

(i) Given : $2x + y = 7$

Putting x=0, we have , $y=7-2\times 0=7$ means $(0,7)$ is a solution.

Putting x=1, we have , $y=7-2\times 1=5$ means $(1,5)$ is a solution.

Putting x=2, we have , $y=7-2\times 2=3$ means $(2,3)$ is a solution.

Putting x=3, we have , $y=7-2\times 3=1$ means $(3,1)$ is a solution.

The four solutions are : $(0,7),(1,5),(2,3),(3,1)$ .

(ii) Given : $\pi x +y = 9$

Putting x=0, we have , $y=9-\pi \times 0=9$ means $(0,9)$ is a solution.

Putting x=1, we have , $y=9-\pi \times 1=9-\pi$ means $(1,9-\pi )$ is a solution.

Putting x=2, we have , $y=9-\pi \times 2=9-2\pi$ means $(2,9-2\pi )$ is a solution.

Putting x=3, we have , $y=9-\pi \times 3=9-3\pi$ means $(3,9-3\pi )$ is a solution.

The four solutions are : $(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi )$ .

(iii) Given: $x = 4y$

Putting x=0, we have , $y=\frac{0}{4}=0$ means $(0,0)$ is a solution.

Putting x=1, we have , $y=\frac{1}{4}$ means $(1,\frac{1}{4})$ is a solution.

Putting x=2, we have , $y=\frac{2}{4}=\frac{1}{2}$ means $(2,\frac{1}{2})$ is a solution.

Putting x=3, we have , $y=\frac{3}{4}$ means $(3,\frac{3}{4})$ is a solution.

The four solutions are : $(0,0)$ , $(1,\frac{1}{4})$ , $(2,\frac{1}{2})$ and $(3,\frac{3}{4})$ .

Question 3: (i) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: ((0,2)

Answer:

(i) Given : $x - 2y = 4$

Putting $(0,2)$ ,

we have , $x - 2y = 0-2(2)=-4\neq 4$

Therefore, $(0,2)$ is not a solution of $x - 2y = 4$ .

Question 3: (ii) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (2,0)

Answer:

Given : $x - 2y = 4$

Putting (2,0),

we have , $x - 2y = 2-2(0)=2\neq 4$

Therefore, (2,0) is not a solution of $x - 2y = 4$.

Question 3: (iii) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (4,0)

Answer:

Given : $x - 2y = 4$

Putting (4,0),

we have , $x - 2y = 4-2(0)=4=4$

Therefore, (4,0) is a solution of $x - 2y = 4$.

Question 3: (iv) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: $(\sqrt2, 4\sqrt2)$

Answer:

Given : $x - 2y = 4$

Putting $(\sqrt2 , 4\sqrt2)$ ,

we have , $x - 2y =\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\neq 4$

Therefore, $(\sqrt2 , 4\sqrt2)$ is not a solution of $x - 2y = 4$ .

Question 3: (v) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (1,1)

Answer:

Given : $x - 2y = 4$

Putting (1,1),

we have , $x - 2y = 1-2(1)=-1\neq 4$

Therefore, (1,1) is not a solution of $x - 2y = 4$ .

Question 4: Find the value of k , if $x = 2$ , $y = 1$ is a solution of the equation $2x + 3y = k$ .

Answer:

Given : $2x + 3y = k$

Putting (2,1),

we have , $k=2x + 3y = 2(2)+3(1)=4+3=7$

Therefore, k=7 for $2x + 3y = k$ putting x=2 and y=1.

Linear Equations in Two Variables Class 9 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Linear Equations in Two Variables Class 9 Maths Chapter 4 are provided in the link below.

• Class 9 Maths NCERT Chapter 4 Exercise 4.1
• Class 9 Maths NCERT Chapter 4 Exercise 4.2

Class 9 Maths NCERT Chapter 4: Extra Question

Question: Find the value of $y$ in the equation $ 2x + 3y = 12 $ when $x = 3 $.

Answer:

First, substitute $x = 3 $ into the equation:

We get, $2(3) + 3y = 12 \Rightarrow 6 + 3y = 12$

$3y = 12 - 6 = 6 \Rightarrow y = \frac{6}{3} = 2$

Thus, the value of $y$ is 2.

Linear Equations in Two Variables Class 9 Chapter 4: Topics

The topics discussed in the NCERT Linear Equations in Two Variables Class 9 solutions are:

• Introduction
• Linear Equations
• Solution of a Linear Equation
• Summary

Linear Equations in Two Variables Class 9 Solutions - Important Formulae

Linear Equations in Two Variables

The equations with two variables are called Linear Equations. The general form of a linear equation in two variables is ax+by=c, where a,b,c are real numbers and $a \neq 0$ and $b \neq 0 $.

Solutions of Linear Equations in Two Variables

The linear equations in two variables have infinitely many solutions. Every solution of linear equations in two variables is a point on the equation's graph.

Approach to Solve Questions of Linear Equations In Two Variables Class 9

These approaches will help students ‌deal with Class 9 Maths Chapter 4 Question Answers efficiently.

1. Understand the standard form: A linear equation consisting of two variables uses the form of $ax+by+c=0$ for all its components to be real numbers and also $a, b \neq 0$.

2. Know that each solution is a pair of values: The equation accepts ordered pairs (x,y) which solve the equation during substitution.

3. Use substitution to find solutions: Executing different x (or y) value substitutions helps determine corresponding values of the opposite variable.

4. Create tables of values: Make value tables that contain 3–5 solutions to reveal the relationship between the x and y variables.

5. Graph the solutions: Draw the pairs of solution points on a Cartesian coordinate system to see that they form a straight line.

6. Interpret the meaning of infinite solutions: The system of a linear equation that includes two variables can have numerous solutions defined through points which exist on the same straight line.

NCERT Solutions for Class 9 Maths Chapter Wise

We at Careers360 compiled all the NCERT class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

Also, read,

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

NCERT Books and NCERT Syllabus

We at Careers360 provide the links for Class 9 NCERT Books and NCERT Syllabus, making it easily accessible to the students. Below are the links for other study resources of NCERT Class 9.

• NCERT Books Class 9 Maths
• NCERT Books for Class 9 Science
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9