PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

 PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables

Ex 4.1

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 and that of a pen to be ₹ y.)
Answer:
Let the cost of a notebook be ₹ x and the cost of a pen be ₹ y.
Then, according to the given data, the cost of a notebook is twice the cost of a pen.
∴ x = 2y
Thus, x = 2y, i.e., x – 2y = 0 is the required linear equation in two variables.

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:

(iii) – 2x + 3y = 6
Answer:
– 2x + 3y = 6
∴ – 2x + 3y – 6 = 0
Here, a = 1, b = – 3 and c = – 6.

(iv) x = 3y
Answer:
x = 3y
∴ x – 3y + 0 = 0
Here, a = 1, b = – 3 and c = 0.

(v) 2x = – 5y
Answer:
2x = – 5y
∴ 2x + 5y + 0 = 0
Here, a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
Answer:
3x + 2 = 0
∴ 3x + oy – 2 = 0
Here, a = 3, b = 1 and c = – 2

(vii) y – 2 = 0
Answer:
y – 2 = 0
∴ 0x + y – 2= 0
Here a = 0, b = 1 and c = – 2

(viii) 5 = 2x
Answer:
5 = 2x
∴ 5 – 2x = 0
∴ – 2x + 0y + 5 = 0
Here a = – 2, b = 0, c = 5

Ex 4.2

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.
Answer:
Option (iii) is true. Since y = 3x + 5 is a linear equation in two variables, it has infinitely many solutions. e.g., (1, 8), (2, 11), (3, 14), (4, 17), (0, 5), (- 1, 2) are all solutions of the given equation y = 3x + 5.

Question 2.
Write four solutions for each of the following equations:
(i) 2x + y = 7
Answer:
2x + y = 7
∴ y = 7 – 2x
Taking x = 0, 1, 2, 3, we get the values of y as 7, 5, 3 and 1 respectively. Thus, (0, 7), (1, 5), (2, 3) and (3, 1) are four solutions of the given equation 2x + y = 7. We can give other answers as well because the given linear equation in two variables has infinitely many solutions.

(ii) πx + y = 9
Answer:
πx + y = 9
∴ y = 9 – πx
For x = 0, y = 9.
For x = 1, y = 9 – π.
For x = – 1, y = 9 + π.
For x = , y = 8.
Thus, (0, 9), (1, 9 – π), (- 1, 9 + π) and (1/π, 8) are four of the infinitely many solutions of the given equation πx + y = 9.

(iii) x = 4y
Answer:
x = 4y
For y = 0, x = 0.
For y = 1, x = 4.
For y = – 1, x = – 4.
For y = 2, x = 8.
Thus, (0, 0), (4, 1), (- 4, – 1) and (8, 2) are four of the infinitely many solutions of the given equation x = 4y.

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
Answer:
Substituting x = 0 and y = 2, we get x – 2y = 0 – 2(2) = – 4, which is not equal to 4. Hence, (0, 2) is not a solution of x – 2y = 4.

(ii) (2, 0)
Answer:
Substituting x = 2 and y = 0, we get x – 2y = 2 – 2 (0) = 2, which is not equal to 4. Hence, (2, 0) is not a solution of x – 2y = 4.

(iii) (4, 0)
Answer:
Substituting x = 4 and y = 0, we get x – 2y = 4 – 2 (0) = 4. Hence, (4, 0) is a solution of x – 2y = 4.

(iv) (√2, 4√2)
Answer:
Substituting x = √2 and y = 4√2, we get x – 2y = √2 – 2(4√2) = – 7√2, which is not equal to 4. Hence, (√2, 4√2) is not a solution of x – 2y = 4.

(v) (1, 1)
Answer:
Substituting x = 1 and y = 1, we get x – 2y = 1 -2(1) = – 1, whIch is not equal to 4. Hence, (1, 1) is not a solution of x – 2y = 4.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Answer:
x = 2, y = 1 is a solution of equation 2x + 3y = k. Hence, x = 2 and y = 1 satisfy the equation.
∴ 2(2) + 3(1) = k
∴ 4 + 3 = k
∴ 7 = k
∴ k = 7

Ex 4.3

Question 1.
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
Answer:
To draw the graph of x + y = 4, we need at least two solutions of x + y = 4. And to be on the safer side, we get three solutions of x + y = 4.
For x = 0, we get 0 + y = 4, i.e., y = 4.
For x = 2, we get 2 + y = 4. i.e., y = 2.
For x = 4. we get 4 + y = 4, i.e., y = 0.
We can represent these solutions In the tabular form as below:

Then, we plot these points on the Cartesian plane and draw the line passing through them.
This line is the graph of x + y = 4.

(ii) x – y = 2
Answer:
x – y = 2
To draw the graph of x – y = 2, we
find three solutions of x – y = 2. For convenience, we express the equation in y-form as y = x – 2.
For x = 0, y = – 2.
For x = 2, y = 0.
For x = 4, y = 2.
We represent these solutions in the tabular form as below:

Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of x – y = 2.

(iii) y = 3x
Answer:
y = 3x
To draw the graph of y = 3x, we find three solutions of y = 3x.
For x = 0, y = 3 × 0 = 0.
For x = 1, y = 3 × 1 = 3.
For x = 2, y = 3 × 2 = 6.
We represent these solutions in the tabular form as below:

Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of y = 3x.

(iv) 3 = 2x + y
Answer:
To draw the graph of 3 = 2x + y. we find three solutions of 3 = 2x + y by expressing it as y = 3 – 2x.
For x = 0, y = 3 – 2 × 0 = 3.
For x = 1, y = 3 – 2 × 1 = 1.
For x = 2, y = 3 – 2 × 2 = – 1.
We represent these solutions in the tabular form as below:

Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the 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?
Answer:
Equations y = 7x and x + y = 16 are two equations of lines passing through point (2, 14) as the coordinates of the point satisfy both the equations.
There are infinitely many equations which are satisfied by the coordinates of the point. Few examples of such equations are y – x = 12, y = 6x + 2, x – y = – 12, etc. This happens so because infinitely many lines pass through a point given in a plane.

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Answer:
As the point (3, 4) lies on the graph of the equation 3y = ax + 7, its coordinates, i.e., x = 3 and y = 4 must satisfy the equation.
Hence, we get
3(4) = a(3) + 7
∴ 12 = 3a + 7
∴ 12 – 7 = 3a
∴ 5 = 3a
∴ 3a = 5
∴ a = 5/3

Question 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 a x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
Answer:
Let the total distance covered be x km and the total fare be ₹ y. Now, the fare for the first km is ₹ 8 and for the remaining (x – 1) km, it is ₹ 5 per km. Hence, the total fare will turn out to be ₹ [(8 + 5 (x – 1)]. Hence, we get the equation as
8 + 5(x – 1) = y
∴ 8 + 5x – 5 = y
∴5x – y + 3 = 0
To draw the graph of this equation, we find three solutions of the equation by expressing the equation in the form y = 5x + 3.

Note: Distance travelled cannot be zero or negative. Hence, we choose only positive values of x.
For x = 1, y = 5(1) + 3 = 8.
For x = 2, y = 5(2) + 3 = 13.
For x = 3, y = 5(3) + 3 = 18.
We represent these solutions in the tabular form as below:

Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of the equation 5x – y + 3 = o derived above.

Question 5.
From the choices given below, choose the equation whose graphs are given in figure (1) and figure (2):
For figure (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

Answer:
For figure (1): The graph of equation (ii) x + y = 0 is given in figure (1) as all the three points represented on the line, i.e., (- 1, 1), (0, 0) and (1, – 1) satisfy equation x + y = 0. For other equations, (1) y = x and (iii) y = 2x are satisfied by the point (0, 0), but not by the other two points. Equation (iv) 2 + 3y = 7x is not satisfied by any point.

For figure (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = – x + 2
(iv) x + 2y = 6

Answer:
For figure (2): The graph of equation (iii) y = – x + 2 is given in figure (2) as all the three points represented on the line, i.e., (- 1, 3), (0, 2) and (2, 0) satisfy equation y = – x + 2. Equation (i) y = x + 2 is satisfied by only one point (0, 2). Equation (ii) y = x – 2 is satisfied by only one point (2, 0). Equation (iv) x + 2y = 6 is not satisfied by any point.

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.
Answer:
We know well that
Work done = Force × Distance travelled.
Let the work done be y, the distance travelled be x and here the constant force applied is 5 units.
Then, the relation reduces to y = 5x which is a linear equation in two variables.
To draw the graph of y = 5x. we find three solutions of the equation and represent them in the tabular form.
For x = 1, y = 5 × 1 = 5.
For x = 3, y = 5 × 3 = 15.
For x = 4, y = 5 × 4 = 20.

Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of the equation derived above. The coordinates of any point on the line will satisfy the derived equation.

(i) From the graph, we observe that when the distance travelled (x) is 2 units, the work done (y) is 10 units.
(ii) From the graph, we observe that when the distance travelled (x) is 0 unit, the work done (y) is 0 unit.

Question 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.
Answer:
1et the contribution of Yamini be ₹ x and the contribution of Fatima be ₹ y. Then, their total contribution is ₹ (x + y). Their total contribution is given to be ₹ 100. Hence, we get the linear equation x + y = 100.
Now, to draw the graph, we find three solutions.
For x = 0, y = 100. For x = 50, y = 50, For x = 100, y = 0.
We represent these solution in the tabular form as below:

We plot these three point in the Cartesian plane and draw the line passing through them. This line is the graph of the mathematical representation of the information given in the data.

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:

  

Ex 4.4

Question 1.
Give the geometric representation of y = 3 as an equation
(i) in one variable
Answer:
If the equation y = 3 is treated as an equation in one variable, its graphical representation is a point on the number line as shown below:

(ii) in two variables.
Answer:
If the equation y = 3 is treated as an equation in two variables, it can be written as 0x + y = 3. Here, for any value of x, the value of y remains 3. Hence, we can easily take (0, 3), (2, 3) and (4, 3) as three solutions of the equation 0x + y = 3. Then, we plot these points in the Cartesian plane and draw the line passing through them. This line is the graph of equation y = 3 as an equation in two variables. This graph is perpendicular to the y-axis and parallel to the x-axis.

Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation
(i) In one variable
Answer:
If the equation 2x + 9 = 0, i.e., x = – 9/2 is treated as an equation in one variable. its graphical representation is a point on the number line as shown below:

(ii) In two variables.
Answer:

MCQ

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
If (2, – 2) is a root of 5x – 2y = k, then k = ………………. .
A. – 40
B. 6
C. 14
D. 10
Answer:
C. 14

Question 2.
If x = 2 and y = 1 is one of the solutions of 4x + ky = 11, then k = ……………… .
A. 2
B. 3
C. 5
D. 6
Answer:
B. 3

Question 3.
If (3, – 2) is one of the solutions of kx – 3y = 21, then k = ……………………. .
A. 3
B. – 3
C. 2
D. 5
Answer:
D. 5

Question 4.
The graph of 2x – 3y = 6 passes through points ……………… .
A. (2, – 3) and (- 2, 3)
B. (2, 3) and (3, 2)
C. (0, 2) and (- 3, 0)
D. (0, – 2) and (3, 0)
Answer:
D. (0, – 2) and (3, 0)

Question 7.
For the equation F = (9/5)C + 32, F and C are numerically equal when ……………….. .
A. C = 45
B. C = – 40
C. C = 40
D. C = 32
Answer:
B. C = – 40