In This Post we are providing Chapter 4 Linear Equation In Two Variables NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Linear Equation In Two Variables Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Linear Equation In Two Variables NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
Exercise 4.1
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).
Answer
Let the cost of pen be y and the cost of notebook be x.
A/q,
Cost of a notebook = twice the pen = 2y.
∴2y = x
⇒ x – 2y = 0
This is a linear equation in two variables to represent this statement.
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 = 9.35 (ii) x – y/5 – 10 = 0 (iii) -2x + 3y = 6 (iv) x = 3y
(v) 2x = -5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x
Answer
(i) 2x + 3y = 9.35
⇒ 2x + 3y – 9.35 = 0
On comparing this equation with ax + by + c = 0, we get
a = 2x, b = 3 and c = -9.35
(ii) x – y/5 – 10 = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -1/5 and c = -10
(iii) -2x + 3y = 6
⇒ -2x + 3y – 6 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 3 and c = -6
(iv) x = 3y
⇒ x – 3y = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -3 and c = 0
(v) 2x = -5y
⇒ 2x + 5y = 0
On comparing this equation with ax + by + c = 0, we get
a = 2, b = 5 and c = 0
(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 3, b = 0 and c = 2
(vii) y – 2 = 0
⇒ 0x + y – 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 0, b = 1 and c = -2
(viii) 5 = 2x
⇒ -2x + 0y + 5 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 0 and c = 5
Exercise 4.2
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
Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.
2. Write four solutions for each of the following equations:
(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y
Answer
(i) 2x + y = 7
⇒ y = 7 – 2x
→ Put x = 0,
y = 7 – 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 – 2 × 1 ⇒ y = 5
(1, 5) is the solution.
→ Now, put x = 2
y = 7 – 2 × 2 ⇒ y = 3
(2, 3) is the solution.
→ Now, put x = -1
y = 7 – 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).
(ii) πx + y = 9
⇒ y = 9 – πx
→ Put x = 0,
y = 9 – π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 – π×1 ⇒ y = 9-π
(1, 9-π) is the solution.
→ Now, put x = 2
y = 9 – π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution.
→ Now, put x = -1
y = 9 – π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).
(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution.
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution.
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).
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)
Answer
(i) Put x = 0 and y = 2 in the equation x – 2y = 4.
0 – 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.
(ii) Put x = 2 and y = 0 in the equation x – 2y = 4.
2 – 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.
(iii) Put x = 4 and y = 0 in the equation x – 2y = 4.
4 – 2×0 = 4
⇒ 4 = 4
∴ (4, 0) is a solution of the given equation.
(iv) Put x = √2 and y = 4√2 in the equation x – 2y = 4.
√2 – 2×4√2 = 4 ⇒ √2 – 8√2 = 4 ⇒ √2(1 – 8) = 4
⇒ -7√2 ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.
(v) Put x = 1 and y = 1 in the equation x – 2y = 4.
1 – 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Answer
Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7
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
Answer
(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0
| x | 0 | 4 |
| y | 4 | 0 |
(ii) x – y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0
| x | 0 | 2 |
| y | -2 | 0 |
(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3
| x | 0 | 1 |
| y | 0 | 3 |
(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1
| x | 0 | 1 |
| y | 3 | 1 |
2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Answer
Here, x = 2 and y =14.
Thus, x + y = 1
also, y = 7x ⇒ y – 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y – 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Answer
The point (3, 4) lies on the graph of the equation.
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7
⇒ a = 5/3
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.
Answer
Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3
| x | 0 | -3/5 |
| y | 3 | 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 For Fig. 4.7
(i) y = x (i) y = x + 2
(ii) x + y = 0 (ii) y = x – 2
(iii) y = 2x (iii) y = –x + 2
(iv) 2 + 3y = 7x (iv) x + 2y = 6
Answer
In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.
In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.
Page No: 75
(i) 2 units (ii) 0 unit
Answer
Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit
| x | 2 | 0 |
| y | 10 | 0 |
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 x and Rs y.) Draw the graph of the same.
Answer
Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0
| x | 0 | 50 | 100 |
| y | 100 | 50 | 0 |
(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.
| C | 0 | -10 |
| F | 32 | 14 |
(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30 + 32
⇒ F = 54 + 32
⇒ F = 86
(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C + 32
⇒ (9/5)C = 95 – 32
⇒ C = 63 × 5/9
⇒ C = 35
(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9
Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0 + 32
⇒ F = 32
(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F – 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.
Exercise 4.4
1. Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Answer
(i) in one variable, it is represented as
y = 3
2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Answer
(i) in one variable, it is represented as
x = -9/2
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