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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables 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 = – \(\frac{9}{2}\) 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:
If the equation 2x + 9 = 0 is treated as an equation in two variables, it can be written as 2x + 0y + 9 = 0. Here, for any value of y, the value of x remains – \(\frac{9}{2}\). Hence, we can easily take \(\left(-\frac{9}{2}, 0\right)\), \(\left(-\frac{9}{2}, 2\right)\), and \(\left(-\frac{9}{2}, 4\right)\) as three solutions of the equation 2x + 9 = 0. Then, we plot these points in the Cartesian plane and draw the line passing through them. This line is the graph of the equation 2x + 9 = 0 as an equation in two variables. This graph is perpendicular to the x-axis and parallel to the y-axis.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) √2x² + 7x + 5√2 = 0
(iv) 2 x² – x + \(\frac{1}{8}\) = 0
(v) 100x² – 20x + 1 = 0
Solution:
(i) Given quadratic
x² – 3x – 10 = 0
Or x² – 5x + 2x – 10 = 0
S = -3, p = -10
Or x (x – 5) + 2 (x – 5) = 0
Or (x – 5) (x + 2) = 0
Either x – 5 = 0 Or x + 2 = 0
x = 5 Or x = -2
Hence, 5 and -2 are roots of given Quadratic Equation.

(ii) Given quadratic equation
2x² + x – 6 = 0 =1
0r 2x² + 4x – 3x – 6 = 0
S = 1 P = -6 × 2 = -12
Or 2x (x + 2) -3 (x + 2) = 0
Or (x + 2) (2x – 3) = 0
Either x + 2 = 0 Or 2x – 3 = 0
x = -2 Or x = –\(\frac{3}{2}\)
Hence, – 2 and \(\frac{3}{2}\) are roots of given quadratic equation.

(iii) Given Quadratic Equation,
√2x² + 7x + 5√2 = 0
Or √2x² + 2x + 5x + 5√2 = 0
S = 7, P = √2 × 5√2 = 10
Or √2x (x + √2) + 5 (x + √2) = 0
Or (x + √2) (√2x + 5) = 0
Either x + √2 = 0 Or √2x + 5 = 0
x = -√2 Or x = –\(\frac{-5}{\sqrt{2}}\)
Hence, -√2 and \(\frac{-5}{\sqrt{2}}\) are roots of given quadratic equation.

(iv) Given quadratic equation
2x² – x + \(\frac{1}{8}\) = 0
Or \(\frac{16 x^{2}-8 x+1}{8}\) = 0
Or 16x² – 8x + 1 = 0
S = -8, P = 16 × 1 = 16
Or 16x² – 8x + 1 = 0
Or 16x² – 4x – 4x + 1 = 0
Or 4x(4x – 1) -1(4x – 1) = 0
Or (4x – 1) (4x – 1) = 0
Either 4x – 1 = 0
Or 4x – 1 = 0
x = \(\frac{1}{4}\) Or x = \(\frac{1}{4}\)
Hence, \(\frac{1}{4}\) and \(\frac{1}{4}\) are roots of given quadratic equation.

(v) Given quadratic equation,
100x² – 20x + 1 = 0
Or 100x² – 10x – 10x + 1 = 0
S = -20, P = 100 × 1 = 100
Or 10x(10x – 1) – 1 (10x – 1) = 0
Or (10x – 1)(10x – 1) = 0
Either 10x – 1 = 0 Or 10x – 1 = 0
x = \(\frac{1}{10}\) Or x = \(\frac{1}{10}\)
Hence, \(\frac{1}{10}\) and \(\frac{1}{10}\) are roots of given quadratic equation.

Question 2.
Solve the problems given in Example I. Statements of these problems are given below:
(i) John and Jivanti together have 45 marbles. Both of them lost S marbles each, and the product of the number of marbles they now have is 124. We would lfke to find out how many marbles they had to start with.

(ii) A cottage Industry produces a certain number of toys in a day. The cost of production of each toy (In rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solution:
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x
The number of marbles Íeft withJohn, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= -x² + 45x – 200
According to question,
-x² + 45x – 200 = 124
Or -x² + 45x – 324 = 0
Or x² – 45x + 324 =0
Or x² – 36x – 9x + 324 = 0
S = -45, P = 324
Or x(x – 36) – 9(x – 36) = 0
Or (x – 36)(x – 9) = 0
Either x – 36 = 0, Or x – 9 = 0
x = 36 Or x = 9
∴ x = 36, 9
Hence, number of marbles they had to start with were 36 and 9 or 9 and 36.

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
According to question.
x(55 – x) = 750
Or 55x – x² = 750
Or -x² + 55x – 750 = 0
Or x² – 55x – 750 = 0
Or x² – 30x – 25x + 750=0
S = -33, P = 750
Or x(x – 30) – 25(x – 30) = 0
Or (x – 30)(x – 25) = 0
Either x – 30 = 0 Or x – 25 = 0
x = 30 Or x = 25
∴ x = 30, 25
Hence, number of toys produced on that day were 30 and 25 or 25 and 30.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one number = x
2nd number = 27 – x
Their product = x (27 – x) = 27x – x²
According to question,
27x – x² = 182
Or – x² + 27x – 182 = 0
Or x² – 27x + 182 = 0
S = -27, P = 182
Or x² – 13x – 14x + 182 = 0
Or x(x – 13) – 14(x – 13) = 0
Or (x – 13) (x – 14) = 0
Either x – 13 = 0 Or x – 14 = 0
x = 13 Or x = 14
x = 13, 14
Hence, two numbers are 13 and 14 Or 14 and 13.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let one positive integer = x
2nd positive integer = x + 1
According to question,
(x)² + (x + 1)² = 365
Or x² + x² + 1 + 2x = 365
Or 2x² + 2x + 365 = 0
Or 2x² + 2x – 364 = 0
Or x² + x – 182 = 0
Or x² + 14x – 13x – 182 = 0
S = 1, P = -182
Or x(x + 14) – 13(x + 14) = 0
(x + 14)(x— 13) = O
Either x + 14 = 0
Or x = -14
Or
x – 13 = 0
x = 13
∵ We have positive integers.
So, we reject x = – 14.
∴ x = 13
∴ One positive integer = 13
and 2nd positive integer = 13 + 1 = 14
Hence, required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. 1f the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base of right triangle = x cm
Altitude of right triangle = (x – 7) cm
and hypotenuse of right triangle = 13 cm (Given)
According to Pythagoras Theorem,
(Base)² + (Altitude)² = (Hypotenuse)²
(x)² ÷ (x – 7)² = (13)²
Or x² + x² + 49 – 14x = 169
Or 2x² – 14x + 49 – 169 = 0
Or 2x² – 14x – 120 = 0
Or 2[x² – 7x – 60] = 0
Or x² – 7x – 60 = 0
Or x² – 12x + 5x – 60 = 0
S = – 7 P = – 60
Or x(x – 12) + 5(x – 12) = 0
Or (x – 12) (x + 5) = 0
Either x – 12 = 0 Or x + 5 = 0
x = 12 Or x= – 5
∵ Length of any triangle cannot be negative.
So, we reject x = – 5
∴ x = 12
Hence, base of right triangle = 12 cm
Altitude of right triangle = (12 – 7) cm = 5 cm.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90, find the number of articles produced and the cost of each article.
Solution:
Let, number of pottery articles produced by industry in one day = x
Cost of production of each article = ₹ (2x + 3)
∴ Total cost of production in panicular day = ₹ [x(2x + 3)] = ₹ (2x² + 3x)
According to question,
2x² + 3x = 90
2x² + 3x – 90 = 0
S = 3, P = 2 × -90 = -180
Or 2x² – 12x + 15x – 90 = 0
Or 2x (x – 6) + 15 (x – 6) = 0
Or (x – 6) (2x + 15) = 0
Either x – 6 = 0 Or 2x + 15 = 0
x = 6 Or x = \(\frac{-15}{2}\)
∵ number of articles cannot be negative
So, we reject x = 2
∴ x = 6
Hence, number of articles produced on certain day = 6
and cost of production of each article = ₹ [2 × 6 + 3] = ₹ 15.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables 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 = \(\frac{5}{3}\)

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:
F = \(\left(\frac{9}{5}\right)\)C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
Answer:
F = \(\left(\frac{9}{5}\right)\)C + 32
For C = – 15,
F = \(\frac{9}{5}\)(- 15) + 32 = – 27 + 32 = 5
For C = 10,
F = \(\frac{9}{5}\)(10) + 32 = 18 + 32 = 50
For C = 60,
F = \(\frac{9}{5}\)(60) + 32 = 108 + 32 = 140
Hence, three solutions of F = \(\left(\frac{9}{5}\right)\)c + 32 can be given in the tabular form as below:

To draw the graph of F = \(\left(\frac{9}{5}\right)\)c + 32.
we take Celsius (°C) on the x-axis and Fahrenheit (°F) on the y-axis with scale 1 cm = 10 units on both the axes.
Then, we plot the points (- 15, 5), (10, 50) and (60, 140) and draw the line passing through them which is the graph of the equation F = \(\left(\frac{9}{5}\right)\)C + 32.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
Answer:
If the temperature is 30 °c means the x-coordinate of the point is 30. Now, the point on the graph of F = \(\left(\frac{9}{5}\right)\)C + 32 having x-coordinate 30 has y-coordinate 86.
From the equation for C = 30, we get
F = \(\left(\frac{9}{5}\right)\)3o + 32 = 54 + 32 = 86.
Hence, if the temperature is 30°C, in the Fahrenheit scale it is 86°F.

(iii) If the temperature is 95 °E what is the temperature in Celsius?
Answer:
If the temperature is 95°F means the y-coordinate of the point is 95. Now, the point on the graph of F = \(\left(\frac{9}{5}\right)\)C + 32 having y-coordinate 95 has x-coordinate 35.
From the equation for F = 95, we get
95 = \(\left(\frac{9}{5}\right)\)C + 32
∴ 63 = \(\left(\frac{9}{5}\right)\)C
∴ C = 63 × \(\frac{5}{9}\)
∴ C = 35

(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?
Answer:
As explained in (ii) and (iii) above, if the temperature is 0 °C, in the Fahrenheit scale it is 32°F. And if the temperature is 0 °F, in the Celsius scale it is – 18 °C approximately as observed from the graph. From the equation, for C = 0, we get
F = \(\left(\frac{9}{5}\right)\)0 + 32 = 0 + 32 = 32, and for
F = 0, we get
0 = \(\left(\frac{9}{5}\right)\)C + 32
∴ – 32 = \(\left(\frac{9}{5}\right)\)C
∴ C = – 32 × \(\frac{5}{9}\)
∴ C = – \(\frac{160}{9}\)
∴ C = – 17 \(\frac{7}{9}\)

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer:
Yes, there is a temperature which is numerically the same in both Fahrenheit and Celsius.
As we see, the graph of F = \(\left(\frac{9}{5}\right)\)C + 32 passes through the point (- 40, – 40), i.e., – 40°C = – 40°F.
From the equation for F = C, we get
C = \(\left(\frac{9}{5}\right)\)C + 32
∴ – 32 = \(\frac{9}{5}\)C – C
∴ – 32 = \(\frac{4}{5}\)C
∴ C = – 32 × \(\frac{5}{4}\)
∴ C = – 40
Hence, – 40 is the temperature which is numerically the same in both Fahrenheit and Celsius.