NCERT Solutions Maths Ch-4 Linear Equations in Two Variables for Class 9th
Chapter 4 of the NCERT Mathematics textbook for Class 9, titled “Linear Equations in Two Variables,” introduces the concept of linear equations in the form ax+by+c=0 The chapter emphasizes understanding how to represent these equations graphically, highlighting the importance of the coordinate plane. Students learn to identify solutions of linear equations and how each equation corresponds to a straight line.
The chapter covers methods for solving linear equations, including substitution and elimination, and explains how to find the intersection point of two lines, which represents the solution to a system of equations. Additionally, it illustrates real-life applications of linear equations, enhancing students’ ability to model and solve practical problems. Overall, the chapter builds foundational skills in algebra and prepares students for more advanced topics in mathematics.
NCERT Solutions Ch-4 Linear Equations in Two Variables for Class 9th Exercise 4.1, 4.2, 4.3
We try to teach you all Questions in easy way. We solve all chapter wise sums of maths textbook. In every chapter include NCERT solutions. For solutions of Exercise 4.1, 4.2, 4.3 click on Tabs :
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 x and that of a pen to be y)
Solution:
Let the cost of a notebook = Rs. x
and the cost of a pen = Rs. y
According to the condition, we have
[Cost of a notebook] =2 x [Cost of a pen]
i. e„ (x) = 2 x (y) or, x = 2y
or, x – 2y = 0
Thus, the required linear equation is x – 2y = 0.
Question 2
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y =
(ii)
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = -5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
Solution:
(i) We have 2x + 3y =
or (2)x + (3)y + (
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –
(ii) x –(y/5)–10 = 0
Solution:
The equation x –(y/5)-10 = 0 can be written as,
1x+(-1/5)y +(–10) = 0
Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0
We get,
a = 1
b = -(1/5)
c = -10
(iii) –2x+3y = 6
Solution:
–2x+3y = 6
Re-arranging the equation, we get,
–2x+3y–6 = 0
The equation –2x+3y–6 = 0 can be written as,
(–2)x+3y+(– 6) = 0
Now comparing (–2)x+3y+(–6) = 0 with ax+by+c = 0
We get, a = –2
b = 3
c =6
(iv) x = 3y
Solution:
x = 3y
Re-arranging the equation, we get,
x-3y = 0
The equation x-3y=0 can be written as,
1x+(-3)y+(0)c = 0
Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0
We get, a = 1
b = -3
c =0
(v) 2x = –5y
Solution:
2x = –5y
Re-arranging the equation, we get,
2x+5y = 0
The equation 2x+5y = 0 can be written as,
2x+5y+0 = 0
Now comparing 2x+5y+0= 0 with ax+by+c = 0
We get, a = 2
b = 5
c = 0
(vi) 3x+2 = 0
Solution:
3x+2 = 0
The equation 3x+2 = 0 can be written as,
3x+0y+2 = 0
Now comparing 3x+0+2= 0 with ax+by+c = 0
We get, a = 3
b = 0
c = 2
(vii) y–2 = 0
Solution:
y–2 = 0
The equation y–2 = 0 can be written as,
0x+1y+(–2) = 0
Now comparing 0x+1y+(–2) = 0with ax+by+c = 0
We get, a = 0
b = 1
c = –2
(viii) 5 = 2x
Solution:
5 = 2x
Re-arranging the equation, we get,
2x = 5
i.e., 2x–5 = 0
The equation 2x–5 = 0 can be written as,
2x+0y–5 = 0
Now comparing 2x+0y–5 = 0 with ax+by+c = 0
We get, a = 2
b = 0
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
Solution:
Option (iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation.
Hence, given linear equation has an infinitely many solutions.
Question 2
Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))
(iii) x = 4y
When x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
When x = 1, 4y = 1 ⇒ y =
∴ Solution is (1,
When x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
When x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)
Question 3
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0,2)
(ii) (2,0)
(iii) (4, 0)
(iv) (√2, 4√2)
(v) (1, 1)
Solution:
(i) (0,2) means x = 0 and y = 2
Puffing x = 0 and y = 2 in x – 2y = 4, we get
L.H.S. = 0 – 2(2) = -4.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x =0, y =2 is not a solution.
(ii) (2, 0) means x = 2 and y = 0
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H:S. 2 – 2(0) = 2 – 0 = 2.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (2,0) is not a solution.
(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = o in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.
(iv) (√2, 4√2) means x = √2 and y = 4√2
Putting x = √2 and y = 4√2 in x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (√2 , 4√2) is not a solution.
(v) (1, 1)means x =1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4
∴ LH.S. ≠ R.H.S.
∴ (1, 1) is not a solution.
Question 4
Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.
Question 1
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution:
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:
Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.
Thus, the line AB is the required graph of x + y = 4
(ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:
Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.
Thus, the ime is the required graph of x – y = 2
(iii) y = 3x
If we have x = 0,
then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) = 3
x= -1, then y = 3(-1) = -3
∴ We get the following table:
Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.
Thus, the line LM is the required graph of y = 3x.
(iv) 3 = 2x + y ⇒ y = 3 – 2x
If we have x = 0, then y = 3 – 2(0) = 3
x = 1,then y = 3 – 2(1) = 3 – 2 = 1
x = 2,then y = 3 – 2(2) = 3 – 4 = -1
∴ We get the following table:
Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.
Thus, the line CD is the required graph of 3 = 2x + y.
Question 2
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
(2, 14) means x = 2 and y = 14
Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0
There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.
Question 3
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
The equation of the given line is 3y = ax + 7
∵ (3, 4) lies on the given line.
∴ It must satisfy the equation 3y = ax + 7
We have, (3, 4) ⇒ x = 3 and y = 4.
Putting these values in given equation, we get
3 x 4 = a x 3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7 = 5 ⇒ a =
Thus, the required value of a is
Question 4
The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.
Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table:
Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.
Thus, the line PQ is the required graph of the linear equation y = 5x + 3.
Question 5
From the choices given below, choose the equation whose graphs are given ¡n Fig. (1) and Fig. (2).
For Fig. (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
For Fig. (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6
Solution:
For Fig. (1), the correct linear equation is x + y = 0
[As (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]
For Fig.(2), the correct linear equation is y = -x + 2
[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)
⇒ 2 = -(0) + 2 ⇒ 2 = 2]
Question 6
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 unit
Solution:
Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force x Distance
⇒ y = 5 x x ⇒ y = 5x
For drawing the graph, we have y = 5x
When x = 0, then y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = -1, then y = 5(-1) = -5
∴ We get the following table:
Plotting the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.
From the graph, we get
(i) Distance travelled =2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.
(ii) Distance travelled = 0 unit i.e., x = 0
∴ If x = 0 ⇒ y = 5(0) – 0
⇒ Work done = 0 unit.
Question 7
Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs.xand Rs.y.) Draw the graph of the same.
Solution:
Let the contribution of Yamini = Rs. x
and the contribution of Fatima Rs. y
∴ We have x + y = 100 ⇒ y = 100 – x
Now, when x = 0, y = 100 – 0 = 100
x = 50, y = 100 – 50 = 50
x = 100, y = 100 – 100 = 0
∴ We get the following table:
Thus, the line PQ is the required graph of the linear equation x + y = 100.
Question 8
In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here Is a
linear equation that converts Fahrenheit to Celsius:
F = (
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature Is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what Is the temperature In Fahrenheit and If the temperature is 0°F, what Is the temperature In Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.
Solution:
(i) We have
F = (
When C = 0 , F = (
When C = 15, F = (
When C = -10, F =
We have the following table:
(ii) From the graph, we have 86°F corresponds to 30°C.
(iii) From the graph, we have 95°F corresponds 35°C.
(iv) We have, C = 0
From (1), we get
F = (
Also, F = 0
From (1), we get
0 = (
(V) When F = C (numerically)
From (1), we get
F =
⇒ 
∴ Temperature is – 40° both in F and C.