In section 4.4 of Chapter 4 of NCERT Maths textbook for Class 9, under the headline “Graph of a Linear Equation in Two Variables”, it is concluded that every point on the line satisfies the equation of the line and every solution of the equation is a point on the line. The chapter deals with the concept of Linear Equations more in-depth. The exercises from page number 74 of the textbook are explained in clear and easy-to-understand steps by our subject matter experts in the NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equation in Two Variables Exercise 4.3. The solutions, based on the NCERT syllabus, are prepared as per the guidelines.
Each concept is explained properly and clearly, with easy-to-understand examples and maths problems. We have devised the most accurate NCERT solutions for all the questions under each exercise after thorough research. It helps students to prepare well for the board exams.
NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.3
Access answers of Maths NCERT Class 9 Chapter 4 – Linear Equations in Two Variables Exercise 4.3
Access other exercise solutions of Class 9 Maths Chapter 4 – Linear Equations in Two Variables
NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.3
1. Draw the graph of each of the following linear equations in two variables:
(i) x+y = 4
Solution:
To draw a graph of linear equations in two variables, let us find the points to plot.
To find the points, we have to find the values that x and y can have, satisfying the equation.
Here,
x+y = 4
Substituting the values for x,
When x = 0,
x+y = 4
0+y = 4
y = 4
When x = 4,
x+y = 4
4+y = 4
y = 4–4
y = 0
| x | y |
| 0 | 4 |
| 4 | 0 |
The points to be plotted are (0, 4) and (4,0).
(ii) x–y = 2
Solution:
To draw a graph of linear equations in two variables, let us find the points to plot.
To find the points, we have to find the values that x and y can have, satisfying the equation.
Here,
x–y = 2
Substituting the values for x,
When x = 0,
x–y = 2
0 – y = 2
y = – 2
When x = 2,
x–y = 2
2–y = 2
– y = 2–2
–y = 0
y = 0
| x | y |
| 0 | – 2 |
| 2 | 0 |
The points to be plotted are (0, – 2) and (2, 0).
(iii) y=3x
Solution:
To draw a graph of linear equations in two variables, let us find the points to plot.
To find the points, we have to find the values that x and y can have, satisfying the equation.
Here,
y = 3x
Substituting the values for x,
When x = 0,
y = 3x
y = 3×0
y = 0
When x = 1,
y = 3x
y = 3×1
y = 3
| x | y |
| 0 | 0 |
| 1 | 3 |
The points to be plotted are (0, 0) and (1, 3).
(iv) 3 = 2x+y
Solution:
To draw a graph of linear equations in two variables, let us find out the points to plot.
To find out the points, we have to find the values which x and y can have, satisfying the equation.
Here,
3 = 2x+y
Substituting the values for x,
When x = 0,
3 = 2x+y
3 = 2×0+y
3 = 0+y
y = 3
When x = 1,
3= 2x+y
3 = 2×1+y
3 = 2+y
y = 3–2
y = 1
| x | y |
| 0 | 3 |
| 1 | 1 |
The points to be plotted are (0, 3) and (1, 1).
2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
We know that an infinite number of lines pass through a point.
The equation of 2 lines passing through (2,14) should be in such a way that it satisfies the point.
Let the equation be 7x = y
7x–y = 0
When x = 2 and y = 14
(7×2)-14 = 0
14–14 = 0
0 = 0
L.H.S. = R.H.S.
Let another equation be 4x = y-6
4x-y+6 = 0
When x = 2 and y = 14
(4×2–14+6 = 0
8–14+6 = 0
0 = 0
L.H.S. = R.H.S.
Since both the equations satisfy the point (2,14), then we can say that the equations of two lines passing through (2, 14) are 7x = y and 4x = y-6
We know that an infinite number of lines pass through one specific point. Since there is only one point (2,14), here, there can be infinite lines that pass through the point.
3. If the point (3, 4) lies on the graph of the equation 3y = ax+7, find the value of a.
Solution:
The given equation is
3y = ax+7
According to the question, x = 3 and y = 4
Now, substituting the values of x and y in the equation 3y = ax+7,
We get,
(3×4) = (a×3)+7
⟹ 12 = 3a+7
⟹ 3a = 12–7
⟹ 3a = 5
⟹ a = 5/3
The value of a, if the point (3,4) lies on the graph of the equation 3y = ax+7 is 5/3.
4. The taxi fare in a city is as follows: For the first kilometre, the fare is ₹8, and for the subsequent distance, it is ₹5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
Solution:
Given,
Total distance covered = x
Total fare = y
Fare for the first kilometre = 8 per km
Fare after the first 1km = 5 per km
If x is the total distance, then the distance after one km = (x-1)km
i.e., fare after the first km = 5(x-1)
According to the question,
The total fare = Fare of first km+ fare after the first km
y = 8+5(x-1)
y = 8+5(x-1)
y = 8+5x – 5
y = 5x+3
Solving the equation,
When x = 0,
y = 5x+3
y = 5×0+3
y = 3
When y = 0,
y = 5x+3
o = 5x+3
5x = -3
x = -3/5
| x | y |
| 0 | 3 |
| -3/5 | 0 |
The points to be plotted are (0, 3) and (-3/5, 0)
5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
For Fig. 4. 6
(i) y = x
(ii) x+y = 0
(iii) y = 2x
(iv) 2+3y = 7x
Solution:
The points given in figure 4.6 are (0,0), (-1,1), (1,-1).
Substituting the values for x and y from these points in the equations, we get,
(i) y = x
(0,0) ⟹ 0 = 0
(-1, 1) ⟹ -1 ≠ 1 ————————— equation not satisfied
(1, -1) ⟹ 1≠ -1 ————————— equation not satisfied
(ii) x+y = 0
(0,0) ⟹ 0+0 = 0
(-1, 1) ⟹ -1+1 = 0
(1, -1) ⟹ 1+(-1) =0
(iii) y = 2x
(0,0) ⟹ 0 = 2×0
0 = 0
(-1, 1) ⟹ 1 = 2×(-1)
1≠ -2 ————————— equation not satisfied
(1, -1) ⟹ -1 = 2×1
-1 ≠ 2 ————————— equation not satisfied
(iv) 2+3y = 7x
(0,0) ⟹ 2+(30) = 7×0
2 ≠ 0 ————————— equation not satisfied
(-1, 1) ⟹ 2+(3×1) = 7×-1
5 ≠ -7 ————————— equation not satisfied
(1, -1) ⟹ 2+(3×-1) = 7×1
-1 ≠ 7 ————————— equation not satisfied
Since only equation x+y = 0 satisfies all the points, the equation whose graphs are given in Fig. 4.6 is
x+y = 0
For Fig. 4. 7
(i) y = x+2
(ii) y = x–2
(iii) y = –x+2
(iv) x+2y = 6
Solution:
The points given in figure 4.7 are (0,2), (2,0), (-1,3)
Substituting the values for x and y from these points in the equations, we get,
(i) y = x+2
(0,2) ⟹2 = 0+2
2 = 2
(2, 0) ⟹ 0= 2+2
0 ≠ 4 ————————— equation not satisfied
(-1, 3) ⟹ 3 = -1+2
3 ≠ 1 ————————— equation not satisfied
(ii) y = x–2
(0,2) ⟹ 2 = 0–2
2 ≠ -2 ————————— equation not satisfied
(2, 0) ⟹ 0 = 2–2
0= 0
(-1, 3) ⟹ 3= –1–2
3 ≠ –3 ————————— equation not satisfied
(iii) y = –x+2
(0,2) ⟹ 2 = -0+2
2 = 2
(2, 0) ⟹ 0 = -2+2
0 = 0
(-1, 3) ⟹ 3= -(-1)+2
3 = 3
(iv) x+2y = 6
(0,2) ⟹ 0+(2×2) = 6
4 ≠ 6 ————————— equation not satisfied
(2, 0) ⟹ 2+(2×0) = 6
2 ≠ 6 ————————— equation not satisfied
(-1, 3) ⟹ -1+(2×3) = 6
5 ≠ 6 ————————— equation not satisfied
Since only equation y = –x+2 satisfies all the points, the equation whose graphs are given in Fig. 4.7 is
y = –x+2
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:
Let the distance travelled by the body be x and the force applied on the body be y.
It is given that,
The work done by a body is directly proportional to the distance travelled by the body.
According to the question,
y ∝ x
y = 5x (5 is a constant of proportionality)
Solving the equation,
(i) when x = 2 units,
then y = 5×2 = 10 units
(2, 10)
(ii) when x = 0 units,
then y = 5×0 = 0 units
(0, 0)
The points to be plotted are (2, 10) and (0, 0).
7. Yamini and Fatima, two students of Class IX of a school, together contributed ₹ 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 ₹ x and ₹ y.) Draw the graph of the same.
Solution:
Let Yamini’s donation be ₹x and Fatima’s donation be ₹y
According to the question;
x+y = 100
We know that,
when x = 0 , y = 100
when x = 50, y = 50
when x = 100, y = 0
The points to be plotted are (0,100), (50,50), and (100,0).
8. In countries like USA and Canada, the 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:
(i) Draw the graph of the linear equation above using Celsius for the x-axis and Fahrenheit for the 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) According to the question,
F = (9/5)C + 32
Solving the equation,
We get,
When C = 0, F = 32
When C = -10 , F = 14
The points to be plotted are (0, 32), (-10, 14).
(ii) When C = 30,
F = (9/5)C +32
F = (9×30)/5+32
= (9×6)+32
= 54+32
= 86oF
(iii) When F = 95,
95 = (9/5)C +32
(9/5)C = 95-32
(9/5)C =63
C = (63×5)/9
=35oC
(iv) When C = 0,
F = (9/5)C +32
F = (9×0)/5 +32
=0+32
=32oF
When F = 0,
0 = (9/5)C+32
(9/5)C = 0-32
(9/5)C = -32
C = (-32×5)/9
=-17.7777
=-17.8oC
(v) When F = C,
C = (9/5)C+32
C – (9/5)C = 32
(5-9)C/5 =32
(-4/5)C = 32
(-4/5)C = (-32×5)/4
= – 40oC
Hence, -40o is the temperature which is numerically the same in both Fahrenheit and Celsius.
In this exercise, there are 8 questions, where question 1 has the main question with 3 sub-questions under it, where students could be asked to draw a graph of the given linear equations. Additionally, question numbers 2 and 3 are short answer questions, whereas question numbers 4, 5, 6 and 7 are long answer questions. Finally, question number 8 from the exercise has 1 main question and 8 sub-questions under it.
We have provided detailed and chapter-wise solutions for the questions under each exercise. Students can access the detailed exercise solutions of NCERT Solutions For Class 9 Maths from the PDF link given above. It will help them to get practice solving different types of questions: The Solutions also help to
- Get an overall idea about the topic
- Clear doubts about the Linear Equation Concepts
- Learn the formula related to equations relevant to this concept
- Gain practice from solving questions
I love this aap