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Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 1.
Solve the following pair of linear equations by the substitution method:
(i) x + y = 14
x – y = 4

(ii) s – t = 3
\(\frac{s}{3}+\frac{t}{2}\) = 6

(iii) 3x – y = 3
9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

(v) √2x + √3y = 0
√3x – √8y = 0

(vi) \(\frac{3 x}{2}-\frac{5 y}{3}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)

Solution:
(i) Given pair of linear equati3ns
x + y = 14 …………(1)
and x – y = 4
From (2) x = 4 ………….(3)
Substitute this value of x in Equation we get: .
4 + y + y = 14
2y = 14 – 4
2y = 10
y = \(\frac{10}{2}\) = 5
Substitute this value of y in equation (3), we get:
x = 4 + 5 = 9
Hence x = 9 and y = 5

(ii) Given pair of linear equations
s – t = 3 …………….(1)
and
\(\frac{s}{3}+\frac{t}{2}\) = 6
\(\frac{2 s+3 t}{6}\) = 6
2s + 3t = 36 …………….(2)
From (1), s = t + 3 ……………….(3)
Substitute this value of s in equation (1), we get:
2(3 + t) + 3t = 36
Or 6 + 2t + 3t = 36
Or 6 + 5t = 36
Or 5t = 36 – 6
Or 5t = 30
Or t = \(\frac{30}{5}\) = 6
Substitute this value of r in equation (3), we get:
s = 3 + 6 = 9
Hence, s = 9 and t = 6

(iii) Given pair of linear equation is:
3x – y = 3 .
and 9x – 3y = 9
From (1),
3x – 3 = y
Or y = 3x – 3 ………….(3)
Substitute this value of y in equation (2), we get :
9x – 3(3x – 3) = 9
Or 9x – 9x + 9 = 9
Or 9 = 9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. Therefore we cannot obtain a specific value of y. This situation has arises because both the given equations are same. Therefore, equations (1) and (2) have infinitely many solutions.

(iv) Given pair of linear equation is:
0.2x + 0.3y = 1.3
or \(\frac{2}{10} x+\frac{3}{10} y=\frac{13}{10}\)
or 2x + 3y = 13 ……………..(1)
0.4x + 0.5y = 2.3
or \(\frac{4}{10} x+\frac{5}{10} y=\frac{23}{10}\)
Or 4x + 5y = 23 ……………(2)
From (1),
2x = 13 – 3y
x = \(\frac{13-3 y}{2}\) …………..(2)
Substitrne this value of x in (2), we get:
4[latex]\frac{13-3 y}{2}[/latex] + 5y = 23
26 – 6y + 5y = 23
-y = 23 – 26 = -3
y = 3
Substitute this value of y in (3), we get:
x = \(\frac{13-3 \times 3}{2}\)
= \(\frac{13-9}{2}=\frac{4}{2}\) = 2
Hence, x = 2 and y = 3.

(v) Given pair of linear equation is:
√2x + √3y = 0 ……………..(1)
√3x – √8y = 0 …………..(2)
From (2), √3x = √8y
or x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y …………..(3)
Substitute this value of x in (1), we get
√2(\(\frac{\sqrt{8}}{\sqrt{3}}\)y) + √3y = 0
or [\(\frac{4}{\sqrt{3}}\) + √3]y = 0
y = 0
Substitute this value of y in (3), we get:
x = \(\frac{\sqrt{8}}{\sqrt{3}}\) × 0 = 0
Hence x = 0 and y = 0

(vi) Given pair of linear equation is:
\(\frac{3 x}{2}-\frac{5 y}{3}=-2\)

or \(\frac{9 x-10 y}{6}\) = -2
or 9x – 10y = -12 …………..(1)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
\(\frac{2 x+3 y}{6}=\frac{13}{6}\)
or 2x + 3y = \(\frac{13}{6}\) × 6
or 2x + 3y = 13 …………..(2)
From (1), 9x = 10y – 12
or x = \(\frac{10 y-12}{9}\) …………….(3)
2[latex]\frac{10 y-12}{9}[/latex] + 3y = 13

or \(\frac{20 y-24}{9}\) + 3y = 13

or \(\frac{20 y-24+27 y}{9}\) = 13

or 47y – 24 = 13 × 9 = 117
47y = 117 + 24 = 141
or y = \(\frac{141}{47}\) = 3
substitute this value of y in (3), we get
x = \(\frac{10 \times 3-12}{9}=\frac{30-12}{9}\)

= \(\frac{18}{9}\) = 2
Hence, x = 2 and y = 3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
Given pair of linear equations is:
2x + 3y = 11
and 2x – 4y = -24 …………(2)
From (2),
2x = 4y – 24
2x = 2 [2y – 12]
Or x = 2y – 12 …………..(3)
Substitute this value of x in (1), we get:
2 (2y – 12) + 3y = 11
Or 4y – 24 + 3y = 11
Or 7y = 11 + 24
Or 7y = 35
y = \(\frac{35}{7}\) = 5
Substitute this value of y in (3), we get:
x = 2(5) – 12 = 10 – 12 = -2
Now, consider y = mx + 3
Substitute the value of x = -2, y = 5, we get:
5 = m(-2) + 3
Or 5 – 3 = -2m
Or 2 = – 2m
Or -2m = 2
Or m = -1
Hence, x = -2, y = 5 and m = -1

Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per kilometre? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If 3 ¡s added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(I) Let two number be x and y.
According to 1st condition,
x – y=26 ……….(1)
According to 2nd condition,
x = 3y …………(2)
Substitute this value of x in (1), we get :
3y – y = 26
Or 2y = 26
y = \(\frac{26}{2}\) = 13
Substitute this value of y in (2), we get:
x = 3 × 13 = 39
Hence, two numbers are 39, 13.

(ii) Let, required two supplementary angles are x, y and x > y
According to 1st condition,
x + y = 180 ………..(1)
According to 2nd condition,
x = y + 18 …………..(2)
Substitute this value of x in (1), we get:
y + 18 + y = 180
Or 2y = 180 – 18
or 2y =162
Or y = \(\frac{162}{2}\) = 81
Substitute this value of y in (2), we get:
x = 81 + 18 = 99
Hence, required angles are 99, 81.

(iii) Let cost of one bat = ₹ x
and cost of one ball = ₹ y
According to 1st condition,
7x + 6y = ₹ 3800 ……………(1)
According to 2nd condition,
3x + 5y = ₹ 1750 …………….(2)
From(1), 7x = 3800 – 6y
Or x =\(\frac{3800-6 y}{7}\) ………….(3)
substitute this value of x in (2), we get:
3[latex]\frac{3800-6 y}{7}[/latex] + 5y = 1750
Or \(\frac{11400-18 y+35 y}{7}\) = 1750
Or 11400 + 17y = 1750 × 7
0r 11400 + 17y = 12250
Or 17y = 12250 – 11400
Or 17y = 850
or y = \(\frac{850}{17}\) = 50
Substitute this value of y in (3), we get:
x = \(\frac{3800-6 \times 50}{7}\)
= \(\frac{3800-300}{7}=\frac{3500}{7}\)
x = 500
Hence cost of one bat = ₹ 500
and cost of one ball = ₹ 50.

(iv) Let the fixed charges for the taxi = ₹ x
and charges for travelling one km = ₹ y
According to 1st condition,
x + 10y = 105 …………..(1)
According to 2nd condition,
x + 15y = 155 ……………(2)
From (1),
x = 105 – 10y …………..(3)
Substitute the value of x in (2), we get
105 – 10y + 15y = 155
Or 5y = 155 – 105
Or 5y = 50
Or y = \(\frac{50}{5}\) = 10
Substitute the value of y in (3), we get:
x = 105 – 10 × 10
= 105 – 100 = 5
Hence, fixed charges for the taxi = ₹ 5
and charges for travelling one km = ₹ 10
Also, charges for travelling 25 km = ₹(10 × 25) + ₹ 5
= ₹[250 + 5] = ₹ 255

(v) Let numerator of given fraction = x
Denominator of given fraction = y
∴ Required fraction = \(\frac{x}{y}\)
Acccrding to 1st condition,
\(\frac{x+2}{y+2}=\frac{9}{11}\)
Or 11(x + 2) = 9(y + 2)
Or 11x + 22 = 9y + 18
Or 11x = 9y + 18 – 22
Or 11x = 9y – 4
Or x = \(\frac{9 y-4}{11}\) ……………..(1)
Acccrding to 2nd equation
\(\frac{x+3}{y+3}=\frac{5}{6}\)
Or 6 (x + 3) = 5(y + 3)
Or 6x + 18 = 5y + 15
Or 6x – 5y = 15 – 18
Or 6x – 5y = -3
Putting the value of x from (1). we get:
6[latex]\left[\frac{9 y-4}{11}\right][/latex] – 5y = -3
Or \(\frac{54 y-24}{11}\) – 5y = -3
Or \(\frac{54 y-24-55 y}{11}\) = -3
Or -y – 24 = -3 × 11
Or -y = -33 + 24
Or -y = -9
Or y = 9
Substitute the value of y in (1), we get:
x = \(\frac{9 \times 9-4}{11}=\frac{81-4}{11}\)
= \(\frac{77}{11}\) = 7
Hence, required fraction is \(\frac{7}{9}\).

(vi) Let Jacob’s present age = x years
and Jacob son’s present age = y years
Five years hence
Jacob’s age = (x + 5) years
His son’s age = (y + 5)years
According to 1st condition.
x + 5 = 3(y + 5)
Or x + 5 = 3y + 15
Or x = 3y + 15 – 5
Or x = 3y + 10 ……………(1)
Five years ago
Jacobs age = (x – 5) years
His son’s age = (y – 5) years
According to 2nd condition.
x – 5 = 7(y – 5)
Or x – 5 = 7y – 35
Or x – 7y = -35 + 5
Or x – 7y = -30
Substitute the value of x from (1), we get:
3y + 10 – 7y = -30
– 4y = – 30 – 10
-4y = -40
y = 10
Substitute tins value of y in (1), we get:
x = 3 (10) + 10
= 30 + 10 = 40
Hence. Jacob and his son’s ages are 40 years and 10 years respecùveiy.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Am and Biju Is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let Ani’s age = x years
and Biju’s age = y years
Dharam’s age = 2x years
Cathy’s age = years
According to 1st condition,
(Ani’s age) (Biju’s age) = 3
x – y = 3 ……………(1)
According to 2nd condition,
(Dharam’ age) – (Cathy’s age) = 30
2x – \(\frac{y}{2}\) = 30
or \(\frac{4 x-y}{2}\) = 30
or 4x – y = 60 ………….(2)
Now (2) – (1) gives,

Substitute this value of x in (1), we get:
19 – y = 3
or -y = 3 – 19
or -y = -16
or y = 16
Hence, Ani’s age = 19 years
Biju’s age =16 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “if you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital ? [From the Bijaganita of Bhaskara II] [Hint: x + 100 = 2(y – 100), y + 10 = 6 (x – 10)].
Solution:
Let Capital of one friend = ₹ x
and capital of 2nd friend = ₹ y
According to 1st condition
x + 100 = 2(y – 100)
or x + 100 = 2y – 200
or x – 2y = -200 – 100
or x – 2y = -300 …………..(1)
According to 2nd condition
y + 10 = 6(x – 10)
or v-f 10 = 6x – 60
or 6x – y = 10 + 60
or 6x – y = 70 ……………(2)
Multiplying (1) by 6, we get
6x – 12y = – 1800 …………….(3)
Now, (3) – (2) gives

Substitute this value of y in (2), we get:
6x – 170 = 70
or 6x = 70 + 170
or 6x = 240
or x = \(\frac{240}{6}\) = 40
Hence, amount of their capital are 40 and 170 respectively.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/b faster, It would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let speed of train =x km/hour
and time taken by train =y hour
∴ Distance covered by train = (Speed) (Time) = (xy) km
According to 1st condition,
(x + 10)(y – 2) = xy
or xy – 2x + 10y – 20 = y
or -2x + 10y – 20 = 0
or x – 5y + 10 = 0
According to 2nd condition,
(x – 10) (y + 3) = xy
or xy + 3x – 10y – 30 = xy
or 3x – 10y – 30 = 0
Multiplying (1) by 3, we get:
3x – 15y + 30 = 0
Now, (3) – (2) gives

Substitute this value of y in (1), we get:
x – 5 × 12 + 10 = 0
or x – 60 + 10 = 0
or x – 50 = 0
or x = 50
Speed of train = 50 km/hour
Time taken by train = 12 hour
Hence, distance covered by train = (50 × 12) km = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let number of students in each row = x
and number of rows = y
Total number of students in the class = xy
According to 1st condition,
(x + 3) (y – 1) = xy
or xy – x + 3y – 3 = xy
or -x + 3y – 3 = 0
or x – 3y + 3 = 0
According to 2nd condition,
(x -3) (y + 2) = xy
or xy + 2x – 3y – 6= xy
or 2x – 3y – 6 = 0
Now, (2) – (1) gives

x = 9
Substitute this value of x in (1), we get:
9 – 3y + 3 = 0
or -3y + 12 = 0
or -3y = -12
or y = \(\frac{12}{3}\) = 4
∴ Number of students in each row = 9 and number of rows = 4
Hence, total number of students in the class = 9 × 4 = 36.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A +∠B) find the three angles.
Solution:
In ∆ABC,
Given that, ∠C = 3∠B = 2(∠A + ∠B)
I II III
From II and III, we get:
3∠B =2(∠A+∠B)
or 3∠B = 2∠A + 2∠B
or 3∠B – 2∠B = 2∠A
or ∠B = 2∠A ……………..(1)
From I and II, we get:
∠C = 3∠B
or ∠C = 3(2∠A) [using (1)]
or ∠C = 6∠A …………….(2)
Sum of three angles of a triangle is 180°
∠A + ∠B + ∠C = 180°
or ∠A + 2∠A + 6∠A = 180°
or 9∠A = 180°
∠A = \(\frac{180^{\circ}}{9}\) = 20°
Hence, ∠A = 20°: ∠B =2 x 20° = 40°; ∠C = 6 x 20° = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle
formed by these lines and the y axis.
Solution:
Given pair of linear equation are 5x – y = 5 and 3x – y = 3
Consider,
5x – y = 5
or 5x = 5 + y
Putting y = 0 in (1), we get:
x = \(\frac{5+0}{5}=\frac{5}{5}\)
Putting y = – 5 in (1), we get:
x = \(\frac{5-5}{5}=\frac{0}{5}\) = 0
Putting y = 5 in (1), we get:
x = \(\frac{5+5}{5}=\frac{10}{5}\) = 2

Plotting A (1, 0); B (0, — 5); C (2, 5) on the graph, we get the equation of line 5x — y = 5
and 3x – y = 3
or 3x = 3 + y
or x = \(\frac{3+y}{3}\) ……………(2)
Putting y = 0 in (2), we get:
x = \(\frac{3+0}{3}=\frac{3}{3}\) = 1
Putting y = 3 in (2), we get:
x = \(\frac{3-3}{3}=\frac{0}{3}\) = 0
Putting y = 3 in (2), we get:
x = \(\frac{3+3}{3}=\frac{6}{3}\) = 2
Table:

Plotting A (1, 0); D (0, -3); E (2, 3) on the graph. we get the equation of line 3x – y = 3. From the graph, it is clear that given lines intersect at A (1, 0). Triangle formed by these lines and y axis are shaded in the graph i.e. ∆ABD. Coordinates of the vertices of ∆ABD are A(1, 0); B(0, -5) and D(0, -3).

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q

(ii) ax + by = c
bx + ay = 1 + c

(iii) \(\frac{x}{a}-\frac{y}{b}\) = 0
ax + by = a² + b²

(iv) (a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b²

(v) 152x – 378y = 74
-378x + 152y = -604
Solution:
(i) Given pair of linear equation are
px + qy = p – q …………(1)
and qx – py = p + q ………….(2)
Multiplying (1) by q and (2) by p, we get:

or y = -1
Substitute this value of y in (1), we get:
px + q(-1) = p – q
or px – q = p – q
or px = p – q + q
or px = p
or x = 1
Hence, x = 1 and y = -1.

(ii) Given pair of linear equation are
ax + by = c
bx + ay = 1 + c
ax + by – c = 0
bx + ay – (1 + c) = 0

From I and III, we get:
\(\frac{x}{-b-b c+a c}=\frac{1}{a^{2}-b^{2}}\)
or x = \(\frac{a c-b c-b}{a^{2}-b^{2}}\)

From II and III, we get:
\(\frac{y}{-b c+a+a c}=\frac{1}{a^{2}-b^{2}}\)
or y = \(\frac{a+a c-b c}{a^{2}-b^{2}}\)
Hence, x = \(\frac{a c-b c-b}{a^{2}-b^{2}}\) and y = \(\frac{a+a c-b c}{a^{2}-b^{2}}\).

(iii) Given pair of linear equation are
\(\frac{x}{a}-\frac{y}{b}\) = 0
or \(\frac{b x-a y}{a b}\) = 0
or bx – ay = 0
or bx—ay = O …(1)
and ax + by = a² + b²
or ax + by – (a² + b²) = 0 …………..(2)

\(\frac{x}{a\left(a^{2}+b^{2}\right)-0}=\frac{y}{0+b\left(a^{2}+b^{2}\right)}=\frac{1}{b^{2}+a^{2}}\)

or \(\frac{x}{a\left(a^{2}+b^{2}\right)}=\frac{y}{b\left(a^{2}+b^{2}\right)}=\frac{1}{a^{2}+b^{2}}\)

or \(\frac{x}{a}=\frac{y}{b}=\frac{1}{1}\)
I II III
From I and III, we get:
\(\frac{x}{a}=\frac{1}{1}\)
⇒ x = a
From II and III, we get:
\(\frac{y}{b}=\frac{1}{1}\)
⇒ y = b
Hence, x = a, y = b.

(iv) Given pair of linear equation are
(a – b)x + (a + b)y = a² – 2ab – b²
or ax – bx + ay + by = a² – 2ab – b² …………….(1)
and (a + b) (x + y) = a² + b²
or ax + bx + ay + by = a² + b²
Now, (1) – (2) gives

Substitute this value of x in (1), we get:
(a – b) (a + b) + (a + b) y = a² – 2ab – b²
or a² – b² + (a + b) y = a² – 2ab – b²
or (a + b) y = a² – 2ab – b² – a² + b²
or (a + b)y = -2ab
or y = \(\frac{-2 a b}{a+b}\)
Hence, x = a + b and y = a + b

(v) Given pair of linear equation are
152x – 378y = – 74
and -378x + 152y = -604
or 76x – 189y + 37 = 0
and -189x + 76y + 302 = 0

From I and III, we get:
\(\frac{x}{59890}=\frac{1}{29945}\)
⇒ x = \(\frac{59890}{29945}\)
⇒ x = 2

From II and III, we get:
\(\frac{y}{29945}=\frac{1}{29945}\)
⇒ y = \(\frac{29945}{29945}\)
⇒ y = 1
Hence x = 2 and y = 1.

Question 8.
ABCD is a cyclic quadrilateral (see Fig.). Find the angles of the cyclic quadrilateral.

Solution:
In cyclic quadrilateral ABCD,
∠A = (4y + 20); ∠B = 3y – 5; ∠C = 4x and ∠D = 7x + 5
Sum of opposite angles of a cyclic quadrilateral are of measure 180°.
∴ ∠A + ∠C = 180°
or 4y + 20 + (4x) = 180°
or 4x + 4y = 180° – 20
or 4x + 4y = 160
or x + y = 40
or y = 40 – x ……………..(1)
and ∠B + ∠D = 180°
or 3y – 5 + (7x + 5)= 180°
or 3y – 5 + 7x + 5 = 180°
or 7x + 3y = 180° …………….(2)
Substitute the value of y from (1) in (2). we get:
7x + 3(40 – x)= 180°
or 7x + 120 – 3x = 180°
or 4x = 180 – 120
or 4x = 60
x = \(\frac{60}{4}\) = 15
Substitute this value of x in (1 ), we get:
y = 40 – 15 = 25
∴ ∠A = 4y + 20 = 4 × 25 + 20 = 120°
∠B = 3y – 5 = 3 × 25 – 5 = 70°
∠C = 4x =4 × 15 = 60°
∠D = 7x + 5 = 7 × 15 + 5 = 110°
Hence, ∠A = 120°, ∠B = 70°; ∠C = 60° and ∠D = 110°.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitély many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0

(ii) 2x + y = 5
3x + 2y = 8

(iii) 3x – 5y = 20
6x – 10y = 40

(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Solution:
(i) Given pair of linear equation is:
x – 3y – 3 = 0
and 3x – 9y – 2 = 0

Here a₁ = 1, b₁ = -3, c₁ = -3
a₂ = 3, b₂ = -9, c₂ = -2

Now, \(\frac{a_{1}}{a_{2}}=\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}\);
\(\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given system of equations has no solution.

(ii) Given pair of linear equations
2x + y = 5
and 3x + 2y = 8
or 2x + y – 5 = 0
and 3x + 2y – 8=0
Here a₁ = 2, b₁ = 1, c₁ = -5
a₂ = 3,b₂ = 2, c₂ = 8
Now,
\(\frac{a_{1}}{a_{2}}=\frac{2}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}=\frac{-5}{-8}=\frac{5}{8}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ given system of equation have unique solution.

or \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1}\)
I          II         III
From I and III, we get:
\(\frac{x}{2}=\frac{1}{1}\)
⇒ x = 2

From I and III, we get:
\(\frac{y}{1}=\frac{1}{1}\)
⇒ y = 1
Hence, x = 2 and y = 1.

(iii) Given pair of linear equations is :
3x – 5y = 20
and 6x – 10y = 40
or 3x – 5y – 20 = 0
and 6x – 10y – 40 =0
Here a₁ = 3, b₁ = -5, c₁ = -20
a₂ = 6, b₂ = -10, c₂ = -40
Now,
\(\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}=\frac{-5}{-10}=\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}=\frac{-20}{-40}=\frac{1}{2}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Hence, given system have infinite solution.

(iv) Given pair of linear equation is:
x – 3y – 7 = 0
and 3x – 3y – 15 = 0
Here a₁ = 1, b₁ = -3, c₁ = -7
a₂ = 3, b₂ = -3, c₂ = -15
Now,
\(\frac{a_{1}}{a_{2}}=\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{-3}{-3}=1\);
\(\frac{c_{1}}{c_{2}}=\frac{-7}{-15}=\frac{7}{15}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ given system have unique solution.

From I and III, we get:
\(\frac{x}{24}=\frac{1}{6}\)
⇒ x = 4

From I and III, we get:
\(\frac{y}{-6}=\frac{1}{6}\)
⇒ y = -1
Hence, x = 4, y = -1.

Question 2.
(i) For which values of a and b does the following pair of linear eqU1tions have an infinite number of solutions?
2x + 3y = 7
(a – b)x ÷ (a + b)y = 3a + b – 2
(ii) For which value of k wifi the following pair of linear equations have no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) Given pair of linear equation are
2x + 3y = 7
and (a – b)x + (a + b)y = 3a + b – 2
or 2x + 3y – 7 = 0
and (a – b)x + (a + b)y – (3a + b – 2) = 0
Here a₁ = 2, b₁ = 3, c₁ = -7
a₂ = a – b, b₂ = a + b, c₂ = -(3a + b – 2)
∵ System of equation have an infinite number of solutions.

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
\(\frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-(3 a+b-2)}\)
Fron I and III, we get:
\(\frac{2}{a-b}=\frac{7}{3 a+b-2}\)
or 6a + 2b – 4 = 7a – 7b
or -a + 9b – 4 = 0
or a = 9b – 4 …………..(1)
From II and III. we get:
\(\frac{3}{a+b}=\frac{7}{3 a+b-2}\)
or 9a + 3b – 6 = 7a + 7b
or 2a – 4b – 6 = 0
or a – 2b – 3 = 0
Substitute the value of a from (1) in above, we get:
9b – 4 – 2b – 3 = 0
or 7b – 7 = 0
or 7b = 7
b = 1
Substitute this value of b in (1), we get
a = 9 × 1 – 4 = 9 – 4
a = 5
Hence a = 5 and b = 1

(ii) Given pair of linear equation are
3x + y = 1
and (2k – 1)x + (k – 1)y = 2k + 1
or 3x + y – 1 = 0
and(2k – 1)x + (k – 1)y – (2k + 1) = 0
Here a₁ = 3, b₁ = -1, c₁ = -1
a₂ = (2k – 1), b₂ = k – 1, c₂ = -(2k + 1)
∵ system of equations have no solution.
∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
\(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{-1}{-(2 k+1)}\)
I II III
From I and III, we get:
\(\frac{3}{2 k-1} \neq \frac{1}{(2 k+1)}\)
⇒ 6k + 3 ≠ 2k – 1
⇒ 4k ≠ -4
⇒ k ≠ –\(\frac{4}{4}\)
⇒ k ≠ -1
From II and III, we get:
\(\frac{3}{2 k-1}=\frac{1}{k-1}\)
⇒ 3k – 3 = 2k – 1
⇒ k = 2
Hence k = 2 and k ≠ -1.

Question 3.
Solve the following pair of linear equations by the substitution and cross- multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
Given pair of linear equation is:
8x + 5y = 9 ………….(1)
3x + 2y = 4 …………..(2)
Substitution Method:
From (2), 2y = 4 – 3x
y = \(\frac{4-3 x}{2}\) …………….(3)
Substitute this value of y in (1), we get:
8x + 5\(\frac{4-3 x}{2}\) = 9
or \(\frac{16 x+20-15 x}{2}\) = 9
or x + 20 = 18
or x = 18 – 20 = -2
Substitute this value of x in (3), we get:
y = \(\frac{4-3(-2)}{2}=\frac{4+6}{2}\)
= \(\frac{10}{2}\) = 5
Hence, x = -2 and y = 5.

Cross-multiplication Method:

Given pair of linear equation is:
8x + 5y – 9 = o
and 3x + 2y – 4= 0
Here a₁ = 8, b₁ = 5, c₁ = -9
a₂ = 3, b₂ = 2, c₂ = -4
Now,
\(\frac{a_{1}}{a_{2}}=\frac{8}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{5}{2}\);
\(\frac{c_{1}}{c_{2}}=\frac{-9}{-4}=\frac{9}{4}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ system have unique solution.

From I and III, we get:
\(\frac{x}{-2}=\frac{1}{1}\)
⇒ x = -2

From II and III, we get:
\(\frac{y}{5}=\frac{1}{1}\)
⇒ y = 5
Hence, x = -2 and y = 5.

Question 4.
Form the pair of linear equations in the following problems and find their solutions (If they exist) by any algebraic method.

(i) A part of monthly hostel charges Is fixed and the remaining depends on the number of days one has taken food
in the mess. When a student A takes food for 20 days she has to pay 1000 as hostel charges whereas a student B, who takes food for 26 days, pays 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash ‘would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time, If the cars travel in the same direction at differ it speeds they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units if Its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 units. Find the dimensions of the rectangle.

Solution:
(i) Let monthly fixed hostel charges = ₹ x
and cost of food per day = ₹ y
According to 1st condition
x + 20y = 1000 ………….(1)
According to 2nd condition
x + 26y = 1180 …………..(2)

From I and III, we get:
\(\frac{x}{2400}=\frac{1}{6}\)
⇒ x = \(\frac{2400}{6}\) = 400

From II and III, we get:
\(\frac{y}{180}=\frac{1}{6}\)
⇒ y = \(\frac{180}{6}\) = 30

Hence, monthly fixed hostel charges and cost of food per day are 400 and 30 respectively.

(ii) Let numerator of fraction = x
Denominator of fraction = y
∴ required fraction = \(\frac{x}{y}\)
According to 1st condition,
\(\frac{x-1}{y}=\frac{1}{3}\)
or 3x – 3 = y
or 3x – y – 3 = 0 …………..(1)
According to 2nd condition,
\(\frac{x}{y+8}=\frac{1}{4}\)
or 4x = y + 8
or 4x – y – 8 = 0 …………….(2)

From I and III, we get:
\(\frac{x}{5}=\frac{1}{1}\)
⇒ x = 5

From II and III, we get:
\(\frac{y}{12}=\frac{1}{1}\)
⇒ y = 12
Hence, required fraction is \(\frac{5}{12}\).

(iii) Let, number of right questions attempted by Yash = x
and Number of wrong questions attempted by Yash = y
According to 1st condition,
3x – y = 40
or 3x – y – 40 = 0 …………….(1)
According to 2nd condition,
4x – 2y = 50
or 4x – 2y – 50 = 0 ……………(2)

\(\frac{x}{50-80}=\frac{y}{-160-(-150)}=\frac{1}{-6-(-4)}\)
or \(\frac{x}{-30}=\frac{y}{-10}=\frac{1}{-2}\)

From I and III, we get:
\(\frac{x}{-30}=\frac{1}{-2}\)
x = \(\frac{-30}{-2}\)
⇒ x = 15

From II and III, we get:
\(\frac{y}{-10}=\frac{1}{-2}\)
y = \(\frac{-10}{-2}\)
⇒ y = 5

∴ Number of right questions = 15
Number of wrong questions = 5
Hence, total number of questions = [No. of right questions] + [No. of wrong questions]
=15 + 5 = 20.

(iv) Let speed of car at place A = x km/hour
and speed of car at place B = y km/hour
Distance between places A and B = 100 km
In case of 5 hours
Distance covered by car A = 5x km
[∵ Distance = Speed × Time]
Distance covered by car B = 5y km
According to I st condition,
5x – 5y = 100
or x – y = 20
or x – y – 20 = 0
In case of one hour
Distance covered by car A = x km
[∵ Distance = Speed × Time]
Distance covered by car B = y km
According to 2nd condition,
x + y = 100
or x + y – 100 = 00 ……………….(2)

From I and III, we get:
\(\frac{x}{120}=\frac{1}{2}\)
x = \(\frac{1}{2}\) × 120
⇒ x = 60

From II and III, we get:
\(\frac{y}{171}=\frac{1}{19}\)
y = \(\frac{171}{19}\)
⇒ y = 9

Hence, length and breadth of rectangle are 17 units and 9 units respectively.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) \(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1

(iii) \(\frac{4}{x}\) + 3y = 14
\(\frac{3}{x}\) – 4y = 23

(iv) \(\frac{5}{x-1}+\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}-\frac{3}{y-2}\) = 1

(v) \(\frac{7 x-2 y}{x y}\) = 5
\(\frac{8 x+7 y}{x y}\) = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) \(\frac{10}{x+y}+\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}-\frac{5}{x-y}\) = -2

(viii) \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)

Solution:
(i) Given pair of linear equations are;
\(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
putting \(\frac{1}{x}\) = u and \(\frac{1}{x}\) = v, then equations reduces to

Substitute this value of v in (1), we get:
3u + 2 (3) = 12
or 3u + 6 = 12
or 3u = 12 – 6 = 6
or u = \(\frac{6}{3}\) = 2
But \(\frac{1}{x}\) = u
or x = \(\frac{1}{u}\)
∴ x = \(\frac{1}{2}\)

and \(\frac{1}{y}\) = v
y = \(\frac{1}{v}\)
or y=—\(\frac{1}{3}\)
Hence x = \(\frac{1}{2}\) and y = \(\frac{1}{3}\)

(ii) Given pair of linear equation are
\(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1
Putting \(\frac{1}{\sqrt{x}}\) = u and \(\frac{1}{\sqrt{y}}\) = v then equations reduces to
2u + 3v = 2 ……………(1)
and 4u – 9v = -1 ……………(2)
Multiplying (1) by 2, we get:
4u + 6v = 4
Now, (2) – (3) gives

Substitute this value of v in (1), we get:
2u + 3(\(\frac{1}{3}\)) = 2
or 2u + 1 = 2
or 2u = 2 – 1 = 1
or u = \(\frac{1}{2}\)

But \(\frac{1}{\sqrt{x}}\) = u
or (\(\frac{1}{\sqrt{x}}\))² = u²
or \(\frac{1}{x}\) = u²
or \(\frac{1}{x}\) = (\(\frac{1}{2}\))²
or x = 4

and \(\frac{1}{\sqrt{y}}\) = v
or (\(\frac{1}{\sqrt{y}}\))² = v²
or \(\frac{1}{y}\) = v²
or \(\frac{1}{y}\) = (\(\frac{1}{3}\))²
or y = 9
Hence x = 4 and y = 9.

(iii) Given pair of linear equations are
\(\frac{4}{x}\) + 3y = 14 and \(\frac{3}{x}\) – 4y = 23
putting \(\frac{1}{y}\) = v then equations reduces to
4v + 3y = 14 ………..(1)
and 3v – 4y = 23 ……………(2)
Multiplying (1) by 3 and (2) by 4, we get:
12v + 9y = 42 ………..(3)
and 12v – 16y = 92 ……………..(4)
Now, (4) – (3) gives

Substitute this value of y in (1), we get:
4v + 3(-2) = 14
or 4v – 6 = 14
or 4v = 14 + 6 = 20
or v = \(\frac{20}{4}\) = 5
But \(\frac{1}{x}\) = v
x = \(\frac{1}{v}\) = \(\frac{1}{5}\)
Hence x = \(\frac{1}{5}\) and y = -2.

(iv) Given pair of linear equation are
\(\frac{5}{x-1}+\frac{1}{y-2}\) = 2 and \(\frac{6}{x-1}-\frac{3}{y-2}\) = 12
Putting \(\frac{1}{x-1}\) = u and \(\frac{1}{y-2}\) = v then equations reduces to
5u + v = 2 ……………(1)
and 6u – 3v = 1 ……………(2)
Multiplying (1) by 3, we get:
15u + 3v = 6 ………..(3)
Now, (3) + (2) gives

u = \(\frac{7}{21}=\frac{1}{3}\)
Substitute this value of u in (1), we get:
5 × \(\frac{1}{3}\) + v = 2
or v = 2 – \(\frac{5}{3}\) = \(\frac{6-5}{3}\)
or v = \(\frac{1}{3}\)

But \(\frac{1}{x-1}\) = u
or \(\frac{1}{x-1}\) = \(\frac{1}{3}\)
or x – 1 = 3
or x = 3 + 1
or x = 4

and \(\frac{1}{y-2}\) = v
or \(\frac{1}{y-2}\) = \(\frac{1}{3}\)
or y – 2 = 3
or y = 3 + 2
or y = 5
Hence x = 4 or y = 5.

(v) Given pair of linear equation are
\(\frac{7 x-2 y}{x y}\) = 5
or \(\frac{7 x}{x y}-\frac{2 y}{x y}\) = 5
or \(\frac{7}{y}-\frac{2}{x}\) = 5
or \(-\frac{2}{x}+\frac{7}{y}\) = 5

and \(\frac{8 x+7 y}{x y}\) = 15
or \(\frac{8 x}{x y}+\frac{7 y}{x y}\) = 15
or \(\frac{8}{y}+\frac{7}{x}\) = 15
or \(\frac{7}{x}+\frac{8}{y}\) = 15

Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v, then equations reduces to
-2u + 7v = 5 ………….(1)
and 7u + 8v = 15 ………….(2)
Multiplying (1) by 7 and (2) by 2, we get:
-14v + 49u = 35
and 14v + 16u = 30
Now, (3) + (4) gives,

Substitute thuis value of u in (1), we get:
-2(1) + 7v = 5
or 7v = 5 + 2
or 7v = 7
or v = \(\frac{7}{7}\) = 1

But \(\frac{1}{x}\) = u
or x = \(\frac{1}{u}\)
or x = \(\frac{1}{1}\)
or x = 1

and \(\frac{1}{y}\) = v
or y = \(\frac{1}{v}\)
or y = \(\frac{1}{1}\)
or y = 1
Hence x = 1 and y = 1.

(vi) Given pair of linear equations are
6x + 3y = 6xy
or \(\frac{6 x+3 y}{x y}=\frac{6 x y}{x y}\)
or \(\frac{6}{y}+\frac{3}{x}=6\)
or \(3\left[\frac{1}{x}+\frac{2}{y}\right]=6\)
or \(\frac{1}{x}+\frac{2}{y}=2\)

and 2x + 4y = 5xy
or \(\frac{2 x+4 y}{x y}=\frac{5 x y}{x y}\)
or \(\frac{2}{y}+\frac{4}{x}=5\)
or \(\frac{4}{x}+\frac{2}{y}=5\)
Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v, then equations reduces to
u + 2v = 2 ……………(1)
and 4u + 2v = 5 ……….(2)
Now, (2) – (1) gives

or u = \(\frac{3}{3}\) = 1
Substitute this value of u in (1), we get:
1 + 2v = 2
or 2v = 2 – 1
or v = \(\frac{1}{2}\)

But \(\frac{1}{x}\) = u
or \(\frac{1}{x}\) = 1
or x = 1

and \(\frac{1}{y}\) = v
or \(\frac{1}{y}\) = \(\frac{1}{2}\)
or y = 2
Hence x = 1 and y = 2.

(vii) Given pair of linear equations are
\(\frac{10}{x+y}+\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}-\frac{5}{x-y}\) = -2
Putting \(\frac{1}{x+y}\) = u and \(\frac{1}{x-y}\) = v, then equations reduces to
10u + 2v = 4 or
5u + v = 2 …………(1)
15u – 5v = -2 ………….(2)
Multiplying (1) by 5, we get
25u + 5v = 10 ………….(3)
Now, (3) + (2) gives

Substitute this value of u in (1), we get:
5(\(\frac{1}{5}\)) + v = 2
or 1 + v = 2
or v = 1
But \(\frac{1}{x+y}\) = u
or \(\frac{1}{x+y}\) = \(\frac{1}{5}\)
or x + y = 5 ……….(4)
and \(\frac{1}{x-y}\) = v
or \(\frac{1}{x-y}\) = 1
or x – y = 1 ………..(5)
Now, (4) + (5) gives

Substitute this value of x in (4), we get:
3 + y = 5
y = 5 – 3 = 2
Hence x = 3 and y = 2.

(viii) Given pair of linear equations are
\(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) and \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)
Putting \(\frac{1}{3 x+y}\) = u and \(\frac{1}{3 x-y}\) = y, then Equations reduces to
u + v = \(\frac{3}{4}\)
or 4u + 4v = 3
or 4u + 4v = 3 ………….(1)

and \(\frac{u}{2}-\frac{v}{2}=\frac{-1}{8}\)
or u – v = \(\frac{-1}{4}\)
or 4u – 4v = -1 ………………(2)
Now, (1) + (2) gives

Substitute this value of u in (1), we get:
4(\(\frac{1}{4}\)) + 4v = 3
or 4v = 2
or v = \(\frac{2}{4}=\frac{1}{2}\)
But \(\frac{1}{3 x+y}\) = \(\frac{1}{4}\)
or 3x + y = 4 …………(3)
and \(\frac{1}{3 x-y}\) = \(\frac{1}{2}\)
or 3x – y = 2 …………….(4)
Now, (3) + (4) gives

x = 1
Substitute this value of x in (3), we get:
3(1) + y = 4
or 3 + y = 4
or y = 4 – 3 = 1
Hence, x = 1 and y = 1.

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions.
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km In 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home party by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:
(i) Let the speed of Ritu in still water = x km/hour
and the speed of current = y km/hour
∴ speed in upstream = (x – y) km/hour
and speed in downstream = (x + y) km/hour
Distance covered by Ritu in downstream in 2 hours = Speed × Time
= (x + y) × 2 km
According to 1st condition
2(x + y) = 20
x + y = 10 ……………..(1)
Distance covered by Rim in upstream in 2 hours
= Speed × Time
= 2(x – y)km
According to 2nd condition,
2(x – y) =4
x – y = 2 ……………….(2)
Now, (1) + (2) gives

Substitute this value of x in (1), we get:
6 + y = 10
y = 10 – 6 = 4
Hence, Ritu’s speed in still water = 6 km/hour
and speed of current = 4 km/hour.

(ii) Let one woman can fmish the work = x days
One man can finish the work = y days
then one woman’s one day’s work = \(\frac{1}{x}\)
One man’s one day’s work = \(\frac{1}{y}\)
According to 1st condition,
\(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\) ……………(1)
According to 2nd equation
\(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\) ……………(2)
Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v, then equations (1) and (2) reduces to
2u + 5v = \(\frac{1}{4}\)
8u + 20v = 1 …………….(3)
and 3u + 6v = \(\frac{1}{3}\)
9u + 18v = 1 ……………..(4)
Multiplying (3) by 9 and (4) by 8, we get:
72u + 180v = 9 ……………..(5)
and 72u + 144v = 8 …………..(6)
Now, (5) – (6) gives

or 9u + \(\frac{1}{2}\) = 1
or 9u = 1 – \(\frac{1}{2}\) = \(\frac{2-1}{2}\)
or 9u = \(\frac{1}{2}\)
or u = \(\frac{1}{2 \times 9}=\frac{1}{18}\)
But \(\frac{1}{x}\) = u
or \(\frac{1}{x}\) = \(\frac{1}{18}\)
or x = 18
and \(\frac{1}{y}\) = v
\(\frac{1}{y}\) = \(\frac{1}{36}\)
or y = 36
Hence, one woman and one man alone can finish work in 18 days and 36 days respectively.

(iii) Let speed of train = x km/hour
and speed of bus = y km/hour
Total distance = 300 km
Case I:
Time taken by train to cover 60 km = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac{60}{x}\) hours
Time taken by bus to cover 240 km = (300 – 60) = \(\frac{240}{y}\) hours
Total time = \(\left(\frac{60}{x}+\frac{240}{y}\right)\) hours
According to 1st condition,
\(\frac{60}{x}+\frac{240}{y}\) = 4
\(\frac{15}{x}+\frac{60}{y}\) = 1 ……………….(1)

Case II:
Time taken by train to cover 100 km = \(\frac{100}{x}\) hours
Time taken by bus to cover 200 km = 300 – 100 = \(\frac{200}{y}\) hours
∴ Total time = \(\left(\frac{100}{x}+\frac{200}{y}\right)\) hours
According to 2nd condition,
\(\left(\frac{100}{x}+\frac{200}{y}\right)\) = 4 hours 10 minutes
or \(\left(\frac{100}{x}+\frac{200}{y}\right)\) = \(\frac{25}{6}\)
\(\left(\frac{24}{x}+\frac{48}{y}\right)\) = 1 ……….(2)
Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v in equations
(1) and (2) then equations reduces to
15u + 60v = 1
and 24u + 48v = 1
15u + 60v – 1 = 0
24u + 48v – 1 = 0

From I and III, we get:
\(\frac{u}{-12}=\frac{1}{-720}\)
⇒ u = \(\frac{12}{720}=\frac{1}{60}\)
From II and III, we get:
\(\frac{v}{-9}=\frac{1}{-720}\)
⇒ v = \(\frac{9}{720}=\frac{1}{80}\)
But \(\frac{1}{x}\) = u
or \(\frac{1}{x}\) = \(\frac{1}{u}\)
or x = 60
and \(\frac{1}{y}\) = v
or \(\frac{1}{y}\) = \(\frac{1}{80}\)
or y = 80
Hence, speed of train and bus are 60 km/hour and 80 km/hour respectively.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of equations by the elimination method and the substitution method.
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4= 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}\) and x – \(\frac{y}{3}\) = 3
Solution:
(i) Given pair of linear equations
x + y = 5 …………..(1)
and 2x – 3y = 4 ………….(2)
Elimination Method
Multiplying (1) by 2, we get:
2x + 2y = 10 …………..(3)

Now, (3) – (2) gives

Substitute this value of y in (1), we get:
x + \(\frac{6}{5}\) = 5
or x = 5 – \(\frac{6}{5}\)
= \(\frac{25-6}{5}\) = \(\frac{19}{5}\)
Hence, x = \(\frac{19}{5}\) and \(\frac{6}{5}\)

Substitution Method:

From(2), 2x =4 + 3y
or x = \(\frac{4+3 y}{2}\) ……………..(4)
Substitute this value of x in (I), we get :
\(\frac{4+3 y}{2}\) + y = 5
Or \(\frac{4+3 y+2 y}{2}\)
Or 4 + 5y = 10
Or 5y = 10 – 4 = 6
Or y = \(\frac{6}{5}\)
Substitute this value of y in (4). we get:
x = \(\frac{4+\left(3 \times \frac{6}{5}\right)}{2}=\frac{4+\frac{18}{5}}{2}\)
= \(\frac{20+18}{5 \times 2}=\frac{38}{5 \times 2}\) = \(\frac{19}{5}\)
Hence x = \(\frac{19}{5}\) and y = \(\frac{6}{5}\)

(ii) Given pair of linear equation is :
3x + 4y = 10 …………….(1)
and 2x – 2y = 2 …………………(2)
Elimination Method
Multiplying equation (2) by 2, we get:
4x – 4y = 4 ……………(3)
Now, (3) + (1) gives

Substitute this value of x in (1), we get:
3(2) + 4y = 10
or 6 + 4y = 10
or 4y = 10 – 6
or 4y = 4
or y = \(\frac{4}{4}\) = 1
Hence x = 2 and y = 1.

Substitution method:

From (2),
2x = 2 + 2y
or x = y + 1 ………..(3)
Substitute this value of x in (1), we get :
3(y + 1) +4y = 10
or 3y + 3 + 4y = 10
or 7y = 10 – 3
or 7y = 7
or y = 1
Substitute this value of y in (3), we get :
x = 1 + 1 = 2
Hence, x = 2 and y = 1.

(iii) Given pair of linear equation is :
3x – 5y – 4 = 0 ……..(1)
and 9x = 2y + 7
or 9x – 2y – 7 = 0
Elimination Method:

Multiplying (1) by 3, we get:
9x – 15y – 12 = 0 ……………(3)
Now, (3) – (2) gives

Substitute this value of y in (1), we get:
3x – 5(-\(\frac{5}{13}\)) – 4 = 0
or 3x + \(\frac{25}{13}\) – 4 = 0
or 3x = 4 – \(\frac{25}{13}\)
or 3x = \(\frac{52-25}{13}\) = \(\frac{27}{13}\)
or x = \(\frac{27}{13} \times \frac{1}{3}\)
= \(\frac{9}{13}\)
Hence, x = \(\frac{9}{13}\) and y = – \(\frac{5}{13}\).

Substitution Method:

From(2), x = \(\frac{2 y+7}{9}\) …………..(4)
Substitute this value of x in (1), we get:
3\(\frac{2 y+7}{9}\) – 5y – 4 = 0
or \(\frac{2 y+7-15 y-12}{3}\) = 0
or – 13y – 5 = 0
or -13y = 5
or y = \(-\frac{5}{13}\)
Substitute this value of y in (4), we get:

Hence x = \(\frac{9}{13}\) and y = –\(\frac{5}{13}\)

(iv) Given pair of linear equation is:
\(\frac{x}{2}+\frac{2 y}{3}\) = -1
or \(\frac{3 x+4 y}{6}\) = -1
or 3x + 4y = -6 ……………(1)
x – \(\frac{y}{3}\) = 3
or \(\frac{3 x- y}{3}\) = 3
or 3x – y = 9 ……………(2)

Elimination Method:

Substitute this value of y in (1), we get :
3x + 4(-3) = -6
or 3x – 12 = -6
or 3x = -6 + 12
or 3x = 6
x = \(\frac{6}{3}\) = 2
Hence x = 2, y = – 3.

Substitution Method:

From (2), y = 3x – 9 …………..(4)
Substitute this value of y in (1), we get :
3x + 4(3x – 9) = -6
or 3x + 12x – 36 = -6
or 15x = -6 + 36
or 15x = 30
or x = \(\frac{30}{15}\) = 2
Substitute this value of x in (4), we get :
y = 3 (2) – 9
= 6 – 9 = -3
Hence x = 2, y = -3.

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if
we only add 1 to the denominator. What ¡s the fraction?

(ii) Five years ago, Nun was thrice as old as Sonu. Ten years later, Nun will be twice as old as Sonu. How old are Nun
and Sonti?

(iii) The sum of the digits of a two-digit nunaber ¡s 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got ₹ 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:
(i) Let numerator of fraction = x
Denominator of fraction = y
Required fraction = \(\frac{x}{y}\)
According to 1st condition,
\(\frac{x+1}{y-1}\) = 1
or x + 1 = y – 1
or x – y + 2 = 0 …………….(1)
According to 2nd condition,
\(\frac{x}{y+1}=\frac{1}{2}\)
or 2x = y + 1
or 2x – y – 1 = 0 ……………..(2)
Now, (2) – (1) gives

or x = 3
Substitute this value of x in (2), we get :
2 × 3 – y – 1 = 0
or 6 – y – 1 = 0
or 5 – y = 0
or y = 5
Hence, required fraction is \(\frac{3}{5}\).

(ii) Let Nun’s present age = x years
Sonus present age = y years
Five years ago
Nun’s age = (x – 5) years
Sonus age = (y – 5) years
According to 1st condition,
x – 5 = 3(y – 5)
or x – 5 = 3y – 15
or x – 3y + 10 = 0 …………….(1)
Ten years later
Nun’s age = (x + 10) years
Sonu’s age = (y + 10) years
According to 2nd condition,
x + 10 = 2 (y + 10)
or x + 10 = 2y + 20
or x – 2y – 10 = 0
Now, (1) – (2) gives

or -y = -20
or y = 20
Substitute this value of y in (2), we get:
x – 2(20) – 10 = 0
or x – 40 – 10 = 0
or x = 50
Hence, Nun’s present age = 50 years
Sonu’s present age = 20 years.

(iii) Let unit’s digit = x
Ten’s digit = y
∴ Required Number = 10y + x
According to 1st condition,
x + y = 9 …………..(1)
On reversing
Unit’s digit = y
Ten’s digit = x
∴ Number = 10x + y
According to 2nd condition,
9[10y + x] = 2[10x + y]
or 90y + 9x = 20x + 2y
or 90y + 9x – 20x – 2y = 0
or -11x + 88y = 0
or x – 8y = 0 ………………(2)
Now, (2) – (1) gives

y = 1
Substitute this value of y in (2), we get :
x – 8 × 1 = 0
or x = 8
Hence, required number = 10y + x
= 10 × 1 + 8 = 18.

(iv) Let, Meena received number of Rs. 50 notes = x
also, Meena received number of Rs. 100 notes = y
According to 1st condition,
x + y = 25 ……………(1)
According to 2nd condition,
50x + 100y = 2000
or x + 2y = 40 ………………(2)
Now, (2) – (1) gives

Substitute this value of y in (1), we get:
x + 15 = 25
or x = 25 – 15 = 10
Hence, Meena received number of notes of
Rs. 50 and Rs. 100 are 10 and 15 respectively.

(v) Let fixed charges for first three days = ₹ x
An additional charge for each day thereafter = ₹ y
In case of Saritha,
x + 4y = 27 …………..(1)
In case of Susy,
x + 2y = 21
Now, (1) – (2) gives

or y = \(\frac{6}{2}\) = 3
Substitute this value of y in (2), we get:
x + 2(3) = 21
or x + 6 = 21
or x = 21 – 6 = 15
Hence, fixed charges for first three days and an additional charge for each day thereafter ₹ 15 and ₹ 3.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Apply the division algorithm to find the quotient and remainder on dividing p (x) by g (x) as given below:
(i) p (x) = x³ – 3x² + 5x – 3, g (x) = x² – 2 (Pb. 2018 Set I, II, III)
(ii) p (x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p (x) = x⁴ – 5x + 6, g (x) = 2 – x² [Pb. 2017 Set-B]
Solution:
(i) Given that p (x) = x³ – 3x² + 5x – 3 and g (x) = x² – 2,

By division algorithm,
x³ – 3x² + 5x – 3 = (x – 3) (x² – 2) + (7x – 9)
Hence, quotient = x – 3 and remainder = 7x – 9

(ii) Given that p (x) = x⁴ – 3x² + 4x + 5
or p (x) = x⁴ + 0x³ – 3x² + 4x + 5
and g (x) = x² + 1 – x
or g (x) = x² – x + 1

By Division Algorithm,
x⁴ – 3x² + 4x + 5 = (x² + x – 3) (x² – x + 1) + 8
Hence, Quotient = x² + x – 3 and remainder = 8

(iii) Given that p (x) = x⁴ – 5x + 6
or p (x) = x⁴ + 0x³ + 0x² – 5x + 6
and g (x) = 2 – x²
or g (x) = – x² + 2

By division algorithtm,
x⁴ – 5x + 6 = (- x² – 2) (- x² + 2) + (- 5x + 10)
Hence, quotient = -x² – 2.
remainder = – 5x + 10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution:

∵ remainder is zero
∵ By division algorithm,
t² – 3 is factor of 2t⁴ + 3t³ – 2t² – 9t – 12

(ii)

∵ remainder is zero
∴ By division algorithm, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2

(iii)

∵ remainder is not zero.
∴ By division algorithm, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and – \(\sqrt{\frac{5}{3}}\).
Solution:
Given that two zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\)
∴ (x – \(\sqrt{\frac{5}{3}}\)) [x – (-\(\sqrt{\frac{5}{3}}\))] are factors of given polynomial or
(x – \(\sqrt{\frac{5}{3}}\)) (x + \(\sqrt{\frac{5}{3}}\)) are factors of given polynomial.
or x² – \(\frac{5}{3}\) is a factor of given polynomial.
Now, apply division algorithm to given polynomial and x² – \(\frac{5}{3}\)

∴ 3x⁴ + 6x³ – 2x² – 10x – 5
= (x² – \(\frac{5}{3}\)) (3x² + 6x + 3)
= (x² – \(\frac{5}{3}\)) (3) [x² + 2x + 1]
= 3 (x² – \(\frac{5}{3}\)) [x² + 2x + 1]
[S = 2, P = 1]
= 3 (x² – \(\frac{5}{3}\)) [x(x + 1) + 1 (x + 1)]
= 3 (x² – \(\frac{5}{3}\)) (x + 1) (x + 1)
Now, other zeroes of polynomials are given by
x + 1 = 0 ; x = -1 or
x + 1 = 0 ; x = -1
The zeroes of the given fourth degree polynomial are :
\(\sqrt{\frac{5}{3}}\), –\(\sqrt{\frac{5}{3}}\), -1, -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4 respectively find g(x).
Solution.
Compare given data with division algorithm, we have
p(x)= g (x). q(x) + r (x) or
p(x) – r(x) = g(x).q(x)
or g(x) . q (x) = p(x) – r(x)
or g(x) = \(\frac{p(x)-r(x)}{q(x)}\)
By putting various values, we get:
g(x) = \(\frac{\left(x^{3}-3 x^{2}+x+2\right)-(-2 x+4)}{x-2}\)

g(x) = \(\frac{x^{3}-3 x^{2}+x+2+2 x-4}{x-2}\)

g(x) = \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\) …………….(1)

Now,

∴ \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\) = x² – x + 1
From (1) and (2), we get:
g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the dhriskm algorithm and
(i) deg p (x) = deg q (x)
(ii) deg r (x) = 0
(iii) deg q (x) = deg r (x)
Solution:
(i) Let p(x) = 5x² – 5x +10; g(x) = 5 q(x) = x² – x + 2; r(x) = 0

∴ By division algorithm,
5x² – 5x + 10 = 5(x² – x + 2) + 0
or p(x) = g(x) q(x) + r(x)
Also, deg p(x) = deg q(x) = 2

(ii) Let p(x) = 7x³ – 42x + 53
g(x) = x³ – 6x + 7
q(x) = 7; r(x) = 4

∴ By division algorithm,
7x³ -42x + 53 = 7(x³ – 6x+ 7)+ 4
or p(x) = q(x) g(x) + r(x)
Also, deg q(x) = 0 = deg r(x)

(iii) Let p(x) = 4x³ + x² + 3x + 6;
g(x) = x² + 3x + 1;
q(x) = 4x – 11;
r(x) = 32x + 17

∴ By division algorithm,
4x³ + x² + 3x + 6 = (4x – 11) (x² + 3x + 1) + (32x + 17)
or p(x) = q(x) . g(x) + r(x)
Also, deg q(x) = deg r(x)

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one
pencil and that of one pen.
Solution:
(i) Let the number of boys in the Quiz = X
and the number of girls in the Quiz = y
Total number of students took part in Quiz = 10
x + y = 10
or x + y – 10 = 0
According to Question,
y = x + 4
or x = y – 4
Now, draw the graph of linear equations
x + y = 10
and x – y + 4 = 0
x + y = 10
or x = 10 – y
Putting y = 0 in (1), we get :
x = 10 – 0 = 10
Putting y = 7 in(1), we get:
x = 10 – 7 = 3
Putting y = 10 in (1) we get:
X = 10 – 10 = 0

Plotting the points A (10, 0), B (3, 7), C (0, 10) and drawing a line joining them we get the graph of the equation x + y = 10

Now x – y + 4 = 0
or x = y – 4
Putting y = 0 in (2), we get:
x = 0 – 4 = -4
Putting y = 7 in (2), we get:
x = 7 – 4 = 3
Putting y = 4 in (2), we get:
x = 4 – 4 = 0

Plotting the points D (-4, 0), B (3, 7), E (0, 4) and drawing a line joining them, we get the graph of the equation x – y + 4 = 0

From the graph it is clear that both the linear equations meets at a point B (3, 7).
∴ Point B (3, 7) is the graphic solution.
Hence, number of boys in the Quiz = 3
Number of girls in the Quiz = 7

(ii) Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to 1st condition,
5x + 7y = 50
According to 2nd condition,
7x + 5y = 46
∴ Pairs of linear equations is
5x + 7y = 50
7x + 5y = 46
Now, draw the graph of these linear equations.
5x + 7y = 50
or 5x = 50 – 7y
x = \(\frac{50-7 y}{5}\) ………………..(1)
Putting y = 0 in (1), we get :
x = \(\frac{50-7 \times 0}{5}=\frac{50}{5}\)
Putting y = 5 in (1), we get:
x = \(\frac{50-7 \times 5}{5}=\frac{50-35}{5}\)
= \(\frac{15}{5}\) = 3
Putting y = 7 in (1), we get :
x = \(\frac{50-7 \times 7}{5}=\frac{50-49}{5}\)
= \(\frac{1}{5}\) = 0.2

Table:

Plotting the points A (10, 0), B (3, 5), C (0.2, 7) and drawing a line joining them, we get the graph of the equation
5x + 7y = 50
Now 7x + 5y = 46
0r 7x = 46 – 5y
or x = \(\frac{46-5 y}{7}\) …………….(2)
Putting y = 0 in (2), we get:

x = \(\frac{46-5 \times 0}{7}=\frac{46}{7}\) = 6.5
Putting y = 5 in (2), we get:
x = \(\frac{46-5 \times 5}{7}=\frac{46-25}{7}\)
= \(\frac{21}{7}\) = 3
Putting y = – 4 in (2), we get:
x = \(\frac{46-5 \times(-4)}{7}=\frac{46+20}{7}\)
= \(\frac{66}{7}\) = 9.5

Table:

Plotting the points E (6.5, 0), B (3, 5), F (9.5. -4) and drawing a line joining them, we get the graph of the equation.
7x + 5y = 46

From the graph, it is clear that both the linear equations meets at a point B (3, 5).
∴ point B (3, 5) is the graphic solution.
Hence, cost of one pencil = ₹ 3
Cost of one pen = ₹ 5

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\) and find out whether the lines representing the following pairs of linear
equations intersect at point, are parallel or coincident :
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
(i) Given pairs of linear equation
5x – 4y + 8 = 0
and 7x + 6y – 9 = 0
Here a₁ = 5, b₁ = – 4, c₁ = 8
a₂ = 7, b₂ = 6, c₂ = -9
Now,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{7}\);

\(\frac{b_{1}}{b_{2}}\) = \(-\frac{4}{6}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{-9}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equation intersect at a point.

(ii) Given pairs of linear equation is
9x + 3y + 12 = 0
and 18x + 6y + 24 = 0
Here, a₁ = 9, b₁ = 3, c₁ = 12
a₂ = 18, b₂ = 6, c₂ = 24

Now \(\frac{a_{1}}{a_{2}}=\frac{9}{18}=\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}\)
Hence, given pairs of linear equation intersect at a point.

(iii) Given pairs of linear equation is
6x – 3y + 10 = 0
and 2x – y + 9 = 0
Here a₁ = 6, b₁ = – 3, c₁ = 10
a₂ = 2, b₂ = -1, c₂ = 9
Now,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{6}{2}\) = 3;

\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-1}\) = 3;

\(\frac{c_{1}}{c_{2}}\) = \(\frac{10}{9}\)

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are parallel to each other.

Question 3.
On comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\). find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y =8; 4x – 6y = 9
(iii) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11; -10x + 6y = -22
(v) \(\frac{4}{3}\)x + 2y = 8; 2x +3y = 12
Solution:
(i) Given pair of linear equation is
3x + 2y = 5
and 2x – 3y = 7
or 3x + 2y – 5 = 0
and 2x – 3y – 7 = 0
Here a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7
Now
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{2}{-3}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-5}{-7}\) = \(\frac{5}{7}\)
∴ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are consistent.

(ii) Given pair of linear equation is:
2x – 3y = 8
and 4x – 6y = 9
Or 2x – 3y – 8 = 0
4x – 6y – 9 = 0
Here a₁ = 2, a₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c ₂ = -9
Now
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-8}{-9}\) = \(\frac{8}{9}\)

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are inconsistent.

(iii) Given pair of linear equation is:
\(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7;
and 9x – 10y = 14

Or \(\frac{3}{2}\)x + \(\frac{5}{3}\)y – 7 = 0
and 9x – 10y – 14 = 0
Here a₁ = \(\frac{3}{2}\), b₁ = \(\frac{5}{3}\), c₁ = -7
a₂ = 9, b₂ = -10, c₂ = -14

Hence, given pairs of linear equations are consistent.

(iv) Given pair of linear equations is
5x – 3y= 11
and -10x + 6y = -22
Or 5x – 3y – 11 = 0
and -10x + 6y + 22 = 0
Here a₁ = 5, b₁ = -3, c₁ = -11
a₂ = -10, b₂ = 6, c₂ = 22
Hence given pair of linear equations is consistent.

(v) Given pair of linear equations is
\(\frac{4}{3}\)x + 2y = 8 and 2x + 3y = 12
or \(\frac{4}{3}\)x + 2y – 8 = 0
and 2x + 3y – 12 = 0
Here a₁ = \(\frac{4}{3}\), b₁ = 2, c₁ = -8
a₂ = 2, b₂ = 2, c₂ = -12
Now,

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are consistent.

Question 4.
Which of the following pairs of linear equations are consistent / inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0,4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0,4x – 4y – 5= 0
Solution:
(1) Given pair of linear equations is
x + y = 5
and 2x + 2y = 10
Or x + y – 5 = 0
2x + 2y – 10 = 0
Here a₁ = 1, b₁ = 1, c₁ = -5
a₂ = 2, b₂ = 2, c₂ = -10
Now

\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-5}{-10}\) = \(\frac{1}{2}\)

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are consistent.
Draw the graph of these equations

x + y = 5
x = 5 – y ………….(1)
Putting y = 0 in (1), we get:
x = 5 – 0 = 5
Putting y = 3 in (1), we get
x = 5 – 3 = 2
Putting y = 5 in (1), we get
x = 5 – 5 = 0

Plotting the points A (5, 0), B (2, 3), C (0, 5) and drawing a line joining them, we ge the graph of the equation x + y = 5
2x + 2y = 10 Or 2 (x + y) = 10
Or x + y = 5
Or x = 5 – y
Putting y = 0 in (1), we get :
x = 5 – 0 = 5
Putting y = 2 in (2), we get :
x = 5 – 2 = 3
Putting y = 5 in (2), we get:
x = 5 – 5 = 0
Putting y = 2 in (2), we get :
x = 5 – 2 = 3
Putting y = 5 in (2), we get:
x = 5 – 5 = 0

Plotting the points A (5, 0), D (3, 2), C (0, 5) and drawing a line joining them, we get the graph of the equation 2x + 2y = 10

The graphs of two equations are coincident. Hence system of equations has infinitely many solutions i.e. consistent.

(ii) Given pair of linear equations is :
x – y = 8
and 3x – 3y = 16
Or
x – y – 8 = 0
and 3x – 3y – 16 = 0

Here a₁ = 1, b₁ = -1, c₁ = -8
a₂ = 3, b₂ = -3, c₂ = -16
Now,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{3}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{-1}{-3}\); = \(\frac{1}{3}\)

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-8}{-16}\) = \(\frac{1}{2}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are inconsistent solution.

(iii) Given pailt of linear equations is :
2x + y – 6 = 0
and 4x – 2y – 4 = 0
Here a₁ = 2, b₁ = 1, c₁ = -6
a₂ = 4, b₂ = -2, c₂ = -4
Now
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}=\frac{1}{-2}\);

\(\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}\)
∴ given pair of system is consistent.
Draw the graph of these linear equations
2x + y – 6 = 0
Or 2x = 6 – y
Or x = \(\frac{6-y}{2}\) ………….(1)
Putting y = 0 in (1), we get:
x = \(\frac{6-0}{2}=\frac{6}{2}\) = 3

putting y = 2 in (1), we get:
x = \(\frac{6-2}{2}=\frac{4}{2}\) = 2

Putting y = -2 in (1), we get:
x = \(\frac{6-(-2)}{2}=\frac{6+2}{2}=\frac{8}{2}\) = 4

Plotting the points A (3, 0), B (2, 2), C (4, -2) and drawing a line joining them, we get the graph of the equation.
2x + y – 6 = 0
Now 4x – 2y – 4= 0
or 2[2x – y – 2] = 0
or 2x – y – 2 = 0
or 2x = y + 2
or x = \(\frac{y+2}{2}\) …………..(2)
Putting y = 0 in (2), we get:
x = \(\frac{0+2}{2}=\frac{2}{2}\) = 1

Putting y = 2 in (2), we get :
\(\frac{2+2}{2}=\frac{4}{2}\) = 2

Putting y = – 2 in (2), we get:
x = \(\frac{-2+2}{2}=\frac{0}{2}\) = 0

Plotting the points D (1, 0), B (2, 2), E (0, -2) and drawing a line joining them, we get the graph of the equation
4x – 2y – 4 = 0

From the graph, it is clear that given system of equations meets at a point B (2, 2).
Hence, given pair of linear equations have unique solution.

(iv) Given pair of linear equations is :
2x – 2y – 2 = 0
and 4x – 4y – 5 = 0
Here a₁ = 2, b₁ = -2, c₁ = -2
a₂ = 4, b₂ = -4, c₂ = -5
Now,
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}=\frac{-2}{-4}=\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pair of system have inconsistent solution.

Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m Find the dimensions of the garden.
Solution:
Let length of garden = x m
Width of garden =y m
Perimeter of garden = 2 [x + y] m
Half perimeter of garden = (x + y) m
According to 1st condition x = y +4
According to 2nd condition
x + y = 36
∴ Pair of linear equations is
x = y + 4
and x + y = 36
x = y + 4 ………………(1)
Putting y = 0 in (1), we get :
x = 0 + 4 = 4
Putting y = – 4 in (1), we get:
x = -4 + 4 = 0
Putting y = 16 in (1), we get :
x = 16 + 4 = 20

Plotting the points A (4, 0) B (0, -4), C (20, 16) and drawing a line joining them.
we get the graph of the equation.
x = y + 4
Now x + y = 36
x = 36 – y
Putting y = 12 in (2), we get:
x = 36 – 12 = 24
Putting y = 24 in (2), we get :
x = 36 – 24 = 12
Putting y = 16 in (2), we get:
x = 36 – 16 = 20

Plotting the points D (24, 12), E (12, 24) C(20, 16) and drawing a line joining them, we get the raph of the equation.
x + y = 36

From the graph. it is clear that pair of linear equations meet at a point C (20, 16).
∴ C (20, 16) i.e. x = 20 and y = 16 is the solution of linear equations.
Hence, length of garden = 20 m
Width of garden = 16 m

Another Method:

Let width of garden = x m
Lengh of garden = (x + 4) m
Perimeter of garden = 2 [Length + Width]
= 2 [x + x + 4] m
= 2[2x + 4]m
∴ Half perimeter of garden = (2x +4) m
According to Question,
2x + 4 = 36
or 2x = 36 – 4
or 2x = 32
or x = \(\frac{32}{16}\) = 2 m
Hence, width of garden = 16 m
and length of garden = (16 + 4)m = 20 m

Question 6.
Given the Linear equation 2x + 3y – 8 = 0, write another linear equadon in two variables such that the geometiical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Case I. For Intersecting Unes
Given linear equation is:
2x + 3y – 8 = 0
There are many another linear equations in two variahes which satisfies the condition of intersecing lines i.e.
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

One of which is as follow:
3x – 2y – 6 = 0
Now, draw the graph of linear equations (1) and (2).
2x + 3y – 8 = 0
2x = 8 – 3y
Putting y = 0 in (1), we get;
x = \(\frac{8-3 \times 0}{2}=\frac{8}{2}\) = 4

Putting y = -2 in (1), we get:
x = \(\frac{8-3(-2)}{2}=\frac{14}{2}\) = 7

Piating y = 2 in (1) we get:
x = \(\frac{8-3 \times 2}{2}=\frac{8-6}{2}=\frac{2}{2}=1\)

Plotting the points A (4, 0), B (7, -2), C (1, 2) and drawing a line joining them, we get the graph of the equation
2x + 3y – 8 = 0
Now 3x – 2y – 6 = 0
3x = 6 + 2y
or x = \(\frac{6+2 y}{3}\)
Putting y = 0 in (1), we get:
x = \(\frac{6+2 \times 0}{3}=\frac{6}{3}\) = 2
Putting y = – 3 in (1), we get :
x = \(\frac{6+2(-3)}{3}\) = \(\frac{6-6}{3}\) = 0
Putting y = 3 in (2), we get :
x = \(\frac{6+2 \times 3}{3}=\frac{6+6}{3}=\frac{12}{3}\) = 4

From the graph it is clear that linear equations intersect at a point G.

Case II:
For Parallel Lines
Given linear equation is
2x + 3y – 8 = 0 ……………(1)
There are many other linear equation in two variables which satisfies the condition of parallel lines i.e.
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
One of which is as follow :
2x + 3 – 5 = 0
Now, draw the graph of linear equations (1) and (3)
Table for linear equation 2x + 3y – 8 = 0 is follow:

Consider, 2x + 3y – 5 = 0
or 2x = 5 – 3y
or x = \(\frac{5-3 y}{2}\) …………..(3)
Putting y = 0 in (3), we get :
x =\(\frac{5-3 \times 0}{2}=\frac{5}{2}\) = 2.5

Putting y = 3 in (3), we get:
x = \(\frac{5-3 \times 3}{2}=\frac{5-9}{2}=\frac{-4}{2}\) = -2

Putting y = 3 in (3), we get:
x = \(\frac{5-3(-3)}{2}=\frac{5+9}{2}=\frac{14}{2}\) = 7

Plotting the points G (2.5, 0), H (-2, 3), I (7, -3) and drawing a line joining them,
we get the graph of the equation
2 + 3y – 5 = 0.

Case III. For Coincident Lines
Given linear equation is
2x + 3y – 8 = 0
There are many other linear equations ¡n two variables which satisfies the condition of coincident lines i.e.
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
One of which is as follow :
6x+ 9y – 24 = 0
Now, draw the graph of linear equations (1) and (4).
Consider linear equation (4)
6x + 9y – 24 = o
or 3[2x + 3 – 8] = 0
or 2x + 3y – 8 = 0
∴ The points of both are same and line of both equations are same.

Question 7.
Drab tht graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed b these lines and the x-axis and shade the triangular region. (Pb. 2018 Set I, II, III)
Solution:
Consider the pair of linear equation
x – y + 1 =0
and 3x + 2y – 12 = 0
x – y + 1 = 0
or x = y – 1
Putting y = 0 in (1), we get:
x = 0 – 1 = -1
Puning y = 3 in (1) we get:
x = 3 – 1 = 2
Putting y = 1 in (1) , we get:
x = 1 – 1 = 0

PloningthepointsA(-1, 0), B(2, 3), C(0, 1) and drawing a line ioining them. we get the graph of the equation x – y + 1 = 0
3x + 2y – 12 = 0
or 3x = 12 – 2y
or x = \(\frac{12-2 y}{3}\) …………..(2)
Putting y = 0 in (2), we get:
x = \(\frac{12-2 \times 0}{3}=\frac{12}{3}\) = 4
Putting y = 3 in (2), we get:
x = \(\frac{12-2 \times 3}{3}=\frac{12-6}{3}=\frac{6}{3}\) = 2
Putting y = 6 in (2), we get:
x = \(\frac{12-2 \times 6}{3}=\frac{12-12}{3}=0\)

Table:

Plotting the points D(4, 0) B (2, 3), E (0, 6) and drawing a line oinin them, we get the graph of the equation 3x – 2y – 12 = 0

The vertices of the triangle formed by pair of linear equations and the x-axis are shaded in the graph. The triangle so formed is ∆ABD.Coordinates of the vertices of ∆ABD are
A(-1, 0), B(2, 3) and D(4, 0).
Now, length of Base AD = AO + OD = 1 + 4 = 5 units
Length of perpendicular BF = 3 units
∴ Area of ∆ABD = \(\frac{1}{2}\) × Base × altitude
= \(\frac{1}{2}\) × AD × BF
= (\(\frac{1}{2}\) × 5 × 3) sq. units
= \(\frac{15}{2}\) = 7.5 sq. units

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x -4
Solution:
(i) Given Quadratic polynomial,
x² – 2x – 8
∵ [S = -2, P = -8]
= x² – 4x + 2x – 8
= x (x – 4) + 2 (x – 4)
= (x – 4) (x + 2)
The value of x² – 2x – 8 is zero,
iff (x – 4) = 0 or (x + 2) = 0
iff x = 4 or x = – 2
Therefore, zeroes of x2 – 2x – 8 are – 2 and 4.
Now, Sum of zeroes = (- 2) + (4) = 2
= \(\frac{-(-2)}{1}=-\frac{(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)
Product of zeroes = (- 2) (4) = – 8
= \(\frac{-8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficient are verified.

(ii) Given quadratic polynomial,
4s² – 4s + 1
= 4s² – 2s – 2s + 1
∵ [S = -4, P = 4 × 1]
= 2s (2s – 1) – 1 (2s – 1)
= (2s – 1) (2s – 1)
The value of 4s² – 4s + 1 is zero
iff (2s – 1) = 0 or (2s – 1) = 0
iff s = \(\frac{1}{2}\) or s = \(\frac{1}{2}\)
Therefore, zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\)

Now, sum of zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
= \(\frac{-(-4)}{4}=\frac{-(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}\)

Product of Zeroes = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

(iii) Given quadratic polynomial,
6x² -3 – 7x
= 6x² – 7x – 3
∵ [S = – 7, P = 6x-3=-18]
= 6x² -9x + 2x-3
= 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
The value of 6x² – 3 – 7x is zero
iff (2x – 3) = 0 or 3x + 1 = 0
iff x = \(\frac{3}{2}\) or x = –\(\frac{1}{3}\)
Therefore, zeroes of 6x² – 3 – 7x are \(\frac{3}{2}\) and –\(\frac{1}{3}\)
Now, Sum of zeroes = \(\frac{1}{3}\)

= \(\frac{3}{2}+\left(\frac{-1}{3}\right)\)

= \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}\)

= \(\frac{7}{6}=\frac{-(-7)}{6}\)

= \(\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)

Product of zeroes = \(\left(\frac{3}{2}\right)\left(\frac{-1}{3}\right)\)

= \(\frac{-3}{6}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

(iv) Given quadratic polynomial,
4u² + 8u = 4u(u + 2)
The value of 4u² + 8 u is zero
iff 4u = 0 or u + 2 = 0
iff u = 0 or u = – 2
Therefore, zeroes of 4u² + 8M are 0 and – 2
Now, Sum of zeroes = 0 + (- 2) = -2
= \(\frac{-8}{4}\)
= \(-\frac{(\text { Coefficient of } u)}{\text { Coefficient of } u^{2}}\)

Product of zeroes = (0) (- 2) = 0
= \(\frac{0}{4}=\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

(v) Given quadratic polynomial,
t² – 15 = t² – (√15)²
= (t – √15) (t + √15)
The value of t² – 15 is zero
iff t – √15 = 0 or t + √15 = 0
iff t = √15 or t = – √15
Therefore, zeroes of t² – 15 are – √15 and √15.
Now, Sum of zeroes = -√15 + √15 = 0 = \(\frac{0}{1}\)
= \(\frac{-(\text { Coefficient of } t)}{\text { Coefficient of } t^{2}}\)

Product of zeroes = (-√15) (√15) = – 15 = \(-\frac{15}{1}\)

= \(\frac{0}{1}\)
= \(=\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

(vi) Given quadratic polynomial,
3x² – x – 4
= 3x² + 3x – 4x – 4
= 3x (x + 1) – 4 (x + 1)
∵ [S = – 1, P = 3 x – 4 = – 12]
= (x + 1) (3x – 4)
The value of 3x² – x – 4 is zero
iff (x + 1) = 0 or 3x – 4 = 0
iff x = -1 or x = \(\frac{4}{3}\)
Therefore, zeroes of 3x² – x – 4 are – 1 and \(\frac{4}{3}\)

Now, Sum of zeroes = – 1 + \(\frac{4}{3}\)
= \(\frac{-3+4}{3}=\frac{1}{3}\)
= \(\frac{-(-1)}{3}=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\)
Product of zeroes = (- 1) \(\frac{4}{3}\)
= \(\frac{-4}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), -1
(ii) √2, \(\frac{1}{3}\)
(iii) 0, √5
(iv) 1, 1
(v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
(vi) 4, 1. [MQP 2015]
Solution:
(i) Given that, sum of zeroes and products of zeroes of given polynomial \(\frac{1}{4}\) are -1 respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = \(\frac{1}{4}\)
and αβ = Product of zeroes = – 1
Now, ax² + bx + c = k (x – α) (x – β) where k is any constant
= k [x² – (α + β)x + αβ]
= k[x² – \(\frac{1}{4}\)x + (-1)]
= k[x² – \(\frac{1}{4}\)x – 1]
for different value of k, we get different quadratic polynomials.

(ii) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are √2 and \(\frac{1}{3}\) respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = √2
and αβ = Product of zeroes = \(\frac{1}{3}\)
Now, ax² + bx + c = k (x – α) (x – β) where k is any constant
= k[x² – (α + β) x + αβ]
= k[x² – √2x + \(\frac{1}{3}\)]of k, we get different quadratic polynomial.
for different values of k, we get different quadratic polynomial.

(iii) Given that, sum of zeroes and products of zeroes of given quadratic polynomial are 0 and √5 respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 0
and αβ = Product of zeroes = √5
Now, ax² + bx + c = k(x – α) (x – β)
where k is any constant
= k [x² – (a + (α+ β)x + αβ)
= k[x² – 0x + √5]
= k[x² + √5]
for different values of k, we get different quadratic polynomial.

(iv) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 1 and 1 respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 1 and
αβ = Product of zeroes = 1
Now, ax² + bx + c = k(x – α) (x – β)
where k is any constant.
= k [x² – (α + β)x + αβ]
= k [x² – 0x + √5]
= it [x² – x + √5]
for different values of k, we get different quadratic polynomial.

(v) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are \(-\frac{1}{4}\) and \(\frac{1}{4}\) respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = –\(\frac{1}{4}\)
and αβ = Product of zeroes = \(\frac{1}{4}\)
Now ax² + bx + c = k (x – α) (x – β) where k is any constant
= k[x² – (α + β) x + αβ]
= k[x² – (-\(\frac{1}{4}\))x + \(\frac{1}{4}\)]
= k[x² + \(\frac{1}{4}\)x + \(\frac{1}{4}\)]
For different values of k, we get different quadratic polynomial.

(vi) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 4 and 1 respectively. Let the quadratic polynomial be
ax² + bx + c and its zeroes be α and β
α + β = sum of zeroes = 4 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β) where k is any constant
= k [x² – (α + β) x + αβ]
= k [x² – 4x + 1]
For different values of k, we get different quadratic polynomials.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Question 1.
Express each number as a product of its prime factors :
(i) 140 (Pb, 2019)
(ii) 156
(iii) 3825
(iv) 5005 (Pb. 2019, Set-1, II, III)
(v) 7429.
Solution:
(i) Prime factorisation of 140 = (2)² (35) = (2)² (5) (7)

(ii) Prime factorisation of 156 = (2)² (39) = (2)² (3) (13)

(iii) Prime factorisation of 3825 = (3)² (425)
= (3)² (5) (85)
= (3)² (5)² (17)

(iv) Prime factorisation of 5005
= (5) (1001)
= (5) (7) (143)
= (5) (7) (11) (13)

(v) Prime factorisation of 7429
= (17) (437)
= (17) (19) (23)

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers.
(i) 26 and 91 [Pb. 2017 Set-C]
(ii) 510 and 92
(iii) 336 and 54.
Solution:
(i) Given numbers are 26 and 91 Prime factorisation of 26 and 91 are
26 = (2) (13) and
91 = (7) (13)
HCF (26, 91)
Product of least powers of common factors
∴ HCF (26, 91) = 13
and LCM (26, 91) = Product of highest powers of all the factors
= (2) (7) (13) = 182
Verification :
LCM (26, 91) × HCF (26, 91)
= (13) × (182) = (13) × (2) × (91)
= (26) × (91)
= Product of given numbers.

(ii) Given numbers are 510 and 92 Prime factorisation of 510 and 92 are
510 = (2) (255)
= (2) (3) (85)
= (2) (3) (5) (17)
and 92 = (2) (46) = (2)² (23)
HCF (510, 92) = Product of least powers of common factors = 2
LCM (510, 92) = Product of highest Powers of all the factors
= (2)² (3) (5) (17) (23) = 23460

Verification:
LCM (510, 92) × HCF (510, 92)
= (2) (23460)
= (2) × (2)² (3) (5) (17) (23)
= (2) (3) (5) (17) × (2)² (23)
= 510 × 92 = Product of given numbers.

(iii) Given numbers are 336 and 54
Prime factorisation of 336 and 54 are
336 = (2) (168)
= (2) (2) (84)
= (2) (2) (2) (42)
= (2) (2) (2) (2) (21)
= (2)⁴ (3) (7)

and 54 = (2) (27)
= (2) (3) (9)
= (2) (3) (3) (3)
= (2) (3)³
HCF (336, 54) = Product of least powers of common factors = (2) (3) = (6)
LCM (336, 54) = Product of highest powers of all the factors
= (2)⁴ (3)³ (7) = 3024

Verification :
LCM (336, 54) × H.C.F. (336, 54)
= 6 × 3024
= (2) (3) × (2)⁴ (3)³ (7)
= (2)⁴ (3) (7) × (2) (3)³
= 336 × 54
= Product of given numbers.

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 [MQP 2015, Pb. 2015 Set A, Set B, Set C]
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) Given numbers are 12, 15 and 21
Prime factorisation of 12, 15 and 21 are
12 = (2) (6)
= (2) (2) (3)
= (2)² (3)
15 = (3) (5)
21 = (3) (7)
HCF (12, 15 and 21) = 3
LCM (12, 15 and 21) = (2)² (3) (5) (7) = 420

(ii) Given numbers are 17, 23 and 29
Prime factorisation of 17, 23 and 29 are
17 = (17) (1)
23 = (23) (1)
29 = (29) (1)
HCF (17, 23 and 29) = 1
LCM (17, 23 and 29)
= 17 × 23 × 29 = 11339

(iii) Given numbers are 8, 9 and 25
Prime factorisation of 8, 9 and 25 are
8 = (2) (4)
= (2) (2) (2)
= (2)³ (1)
9 = (3) (3) = (3)² (1)
25 = (5) (5) = (5)² (1)
HCF (8, 9 and 25) = 1
LCM (8, 9 and 25) = (2)³ (3)² (5)² = 1800

Question 4.
Prove that HCF (306,657) = 9, find LCM (306,657). [Pb. 2016 Set-B, 2017 Set-B]
Solution:
Given numbers are 306 and 657 Prime factorisation of 306 and 657 are
306 = (2) (153)
= (2) (3) (51)
= (2) (3) (3) (17)
= (2) (3)2 (17)

657 = (3) (219)
= (3) (3) (73)
= (3)² (73)

HCF (306, 657) = (3)² = 9
∵HCF × LCM = Product of given number
∵9 × LCM (306, 657) = 306 × 657

= 34 × 657 = 22338

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n:
Solution:
Let us suppose that 6ⁿ ends with the digit 0 for some n ∈ N.
6ⁿ is divisible by 5.
But, prime factor of 6 are 2 and 3 Prime factor of (6)ⁿ are (2 × 3)ⁿ
⇒ It is clear that in prime factorisation of 6ⁿ there is no place for 5.
∵ By Fundamental Theorem of Arithmetic, Every composite number can be expressed as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.
∴ Our supposition is wrong.
Hence, there exists no natural number n for which 6ⁿ ends with the digit zero.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Consider, 7 × 11 × 13 + 13 = 13 [7 × 11 + 1]
which is not a prime number because it has a factor 13. So, it is a composite number.
Also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5[7 × 6 × 5 × 4 × 3 × 2 × 1 + 1], which is not a prime number because it has a factor 5. So it is a composite number.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for die same. Suppose they both start at the same point and at file same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken by Sonia to drive one round of the field =18 minutes
Time taken by Ravi to drive one round of same field = 12 minutes
They meet again at the starting point = LCM (18, 12)
Now, Prime factorisation of 18 and 12 are 18 = (2) (9)
= (2) (3) (3)
= (2) (3)²

12 = (2) (6)
= (2) (2) (3)
= (2)² (3)
LCM (18, 12) = (2)² (3)² = 4 × 9 = 36
Hence, After 36 minutes Sonia and Ravi will meet again at the starting point.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255.
Solution:
(i) By Euclid’s division Algorithm

Step 1.
Since 225 > 135,
we apply the division Lemma to 225 and 135,
we get 225 = 135 × 1 + 90

Step 2.
Since the remainder 90 ≠ 0,
we apply the division Lemma to 135 and 90,
we get 135 = 90 × 1 + 45

Step 3. Since the remainder 45 ≠ 0,
we apply the division Lemma to 90 and 45,
we get 90 = 45 × 2 + 0

Since the remainder has now become zero, so we stop procedure.
∵ divisor in the step 3 is 45
∵ HCF of 90 and 45 is 45
Hence, HCF of 135 and 225 is 45.

(ii) To find HCF of 196 and 38220
Step 1.
Since 38220 > 196,
we apply the division Lemma to 196 and 38220,
we get 38220 = 196 × 195 + 0
Since the remainder has now become zero so we stop the procedure.
∵ divisor in the step is 196
∵ HCF of 38220 and 196 is 196.
Hence, HCF of 38220 and 196 is 196.

(iii) To find HCF of 867 and 255
Step 1.
Since 867 > 255,
we apply the division Lemma to 867 and 255,
we get 867 = 255 × 3 + 102

Step 2.
Since remainder 102 ≠ 0,
we apply the divison Lemma to 255 and 102,
we get 255 = 102 × 2 + 51

Step 3.
Since remainder 51 ≠ 0,
we apply the division Lemma to 51 and 102, by taking 102 as division,
we get 102 = 51 × 2 + 0
Since the remainder has now become zero, so we stop the procedure.
∵ divisor in step 3 is 51.
HCF of 102 and 51 is 51.
Hence, HCF of 867 and 255 is 51.

Question 2.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let a be any positive odd integer, we apply the division algorithm with a and b = 6.
Since 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5. i.e., a can be 6q or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5 where q is quotient. However, since a is odd ∵ a cannot be equal to 6q, 6q + 2, 6q + 4 ∵ all are divisible by 2. Therefore, any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.

Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution:
Total number of members in army = 616 and 32 (A band of two groups)
Since two groups are to march in same number of columns and we are to find out the maximum number of columns.
∴ Maximum Number of columns = HCF of 616 and 32
Step 1.
Since 616 > 32, we apply the division Lemma to 616 and 32, to get
616 = 32 × 19 + 8

Step 2.
Since the remainder 8 ≠ 0, we apply the division Lemma to 32 and 8, to get
32 = 8 × 4 + 0.
Since the remainder has now become zero
∵ divisor in the step is 8
∵ HCF of 616 and 32 is 8.
Hence, maximum number of columns in which they can march is 8.

Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint. Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + L]
Solution:
Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2.
If x = 3 q
Squaring both sides,
(x)² = (3q)²
– 9q² = 3 (3q²) = 3m
where m = 3 q²
where m is also an integer
Hence x² = 3m ………… (1)
If x = 3q + 1
Squaring both sides,
x² = (3q + 1)²
x² = 9q² + 1 + 2 × 3q × 1
x² = 3 (3 q² + 2q) + 1
x² = 3m + 1 …. (2)
where m = 3q² + 2q where m is also an integer
From (1) and (2),
x² = 3m, 3m + 1
Hence, square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let x be any positive integer and b = 3
x = 3 q + r where q is quotient and r is remainder
If 0 ≤ r < 3
If r = 0 then x = 3 q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2
x is of the form 3q or 3q + 1 or 3q + 2
If x = 3q
Cubing both sides,
x³ = (3q)³
x³ = 27q³ = 9 (3q³) = 9m
where m = 3q³ and is an integer .
x³ = 9m ……….. (1)
If x = 3q + 1 cubing both sides,
x³ = (3 q +1)³
x³ = 27q³ + 27q² + 9q + 1
= 9 (3q³ + 3q² + q) + 1
= 9m + 1
where m = 3q³ + 3q² + q and is an integer
Again x³ = 9m + 1 …………. (2)
If x = 3q +2
Cubing both sides,
(x)³ = (3q + 2)²
= 27 q³ + 54 q² + 36q + 8
x³ = 9 (3 q³ + 6q² + 4q) + 8
x³ = 9m + 8 ………. (3)
where m = 3 q³ + 6q² + 4q
Again x³ = 9m + 8
From (1) (2), & (3), we find that
x³ can be of the form 9m, 9m + 1, 9m + 8.
Hence, x³ of any positive integer can be of he form 9m, 9m + 1 or 9m + 8