NCERT Important questions & Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables | EduGrown

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

You can find Chapter 2 Linear Equations in Two Variables Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.1

1. 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 = 9.35     (ii) 5 = 2x

Solutions

(i) We have 2x + 3y = 9.35
or (2)x + (3)y + (−9.35 ) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –9.35

(ii) We have 5 = 2x ⇒ 5 – 2x = 0
or -2x + 0y + 5 = 0
or (-2)x + (0)y + (5) = 0
Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.2

1.Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9

Solutions:-

(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
∴ Solution is (3, 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))

2.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.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.3

1.Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2

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

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.

3.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.

4.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 = (9/5 )C + 32
(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 = (9/5 )C + 32
When C = 0 , F = (9/5 ) x 0 + 32 = 32
When C = 15, F = (9/5 )(-15) + 32= -27 + 32 = 5
When C = -10, F = 9/5 (-10)+32 = -18 + 32 = 14
We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(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 = (9/5)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (9/5)C + 32 ⇒ −32×59/9 = C ⇒ C = -17.8
(V) When F = C (numerically)
From (1), we get
F = 9/5F + 32 ⇒ F – 9/5F = 32
⇒ −4/5F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.4

1.Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) y = 3
∵ y = 3 is an equation in one variable, i.e., y only.
∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3
We can write y = 3 in two variables as 0.x + y = 3
Now, when x = 1, y = 3
x = 2, y = 3
x = -1, y = 3
∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get aline AB as solution of 0. x + y = 3,
i.e. y = 3.

NCERT Quick revision Notes For Chapter-4 Linear Equation in Two Variables

NCERT Solutions For Chapter-4 Linear Equation in Two Variables

NCERT MCQs For Chapter-4 Linear Equation in Two Variables