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Important Questions Class 9 Mathematics Chapter 4 – Linear Equations in Two Variables
Mathematics Chapter 4 of Class 9 introduces students to Linear Equations In Two Variables. A required linear equation in two variables has two numbers that can satisfy the given equation. These two numbers are called the solution of the required linear equation in two variables.
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ToggleFollowing are the key topics covered in Chapter 4 of CBSE Class 9 Mathematics syllabus:
Graphical representation of a linear equation in 2 variables
 Any given linear equation in the required standard form ax+by+c=0 has a couple of solutions in the required form (x,y) that can be illustrated in the coordinate plane.
 When a needed equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.
Solutions of Linear equation in 2 variables on a graph
 A given linear equation ax+by+c=0 is illustrated graphically as a straight line.
 Each point on the given line is a solution for the linear equation.
 Each solution of the given linear equation is a given point on the required line.
Lines passing through the origin
 Particular linear equations exist such that their required solution is (0, 0). Such equations, when illustrated graphically, pass through the origin.
 The coordinate axes, that is, the xaxis and yaxis, can be defined as y=0 and x=0, respectively.
Lines parallel to coordinate axes
 Linear equations of the given form y=a, when represented graphically, are lines parallel to the xaxis, and a is the needed ycoordinate of the required points in the same line.
 Linear equations of the given form x=a, when represented graphically, are lines parallel to the yaxis, and a is the needed xcoordinate of the required points in the same line.
Extramarks’ credibility lies in providing reliable and trusted study material related to NCERT is the most useful study companion for students and enables students with their comprehensive online study solutions from Class 1 to Class 12. Our qualified Mathematics subject experts have prepared various NCERT explanations to help students in their studies and exam preparation. Students can direct to our Important Questions Class 9 Mathematics Chapter 4 to practise examoriented questions. We have collected questions from diverse sources such as NCERT textbooks and exemplars, CBSE sample papers, CBSE past year question papers, etc. Students can prepare excellently for their exams and quizzes by solving various chapter questions from our Important Questions Class 9 Mathematics Chapter 4.
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Important Questions Class 9 Mathematics Chapter 4 – With Solutions
Our inhouse Mathematics faculty experts have collected an entire list of Important Questions Class 9 Mathematics Chapter 4 by referring to various sources. For every question, the team has prepared a stepbystep explanation that will aid students in understanding the concepts used in each question. Also, the given questions are prepared in a way that will cover the entire chapter.. By practising from our question bank, students should be able to revise the chapter and selfassess their strong and weak points. And improvise by further focusing on weak areas of the chapter and practising harder to maximise their potential.
Following are some of the questions and explanations from our question bank of Mathematics Class 9 Chapter 4 Important Questions:
Question 1: Define the following linear equations in the form ax + by + c = 0 and show the values of a, b and c in every individual case:
(i) x – y/5 – 10 = 0
(ii) 2x+3y = 6
(iii) y – 2 = 0
Answer 1:
(i) The equation xy/510 = 0
(1)x + (1/5) y + (10) = 0
Directly compare the above equation with ax + by + c = 0
Therefore, we get;
a = 1
b = ⅕
c = 10
(ii) –2x + 3y = 6
Rearranging the provided equation, we obtain,
–2x + 3y – 6 = 0
The required equation –2x + 3y – 6 = 0 can be written as,
(–2)x + 3y +(– 6) = 0
Directly comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0
We obtain a = –2
b = 3
c = 6
(iii) y – 2 = 0
y – 2 = 0
The required equation y – 2 = 0 can be written as,
0x + 1y + (–2) = 0
Directly comparing 0x + 1y + (–2) = 0 with ax + by + c = 0
We obtain a = 0
b = 1
c = –2
Question 2: The price of a notebook is twice the cost of a pen. Note a linear equation in two variables to illustrate this statement.
(Taking the price of a notebook to be ₹ x and that of a pen to be ₹ y)
Answer 2: Let the price of one notebook be = ₹ x
Let the price of one pen be = ₹ y
As per the question,
The price of one notebook is twice the cost of one pen.
i.e., the price of one notebook = 2×price of a pen
x = 2×y
x = 2y
x2y = 0
x2y = 0 is the required linear equation in two variables to illustrate the statement, ‘The price of one given notebook is twice the cost of a pen.
Question 3: Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Answer 3:
(i) 2x + 9 = 0
We have, 2x + 9 = 0
2x = – 9
x = 9/2
which is the required linear equation in one variable, that is, x only.
Therefore, x= 9/2 is a unique solution on the number line as shown below:
(ii) 2x +9=0
We can write 2x + 9 = 0 in the two variables as 2x + 0, y + 9 = 0
or x = −9−0.y/2
∴ When y = 1, x = −9−0.(1)/2 = 9/2
y=2 , x = −9−0.(2)/2 = 9/2
y = 3, x = −9−0.(3)/2= 9/2
Therefore, we obtain the following table:
X  9/2  9/2  9/2 
Y  1  2  3 
Now, plotting the ordered pairs (−9/2,3), (−9/2,3) and (−9/2,3) on graph paper and connecting them, we get a line PQ as the solution of 2x + 9 = 0.
Question 4: Note four solutions individually for the following equations:
(i) 2x + y = 7
Answer 4: For the four answers of 2x + y = 7, we replace different values for x and y
Let x = 0
Then,
2x + y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Now,
2x + y = 7
(2×1)+y = 7
2+y = 7
y = 7 – 2
y = 5
(1,5)
Let y = 1
Now,
2x + y = 7
2x+ 1 = 7
2x = 7 – 1
2x = 6
x = 3
(3,1)
Let x = 2
Now,
2x + y = 7
2(2)+y = 7
4+y = 7
y = 7 – 4
y = 3
(2,3)
The answers are (0, 7), (1,5), (3,1), (2,3)
Question 5: The linear equation 2x 5y = 7 has
(A) A unique solution
(B) Two solutions
(C) Infinitely many solutions
Answer 5: (C) Infinitely many solutions
Solution:
Linear equation: The equation of two variables which gives a straight line graph is called a linear equation.
Here the linear equation is 2x – 5y = 7
Let y = 0, then the value of x is:
2x – 5(0)=7
2x =7
x = 7/2
Now, let y = 1, then the value of x is:
2x – 5 (1) =7
2x 5 =7
2x = 7 + 5
2x =12
x = 12/2
x = 6
Here for different values of y, we are getting different values of x
Therefore the equation has infinitely many solutions
Question 6: Represent the following linear equations in the form ax + by + c = 0 and show the required values of a, b and c in every case:
Answer 6: (i) x –(y/5)–10 = 0
The required equation x –(y/5)10 = 0 can be written as,
1x+(1/5)y +(–10) = 0
Comparing the given equation x+(1/5)y+(–10) = 0 with ax+by+c = 0
We obtain,
a = 1
b = (1/5)
c = 10
(ii) –2x+3y = 6
–2x+3y = 6
Rearranging the equation, we obtain,
–2x+3y–6 = 0
The required equation –2x+3y–6 = 0 can be written as,
(–2)x+3y+(– 6) = 0
Comparing the given equation (–2)x+3y+(–6) = 0 with ax+by+c = 0
We obtain a = –2
b = 3
c =6
(iii) x = 3y
x = 3y
Rearranging the equation, we obtain,
x3y = 0
The required equation x3y=0 can be written as,
1x+(3)y+(0)c = 0
Comparing the given equation 1x+(3)y+(0)c = 0 with ax+by+c = 0
We obtain a = 1
b = 3
c =0
(iv) 2x = –5y
2x = –5y
Rearranging the equation, we obtain,
2x+5y = 0
The required equation 2x+5y = 0 can be written as,
2x+5y+0 = 0
Comparing the given equation 2x+5y+0= 0 with ax+by+c = 0
We obtain a = 2
b = 5
c = 0
(v) 3x+2 = 0
3x+2 = 0
The required equation 3x+2 = 0 can be written as,
3x+0y+2 = 0
Comparing the given equation 3x+0+2= 0 with ax+by+c = 0
We obtain a = 3
b = 0
c = 2
(vi) y–2 = 0
y–2 = 0
The required equation y–2 = 0 can be written as,
0x+1y+(–2) = 0
Comparing the given equation 0x+1y+(–2) = 0 with ax+by+c = 0
We obtain a = 0
b = 1
c = –2
(vii) 5 = 2x
5 = 2x
Rearranging the equation, we obtain,
2x = 5
i.e., 2x–5 = 0
The required equation 2x–5 = 0 can be written as,
2x+0y–5 = 0
Comparing the given equation 2x+0y–5 = 0 with ax+by+c = 0
We obtain a = 2
b = 0
c = 5
Question 7: Note four solutions individually for the following equations:
πx + y = 9
Answer 7: For the four answers of πx + y = 9, we replace other values for x and y
Let x = 0
Now,
πx + y = 9
(π × 0)+y = 9
y = 9
(0,9)
Let x = 1
Now,
πx + y = 9
(π×1)+y = 9
π+y = 9
y = 9π
(1,9π)
Let y = 0
Now,
πx + y = 9
πx +0 = 9
πx = 9
x =9/π
(9/π,0)
Let x = 1
Now,
Put x=2, we have
πx + y = 9
π(2) + y = 9
y = 9 – 2π
The answers are (0,9), (1,9π),(9/π,0),(2,9 – 2π)
Question 8: Find out the value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k.
Answer 8: The provided equation is
2x + 3y = k
As per the given question, x = 2 and y = 1.
Then, Replacing the values of x and y in the equation 2x + 3y = k,
We get,
⇒(2 x 2)+ (3 × 1) = k
⇒4+3 = k
⇒7 = k
⇒k = 7
The required value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k, is 7.
Question 9: Establish that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the required linear equation y = 9x – 7.
Answer 9: We include the equation,
y = 9x – 7
For A (1, 2),
Replacing (x,y) = (1, 2),
We obtain,
2 = 9(1) – 7
2 = 9 – 7
2 = 2
For B (–1, –16),
Replacing (x,y) = (–1, –16),
We get,
–16 = 9(–1) – 7
16 = – 9 – 7
16 = – 16
For C (0, –7),
Replacing (x,y) = (0, –7),
We obtain,
– 7 = 9(0) – 7
7 = 0 – 7
7 = – 7
Therefore, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Therefore, A (1, 2), B (–1, –16) and C (0, –7) are answers to the linear equation y = 9x – 7
Thus, points A (1, 2), B (–1, –16), and C (0, –7) lie on the graph of the linear equation y = 9x – 7.
Question 10: Note the linear equation such that every point on its graph has a coordinate 3 times its abscissa.
Answer 10:
As per the question,
A given linear equation such that every point on its graph has a coordinate(y) which is 3 times its
abscissa(x).
So we obtain
⇒ y = 3x.
Therefore, y = 3x is the required linear equation.
Question 11: Illustrate the graph of the given linear equation 3x + 4y = 6. At what points does the graph cut the X and Yaxis?
Answer 11: Given the equation,
3x + 4y = 6.
We need at least 2 points on the graph to illustrate the graph of this equation,
Therefore, the points the graph cuts
(i) xaxis
The given point is on the xaxis. We have y = 0.
Replacing y = 0 in the equation, 3x + 4y = 6,
We get,
3x + 4×0 = 6
⇒ 3x = 6
⇒ x = 2
Therefore, the point at which the graph cuts the xaxis = (2, 0).
(ii) yaxis
Since the point is on the yaxis, we have x = 0.
Replacing x = 0 in the equation, 3x + 4y = 6,
We obtain,
3×0 + 4y = 6
⇒ 4y = 6
⇒ y = 6/4
⇒ y = 3/2
⇒ y = 1.5
Thus, the point at which the graph cuts the xaxis = (0, 1.5).
By plotting the points (0, 1.5) and (2, 0) on the graph.
we obtain,
image source: Online
Question 12: Show that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the given graph of the linear equation y = 9x – 7.
Answer 12: We have the given equation,
y = 9x – 7
For A (1, 2),
Substitute the values of (x,y) = (1, 2),
We obtain,
2 = 9 (1) – 7 = 9 – 7 = 2
For B (–1, –16),
Substitute the values of (x,y) = (–1, –16),
We obtain,
–16 = 9(–1) – 7 = – 9 – 7 = – 16
For C (0, –7),
Substitute the values of (x,y) = (0, –7),
We obtain,
– 7 = 9(0) – 7 = 0 – 7 = – 7
Thus, we locate that points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Thus, A (1, 2), B (–1, –16), and C (0, –7) are required solutions of the linear equation y = 9x – 7
Hence, the given points A (1, 2), B (–1, –16) and C (0, –7) lie on the graph of the required linear equation y = 9x – 7.
Question 13: Show the required geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Answer 13: (i) In one variable
y = 3
∵ y = 3 is the required equation in one variable, that is, y only.
∴ y = 3 is the required unique solution on the number line as displayed below:
(ii) In two variables
When an equation in two variables, it can be expressed as,
0.x + y – 3 = 0
which is a linear equation in the variables x and y.
When x = 0, y = 3
When x = 1, y = 3
Image source: https://cdn1.byjus.com/wpcontent/uploads/2020/07/ncertsolutionsclass9chapter413.png
Question 14: In countries like the USA and Canada, the temperature is measured In the required Fahrenheit, whereas in countries like India, it is calculated in Celsius. Given here is a linear equation that converts Fahrenheit to Celsius:
F = (95 )C + 32
(i) Draw the linear equation graph above using Celsius for the xaxis and Fahrenheit for the yaxis.
(ii) If the required temperature Is 30°C, what is the temperature in Fahrenheit?
(iii) If the required temperature is 95°F, what is the temperature in Celsius?
(iv) If the required temperature is 0°C, what is Fahrenheit? If the required temperature is 0°F, what Is the temperature In Celsius?
(v) Is a required temperature numerically the same in Fahrenheit and Celsius? If yes, find It.
Answer 14: (i)F=9/5C+32
When C=0 then F=32
also, when C=5 then F=41
C  0  5 
F  32  41 
Image source: https://s3uswest2.amazonaws.com/infinitestudentimages/ckeditor_assets/pictures/64720/content_degree.png
(ii) Placing the value of C=30 in F=9/5C+32, we obtain
F=9/5×30+32
F=54+32
F=86
(iii) Putting the value of F=95 in F=9/5C+32, we get
95=9/5C+32
9/5C=95−32
C=63×5/9
C=35
Question 15: If the work accomplished by a body on application of a steady force is directly proportional to the required distance traversed by the body, convey this in the required form of an equation in two variables and sketch the exact graph by taking the steady force as 5 units. Likewise, read from the graph the work done when the distance traversed by the body is
(i) 2 units
(ii) 0 unit
Answer 15: Steady force is 5 units.
Let the distance traversed = x units and work done = y units.
Work done = Force x Distance
y = 5 x x
y = 5x
For sketching the graph, we are having y = 5x
When x = 0,
y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = 1, then y = 5(1) = 5
∴ We obtain the following given table:
x  0  1  1 
y  0  5  5 
Plotting the required ordered pairs (0, 0), (1, 5) and (1, 5) on the graph paper and joining the points, we acquire a straight line AB as shown.
From the required graph, we obtain
(i) The required distance traversed =2 units, that is, x = 2
∴ If x = 2,
then y = 5(2) = 10
The required work done = 10 units when the distance travelled by it is 2 units.
(ii) The required distance traversed = 0 unit ,that is , x = 0
∴ If x = 0
y = 5(0) – 0
The required work done = 0 unit when the distance travelled by it is 0 units.
.
Benefits Of Solving Important Questions Class 9 Mathematics Chapter 4
Practice is the key to scoring good marks in Mathematics. Mathematics taught in Classes 8, 9, and 10 prepares the foundation for higher classes and also many other educational streams. We recommend students register on our website to get access to our Important Questions Class 9 Mathematics Chapter 4. By rigorously solving questions and going through the required solutions, students will get enough confidence by clarifying their doubts to solve any other complicated questions in the given chapter, Linear Equations In Two Variables.
Given below are a few benefits of frequently solving questions from our question bank of Important Questions Class 9 Mathematics Chapter 4:
 Our expert Mathematics teacher faculty has carefully compiled the most important questions in our questionnaire Important Questions Class 9 Mathematics Chapter 4. These questions are picked after referring to many past years’ question papers, NCERT textbooks, and other Mathematics reference books.
 The given questions and solutions provided are completely based on the NCERT book and in accordance with the latest CBSE syllabus and guidelines. So the students can confidently bank on our study resources.
 The given questions covered in our set of important Questions Class 9 Mathematics Chapter 4 are entirely based on several topics covered in the chapter Linear Equations in Two Variables. It is suggested that students revise and clear all their doubts before solving all these important questions.
 By going through all the questions given in our Chapter 4 Class 9 Mathematics Important Questions, students will get an idea of the question paper pattern. . Practising questions comparable to the exam questions would aid students in gaining confidence, scoring better marks, and eventually acing their exams
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Q.1 If the point (4, 3) lies on the graph of the equation 3x ay = 6, find whether (2, 6) also lies on the same graph.
Marks:3
Ans
$\begin{array}{l}\text{Since}\left(\text{4, 3}\right)\text{liesonline3x}\text{ay=6}\\ \text{So,puttingx=4andy=3,weget}\\ \text{3}\left(\text{4}\right)\text{a}\left(\text{3}\right)\text{=6}\\ \text{12}\text{3a=6}\\ \text{12}\text{6=3a}\\ \text{a=}\frac{\text{6}}{\text{3}}\\ \text{=2}\\ \text{So,equationbecomes}\\ \text{3x2y=6}\\ \text{Puttingx=}\text{2andy=}\text{6inLHS,}\\ \text{LHS=3x}\text{y}\\ \text{=3}\left(\text{2}\right)\text{2}\left(\text{6}\right)\\ \text{=}\text{6+12}\\ \text{=6=RHS}\\ \text{Thus,}\left(\text{2,}\text{6}\right)\text{alsolieson the graph of the sameline.}\end{array}$
Q.2 For what value of k, the point (k, 5) lies on the line 4x 5y = 10
Marks:2
Ans
Point (k, 5) lies on the line 4x 5y = 10,
Since the point satisfies the line, so, we have
4(k) 5(5) =10
4k 25 = 10
4k = 35
k =
$\frac{35}{4}$
Q.3 Find four different solutions of the equation x+2y = 6.
Marks:3
Ans
Let us consider x = 0
Then the given equation reduces to 2y = 6
y = 3
So, (0,3) is a solution of x+2y = 6.
At x = 2, the given equation reduces to
2+2y = 6
y = 2
So, (0,2) is a solution of given equation.
At y = 0, the given eqution reduces to x = 6.
So (6,0) is a solution of given equation.
At y = 1 given equation reduces to x+2 = 6
x = 4.
So (4,1) is a solution of given equation.’
Thus (2,2), (0,3), (6,0) and (4,1) are the solutions of given equation.
Q.4 Draw the graph of x+y = 7.
Marks:5
Ans
$\begin{array}{l}\text{Todrawthegraph,weneedatleasttwosolutions.}\\ \text{x+y=7}\text{y=7}\text{x}\\ \text{x=0}\text{y=7}\text{0}\text{y=7}\\ \text{x=7}\text{y=7}\text{7}\text{y=0}\\ \left(\text{0,7}\right)\text{and}\left(\text{7,0}\right)\text{aresolutionsofx+y=7}\\ \text{We\xe2\u20ac\u2039canusethefollowingtabletodrawthegraph:}\\ \begin{array}{ccc}\hline \text{x}& \text{0}& \text{7}\\ \text{y}& \text{7}& \text{0}\\ \hline\end{array}\end{array}$
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FAQs (Frequently Asked Questions)
1. What is the standard equation in Linear Equations In Two Variables?
The standard equation for Equations in two variables is ax + by +c =0
2. How can I get good grades in Class 9 Mathematics examinations?
Mathematics is such a subject that requires plenty of practice. To achieve well in Mathematics, one must have a potent conceptual knowledge of the chapter, be good enough at calculations, practice questions regularly, give mock tests from time to time, get feedback and avoid silly mistakes. The more you practice, the better you will get. Everyday practice with discipline, working diligently and conscientiously towards your ambition, will ensure 100% in your exams.
3. What can I get from the Extramarks website?
Extramarks is one of the best educational platforms and it has its archive of academic resources, which also assists students in accomplishing their exams. You can acquire all the NCERTrelated material like NCERT solutions, solved exemplar solutions, NCERTbased mock tests, CBSE revision notes, and Class 9 Mathematics Chapter 4 important questions on the Extramarks website. Besides this, you can get great suggestions from our subject matter experts during doubtclearing sessions when you sign up on Extramarks official website. .
4. How many total chapters will students study in Class 9 Mathematics?
There are 15 chapters in Class 9 Mathematics. The list is given below:
 Chapter 1 Number System
 Chapter 2 –Polynomials
 Chapter 3 – Coordinate Geometry
 Chapter 4 –Linear Equations In Two Variables
 Chapter 5 – Introduction To Euclid’s Geometry
 Chapter 6 – Lines And Angles
 Chapter 7 –Triangles
 Chapter 8 – Quadrilaterals
 Chapter 9 –Areas Of Parallelograms And Triangles
 Chapter 10 – Circles
 Chapter 11 Constructions
 Chapter 12 Heron’s Formula
 Chapter 13Surface Area And Volumes
 Chapter 14 Statistics
 Chapter 15 Probability
5. What are the important chapters in Class 9 Mathematics Chapter 4?
The NCERT Mathematics book has 15 chapters. When it comes to grasping the fundamentals and taking the test, each and every chapter is equally important. Additionally, because CBSE does not specify marks distribution. Students need to study all the chapters in order to receive a good grade.
6. Where can I get important questions for Class 9 Mathematics Chapter 4 online?
On the Extramarks website, you can find all the important questions for Class 9 Mathematics Chapter 4, along with their answers. You can even find many other NCERTbased study solutions on the website for Classes 1 to 12.