NCERT Solutions Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables

Exercise 3.1

Q.1) Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then.
Also, three years from now, I shall be three times as old as you will be." (Isn't this interesting?) Represent this situation algebraically and graphically
Sol.1) Let present age of Aftab be 𝑥 And,
present age of daughter is represented by 𝑦
Then Seven years ago,
Age of Aftab = 𝑥 − 7
Age of daughter = 𝑦 − 7
According to the question,
(𝑥 − 7) = 7 (𝑦 – 7 )
𝑥 – 7 = 7 𝑦 – 49
𝑥 − 7𝑦 = − 49 + 7
𝑥 – 7𝑦 = − 42 …(i)
𝑥 = 7𝑦 – 42
Putting 𝑦 = 5, 6 and 7, we get
𝑥 = 7 × 5 − 42 = 35 − 42 = − 7
𝑥 = 7 × 6 − 42 = 42 – 42 = 0
𝑥 = 7 × 7 – 42 = 49 – 42 = 7

Three years from now ,
Age of Aftab = x +3
Age of daughter = y +3
According to the question,
(𝑥 + 3) = 3 (𝑦 + 3)
𝑥 + 3 = 3𝑦 + 9
𝑥 − 3𝑦 = 9 − 3
𝑥 − 3𝑦 = 6 …(ii)
𝑥 = 3𝑦 + 6
Putting, 𝑦 = −2, −1 and 0, we get
𝑥 = 3 × − 2 + 6 = −6 + 6 = 0
𝑥 = 3 × − 1 + 6 = −3 + 6 = 3
𝑥 = 3 × 0 + 6 = 0 + 6 = 6

Algebraic representation From equation (i) and (ii)
𝑥 – 7𝑦 = – 42 …(i)
𝑥 − 3𝑦 = 6 …(ii)
Graphical representation

Q.2) The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Sol.2) Let cost of one bat = 𝑅𝑠. 𝑥
Cost of one ball = 𝑅𝑠. 𝑦
3 bats and 6 balls for 𝑅𝑠. 3900
So that 3𝑥 + 6𝑦 = 3900 … (i)
Dividing equation by 3, we get
𝑥 + 2𝑦 = 1300
Subtracting 2𝑦 both side we get
𝑥 = 1300 – 2𝑦
Putting 𝑦 = −1300, 0 and 1300 we get
𝑥 = 1300 – 2 (−1300) = 1300 + 2600 = 3900
𝑥 = 1300 − 2(0) = 1300 − 0 = 1300
𝑥 = 1300 – 2(1300) = 1300 – 2600 = − 1300

Given that she buys another bat and 2 more balls of the same kind for Rs. 1300
So, we get 𝑥 + 2𝑦 = 1300 … (ii)
Subtracting 2𝑦 both side we get
𝑥 = 1300 – 2𝑦
Putting 𝑦 = − 1300, 0 and 1300 we get
𝑥 = 1300 – 2 (−1300)
= 1300 + 2600 = 3900
𝑥 = 1300 – 2 (0)
= 1300 − 0 = 1300
𝑥 = 1300 – 2(1300)
= 1300 – 2600 = −130

Algebraic representation
3𝑥 + 6𝑦 = 3900 … (i)
𝑥 + 2𝑦 = 1300 … (ii)
Graphical representation,

Q.3) The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Sol.3) Let cost each kg of apples = 𝑅𝑠. 𝑥
Cost of each kg of grapes = 𝑅𝑠. 𝑦
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be 𝑅𝑠. 160
So that 2 𝑥 + 𝑦 = 160 … (i)
2𝑥 = 160 − 𝑦
𝑥 = 160 – 𝑦/2
Let 𝑦 = 0 , 80 and 160, we get
𝑥 = 160 – 0/2 = 80
𝑥 = 160− 80/2 = 40

𝑥 = 160 – 2 × 80/2 = 0

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs.300
So we get 4𝑥 + 2𝑦 = 300 … (ii)
Dividing by 2 we get 2𝑥 + 𝑦 = 150
Subtracting 2𝑥 both side, we get
𝑦 = 150 – 2𝑥
Putting 𝑥 = 0 , 50 , 100 we get
𝑦 = 150 – 2 × 0 = 150
𝑦 = 150 – 2 × 50 = 50
𝑦 = 150 – 2 × (100) = −50

Algebraic representation,
2𝑥 + 𝑦 = 160 … (i)
4𝑥 + 2𝑦 = 300 … (ii)
Graphical representation,

Exercise 3.2

Q.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
Sol.1) Let number of boys = 𝑥
Number of girls = 𝑦
Given that total number of student is 10 so that 𝑥 + 𝑦 = 10
Subtract 𝑦 both side we get 𝑥 = 10 – 𝑦
Putting 𝑦 = 0 , 5, 10 we get
𝑥 = 10 – 0 = 10
𝑥 = 10 – 5 = 5
𝑥 = 10 – 10 = 0

Given that If the number of girls is 4 more than the number of boys So that
𝑦 = 𝑥 + 4
Putting 𝑥 = −4, 0, 4, and we get
𝑦 = − 4 + 4 = 0
𝑦 = 0 + 4 = 4
𝑦 = 4 + 4 = 8

Graphical representation
Therefore, number of boys = 3 and number of girls = 7.

ii) 5 pencils and 7 pens together cost Rs.50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Sol. Let cost of pencil = 𝑅𝑠. 𝑥
Cost of pens = 𝑅𝑠. 𝑦
5 pencils and 7 pens together cost 𝑅𝑠. 50, So we get
5𝑥 + 7𝑦 = 50
Subtracting 7𝑦 both sides we get
5𝑥 = 50 – 7𝑦
Dividing by 5 we get

Q.2) On comparing the ratios 𝑎₁/𝑎₂, 𝑏₁/𝑏₂ and 𝑐₁/𝑐₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
Sol.2) (i) 5𝑥 – 4𝑦 + 8 = 0
7𝑥 + 6𝑦 – 9 = 0
Comparing these equation with 𝑎₁𝑥 + 𝑏₁𝑦 + 𝑐₁ = 0
We get 𝑎₁ = 5, 𝑏₁ = −4, and 𝑐₁ = 8
𝑎₂ = 7, 𝑏₂ = 6 and 𝑐₂ = −9

Therefore, both are intersecting lines at one point.
(ii) 9𝑥 + 3𝑦 + 12 = 0
18𝑥 + 6𝑦 + 24 = 0
Comparing these equations with
𝑎₁𝑥 + 𝑏₁𝑦 + 𝑐₁ = 0
𝑎₂𝑥 + 𝑏₂𝑦 + 𝑐₂ = 0
We get 𝑎₁ = 9, 𝑏₁ = 3, and 𝑐₁ = 12
𝑎₂ = 18, 𝑏₂ = 6 and 𝑐₂ = 24

Q.3) On comparing the ratios 𝑎₁/𝑎₂, 𝑏₁/𝑏₂ and 𝑐₁/𝑐₂ 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) 3/2𝑥 + 5/3𝑦 = 7 ; 9𝑥 – 10𝑦 = 14 (iv) 5𝑥 – 3𝑦 = 11 ; – 10𝑥 + 6𝑦 = – 22
(v) 4/3𝑥 + 2𝑦 = 8 ; 2𝑥 + 3𝑦 = 12
Sol.3) (i) 3𝑥 + 2𝑦 = 5 ; 2𝑥 – 3𝑦 = 7

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions.
Hence, the pair of linear equations is consistent.

Q.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
Sol.4) (i) 𝑥 + 𝑦 = 5; 2𝑥 + 2𝑦 = 10

Therefore, these linear equations are parallel to each other and thus, have no possible solution.
Hence, the pair of linear equations is inconsistent.

Q.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
Sol.5) Let length of rectangle = x m
Width of the rectangle = y m
According to the question,
𝑦 − 𝑥 = 4 ... (i)
𝑦 + 𝑥 = 36 ... (ii)
𝑦 − 𝑥 = 4
𝑦 = 𝑥 + 4

Graphical representation
From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

Q.6) Given the linear equation 2𝑥 + 3𝑦 − 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines
Sol.6) (i) Intersecting lines:

Q.7) Draw the graphs of the equations 𝑥 − 𝑦 + 1 = 0 and 3𝑥 + 2𝑦 − 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the xaxis, and shade the triangular region.
Sol.7) 𝑥 − 𝑦 + 1 = 0
𝑥 = 𝑦 – 1

From the figure, it can be observed that these lines are intersecting each other at point
(2, 3) and x-axis at (-1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (-1, 0), and (4, 0).

Exercise 3.3

Q.1) Solve the following pair of linear equations by the substitution method.
(i) 𝑥 + 𝑦 = 14 ; 𝑥 – 𝑦 = 4 (ii) 𝑠 – 𝑡 = 3 ; 𝑠/3 + 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) 3/2𝑥 − 5/3𝑦 = −2 ; 𝑥/3 + 𝑦/2 = 13/6
Sol.1) (i) 𝑥 + 𝑦 = 14                 ... (i)
𝑥 – 𝑦 = 4                         ... (ii)
From equation (i), we get
𝑥 = 14 − 𝑦                         ... (iii)
Putting this value in equation (ii), we get
(14 − 𝑦) − 𝑦 = 4
14 − 2𝑦 = 4
10 = 2𝑦
𝑦 = 5                 ... (iv)
Putting this in equation (iii), we get
𝑥 = 9
∴ 𝑥 = 9 and 𝑦 = 5

(ii) 𝑠 – 𝑡 = 3                 ... (i)
𝑠/3 + 𝑡/2 = 6 .         .. (ii)
From equation (i), we get 𝑡 + 3
Putting this value in equation (ii), we get
𝑡 + 3/3 + 𝑡/2 = 6
2𝑡 + 6 + 3𝑡 = 36
5𝑡 = 30 𝑡 = 30/5          ... (iv)
Putting in equation (iii), we obtain 𝑠 = 9
∴ 𝑠 = 9, 𝑡 = 6

(iii) 3𝑥 − 𝑦 = 3                 ... (i)
9𝑥 − 3𝑦 = 9 .                 .. (ii)
From equation (i), we get
𝑦 = 3𝑥 − 3 .                 .. (iii)
Putting this value in equation (ii), we get
9𝑥 − 3(3𝑥 − 3) = 9
9𝑥 − 9𝑥 + 9 = 9
9 = 9
This is always true. Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by 𝑦 = 3𝑥 − 3
Therefore, one of its possible solutions is 𝑥 = 1, 𝑦 = 0.

(iv) 0.2𝑥 + 0.3𝑦 = 1.3         ... (i)
0.4𝑥 + 0.5𝑦 = 2.3                 ... (ii)
0.2𝑥 + 0.3𝑦 = 1.3
Solving equation (i), we get
0.2𝑥 = 1.3 – 0.3𝑦 Dividing by 0.2, we get
𝑥 = 1.3/0.2 − 0.3/0.2
𝑥 = 6.5 – 1.5 𝑦                 …(iii)
Putting the value in equation (ii), we get
0.4𝑥 + 0.5𝑦 = 2.3
(6.5 – 1.5𝑦) × 0.4𝑥 + 0.5𝑦 = 2.3
2.6 – 0.6𝑦 + 0.5𝑦 = 2.3
−0.1𝑦 = 2.3 – 2.6
𝑦 = (−0.3/−0.1) 𝑦 = 3
Putting this value in equation (iii) we get
𝑥 = 6.5 – 1.5 𝑦
𝑥 = 6.5 – 1.5(3)
𝑥 = 6.5 − 4.5
𝑥 = 2
∴ 𝑥 = 2 and 𝑦 = 3

v) √2𝑥 + √3𝑦 = 0           ….. (i)
√3𝑥 − √8𝑦 = 0              ….. (ii)
From equation (i), we get

Q.2) Solve 2𝑥 + 3𝑦 = 11 and 2𝑥 − 4𝑦 = − 24 and hence find the value of ′𝑚′ for which 𝑦 = 𝑚𝑥 + 3.
Sol.2) 2𝑥 + 3𝑦 = 11 ... (i)
Subtracting 3y both side we get
2𝑥 = 11 – 3𝑦 … (ii)
Putting this value in equation second we get
2𝑥 – 4𝑦 = – 24 … (iii)
11 − 3𝑦 – 4𝑦 = − 24
7𝑦 = − 24 – 11
−7𝑦 = − 35
𝑦 = − (35/−7)
𝑦 = 5
Putting this value in equation (iii) we get
2𝑥 = 11 – 3 × 5
2𝑥 = 11 − 15
2𝑥 = − 4
Dividing by 2 we get 𝑥 = − 2
Putting the value of x and y
𝑦 = 𝑚𝑥 + 3.
5 = −2𝑚 + 3
2𝑚 = 3 – 5
𝑚 = − 2/2
𝑚 = −1

Q.3) i) 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.
Sol.3) i)
Let larger number = x
Smaller number = y
The difference between two numbers is
26 𝑥 – 𝑦 = 26
𝑥 = 26 + 𝑦
Given that one number is three times the other So
𝑥 = 3𝑦
Putting the value of 𝑥 we get,
26𝑦 = 3𝑦
−2𝑦 = − 26
𝑦 = 13
So value of 𝑥 = 3𝑦 Putting value of 𝑦, we get
𝑥 = 3 × 13 = 39
Hence the numbers are 13 and 39.

Q.3) ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Sol. Let first angle = 𝑥
And second number = 𝑦
As both angles are supplementary so that sum will
180 𝑥 + 𝑦 = 180
𝑥 = 180 − 𝑦 ... (i)
Difference is 18 degrees so that
𝑥 – 𝑦 = 18
Putting the value of 𝑥 we get
180 – 𝑦 – 𝑦 = 18
− 2𝑦 = −162
𝑦 = − (162/−2)
𝑦 = 81
Putting the value back in equation (i), we get 𝑥 = 180 – 81 = 99
Hence, the angles are 99° and 81°.

Q.3) iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Sol. Let cost of each bat = 𝑅𝑠. 𝑥
Cost of each ball = 𝑅𝑠. 𝑦
Given that coach of a cricket team buys 7 bats and 6 balls for Rs. 3800.
7𝑥 + 6𝑦 = 3800
6𝑦 = 3800 – 7𝑥 .
Dividing by 6, we get
𝑦 = 3800 – 7𝑥/6 … (i)
Given that she buys 3 bats and 5 balls for Rs. 1750 later.
3𝑥 + 5𝑦 = 1750
Putting the value of 𝑦
3𝑥 + 5 ( 3800 – 7𝑥/6) = 1750
Multiplying by 6, we get
18𝑥 + 19000 – 35𝑥 = 10500
−17𝑥 = 10500 − 19000
−17𝑥 = −8500
𝑥 = − 8500/−17
𝑥 = 500
Putting this value in equation (i) we get
𝑦 = 3800 – 7 × 500
6/𝑦 = 300/6 
𝑦 = 50
Hence cost of each bat = Rs. 500 and cost of each balls = Rs. 50.

Q.3) 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 Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km?
How much does a person have to pay for traveling a distance of 25 km ?
Sol. Let the fixed charge for taxi = 𝑅𝑠. 𝑥
And variable cost per km = 𝑅𝑠. 𝑦
Total cost = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs. 105
𝑥 + 10𝑦 = 105 … (i)
𝑥 = 105 – 10𝑦
Given that for a journey of 15 km, the charge paid is Rs. 155
𝑥 + 15𝑦 = 155
Putting the value of 𝑥 we get
105 – 10𝑦 + 15𝑦 = 155
5𝑦 = 155 – 105
5𝑦 = 50
Dividing by 5, we get
𝑦 = 50/5 = 10
Putting this value in equation (i) we get
𝑥 = 105 – 10 × 10
𝑥 = 5
People have to pay for traveling a distance of 25 𝑘𝑚
= 𝑥 + 25𝑦
= 5 + 25 × 10
= 5 + 250 = 255
A person have to pay Rs 255 for 25 Km.

Q.3) v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Sol. Let Numerator = 𝑥
Denominator = 𝑦
Fraction will = 𝑥/𝑦
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
𝑥 + 2/𝑦+2 = 9/11
By Cross multiplication, we get
11𝑥 + 22 = 9𝑦 + 18
Subtracting 22 both side, we get
11𝑥 = 9𝑦 – 4
Dividing by 11, we get
𝑥 = 9𝑦 – 4/11 … (i)
Given that 3 is added to both the numerator and the denominator it becomes 5/6
If, 3 is added to both the numerator and the denominator it becomes 5/6
𝑥+3/𝑦 + 3 = 5/6 … (ii)
By Cross multiplication, we get
6𝑥 + 18 = 5𝑦 + 15
Subtracting the value of 𝑥, we get
6(9𝑦 – 4)/11
+ 18 = 5𝑦 + 15
Subtract 18 both side we get
6(9𝑦 – 4/11)
= 5𝑦 – 3
54 – 24 = 55𝑦
− 33 − 𝑦 = −9
𝑦 = 9
Putting this value of 𝑦 in equation (i), we get
𝑥 = 9𝑦 – 4/11 … (i)
𝑥 = 81 – 4/77
𝑥 = 77/11
𝑥 = 7
Hence our fraction is 7/9
.
Q.3) 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?
Sol. Let present age of Jacob = 𝑥 𝑦𝑒𝑎𝑟
And present Age of his son is = 𝑦 𝑦𝑒𝑎𝑟
Five years hence,
Age of Jacob will = 𝑥 + 5 𝑦𝑒𝑎𝑟
Age of his son will = 𝑦 + 5
year Given that the age of Jacob will be three times that of his son
𝑥 + 5 = 3(𝑦 + 5)
Adding 5 both side, we get
𝑥 = 3𝑦 + 15 − 5
𝑥 = 3𝑦 + 10 … (i)
Five years ago, Age of Jacob will = 𝑥 − 5 𝑦𝑒𝑎𝑟
Age of his son will = 𝑦 − 5
year Jacob’s age was seven times that of his son
𝑥 – 5 = 7(𝑦 − 5)
Putting the value of 𝑥 from equation (i) we get
3𝑦 + 10 – 5 = 7𝑦 – 35
3𝑦 + 5 = 7𝑦 – 35
3𝑦 – 7𝑦 = −35 – 5
−4𝑦 = − 40
𝑦 = −40/−4𝑦 = 10 𝑦𝑒𝑎𝑟
Putting the value of 𝑦 in equation first we get
𝑥 = 3 × 10 + 10𝑥 = 40 𝑦𝑒𝑎𝑟𝑠
Hence, Present age of Jacob = 40 𝑦𝑒𝑎𝑟𝑠 and present age of his son = 10 𝑦𝑒𝑎𝑟𝑠.

Exercise 3.4

Q.1) Solve the following pair of linear 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 & 9𝑥 = 2𝑦 + 7 (iv) 𝑥/2 + 2𝑦/3 = − 1 and 𝑥 – 𝑦/3 = 3
Sol.1) (i) 𝑥 + 𝑦 = 5 and 2𝑥 – 3𝑦 = 4
By elimination method
𝑥 + 𝑦 = 5 ... (i)
2𝑥 – 3𝑦 = 4 ... (ii)
Multiplying equation (i) by (ii), we get
2𝑥 + 2𝑦 = 10 ... (iii)
Subtracting equation (ii) from equation (iii), we get
5𝑦 = 6 𝑦 = 6/5
Putting the value in equation (i), we get
𝑥 = 5 − (6/5) = 19/5
Hence, 𝑥 = 19/5
and 𝑦 = 6/5
By substitution method
𝑥 + 𝑦 = 5 ... (i) Subtracting 𝑦 both side, we get
𝑥 = 5 − 𝑦 ... (iv)
Putting the value of 𝑥 in equation (ii) we get
2(5 – 𝑦)– 3𝑦 = 4
−5𝑦 = − 6 𝑦
= −6/−5
= 6/5
Putting the value of 𝑦 in equation (iv) we get
𝑥 = 5 – 6/5
𝑥 = 19/5
Hence, 𝑥 = 19/5 and 𝑦 = 6/5 again

(ii) 3𝑥 + 4𝑦 = 10 and 2𝑥 – 2𝑦 = 2
By elimination method
3𝑥 + 4𝑦 = 10 .... (i)
2𝑥 – 2𝑦 = 2 ... (ii)
Multiplying equation (ii) by 2, we get
4𝑥 – 4𝑦 = 4 ... (iii)
Adding equation (i) and (iii), we get
7𝑥 + 0 = 14
Dividing both side by 7, we get
𝑥 = 14/7 = 2
Putting in equation (i), we get
3𝑥 + 4𝑦 = 10
3(2) + 4𝑦 = 10
6 + 4𝑦 = 10
4𝑦 = 10 – 6
4𝑦 = 4
𝑦 = 4/4 = 1
Hence, answer is 𝑥 = 2, 𝑦 = 1
By substitution method
3𝑥 + 4𝑦 = 10 ... (i)
Subtract 3𝑥 both side, we get
4𝑦 = 10 – 3𝑥
Divide by 4 we get
𝑦 = 10 − 3𝑥/4
Putting this value in equation (ii), we get
2𝑥 – 2𝑦 = 2 ... (i)
2𝑥 – 2 (10 − 3𝑥 )/4 = 2
Multiply by 4 we get
8𝑥 − 2(10 – 3𝑥) = 8
8𝑥 − 20 + 6𝑥 = 8
14𝑥 = 28
𝑥 = 28/14
= 2/𝑦 = 10 − 3𝑥 
4/𝑦 =
4/4 = 1
Hence, answer is 𝑥 = 2, 𝑦 = 1 again.

(iii) 3𝑥 – 5𝑦 – 4 = 0 and 9𝑥 = 2𝑦 + 7
By elimination method
3𝑥 – 5𝑦 – 4 = 0
3𝑥 – 5𝑦 = 4 ...(i)
9𝑥 = 2𝑦 + 7 ... (ii)
Multiplying equation (i) by 3, we get
9 𝑥 – 15𝑦 = 11 ... (iii)
Subtracting equation (ii) from equation (iii), we get
−13𝑦 = 5
𝑦 = − 5/13
Putting value in equation (i), we get

Multiplying by 6, we get
9 + 𝑦 + 4𝑦 = − 6
5𝑦 = −15
𝑦 = − 3
Hence our answer is 𝑥 = 2 and 𝑦 = −3.

Q.2) i) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: 
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Sol.2) i) (i) Let the fraction be 𝑥/𝑦
According to the question,
𝑥 + 1/𝑦 − 1 = 1
⇒ 𝑥 − 𝑦 = −2               ... (i)
𝑥/𝑦 + 1 = 1/2
⇒ 2𝑥 − 𝑦 = 1               ... (ii)
Subtracting equation (i) from equation (ii), we get
𝑥 = 3 ... (iii)
Putting this value in equation (i), we get
3 − 𝑦 = −2
−𝑦 = −5
𝑦 = 5
Hence, the fraction is 3/5

Q.2) ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Sol. Let present age of Nuri = 𝑥
and present age of Sonu = 𝑦
According to the given information, question,
(𝑥 − 5) = 3(𝑦 − 5)
𝑥 − 3𝑦 = −10               ... (i)
(𝑥 + 10𝑦) = 2(𝑦 + 10)
𝑥 − 2𝑦 = 10               ... (ii)
Subtracting equation (i) from equation (ii), we get
𝑦 = 20               ... (iii)
Putting this value in equation (i), we get
𝑥 − 60 = −10
𝑥 = 50
Hence, age of Nuri = 50 𝑦𝑒𝑎𝑟𝑠 and age of Sonu = 20 𝑦𝑒𝑎𝑟𝑠

Q.2) iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Sol. Let the unit digit and tens digits of the number be 𝑥 and 𝑦 respectively. Then,
number = 10𝑦 + 𝑥
Number after reversing the digits = 10𝑥 + 𝑦
According to the question,
𝑥 + 𝑦 = 9               ... (i)
9(10𝑦 + 𝑥) = 2(10𝑥 + 𝑦)
88𝑦 − 11𝑥 = 0
− 𝑥 + 8𝑦 = 0               ... (ii)
Adding equation (i) and (ii), we get
9𝑦 = 9 𝑦 = 1               ... (iii)
Putting the value in equation (i), we get 𝑥 = 8
Hence, the number is 10𝑦 + 𝑥 = 10 × 1 + 8 = 18.

Q.2) iv) Meena went to bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs. 100 she received.
Sol. Let the number of Rs. 50 notes and Rs. 100 notes be 𝑥 and 𝑦 respectively.
According to the question,
𝑥 + 𝑦 = 25               ... (i)
50𝑥 + 100𝑦 = 2000               ... (ii)
Multiplying equation (i) by 50, we get
50𝑥 + 50𝑦 = 1250               ... (iii)
Subtracting equation (iii) from equation (ii), we get
50𝑦 = 750
𝑦 = 15
Putting this value in equation (i), we have 𝑥 = 10
Hence, Meena has 10 notes of Rs. 50 and 15 notes of Rs. 100.

Q.2) v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Sol. Let the fixed charge for first three days and each day charge thereafter be 𝑅𝑠. 𝑥 and 𝑅𝑠. 𝑦 respectively.
According to the question,
𝑥 + 4𝑦 = 27               ... (i)
𝑥 + 2𝑦 = 21               ... (ii)
Subtracting equation (ii) from equation (i), we get
2𝑦 = 6
𝑦 = 3               ... (iii)
Putting in equation (i), we get
𝑥 + 12 = 27
𝑥 = 15
Hence, fixed charge = 𝑅𝑠. 15 and Charge per day = Rs. 3.

Exercise 3.5

Q.1) Which of the following pairs of linear equations has unique solution, no solution or infinitely 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
Sol.1) (i) 𝑥 – 3𝑦 – 3 = 0 ; 3𝑥 – 9𝑦 – 2 = 0

Q.2) (i) For which values of 𝑎 and 𝑏 does the following pair of linear equations have an infinite number of solutions?
2𝑥 + 3𝑦 = 7
(𝑎 – 𝑏)𝑥 + (𝑎 + 𝑏)𝑦 = 3𝑎 + 𝑏 – 2
Sol.2) 2𝑥 + 3𝑦 = 7
(𝑎 – 𝑏)𝑥 + (𝑎 + 𝑏)𝑦 = 3𝑎 + 𝑏 – 2

𝑎 − 5𝑏 = 0 ... (ii)
Subtracting equation (i) from (ii), we get
4𝑏 = 4
𝑏 = 1
Putting this value in equation (ii), we get
𝑎 − 5 × 1 = 0
𝑎 = 5
Hence, 𝑎 = 5 and 𝑏 = 1 are the values for which the given equations give infinitely many solutions.

Q.2) ii)For which value of 𝑘 will the following pair of linear equations have no solution?
3𝑥 + 𝑦 = 1
(2𝑘 – 1)𝑥 + (𝑘 – 1)𝑦 = 2𝑘 + 1
Sol. 3𝑥 + 𝑦 − 1 = 0
(2𝑘 – 1)𝑥 + (𝑘 – 1)𝑦 − (2𝑘 + 1) = 0

3𝑘 − 3 = 2𝑘 − 1
𝑘 = 2
Hence, for 𝑘 = 2, the given equation has no solution.

Q.3) Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8𝑥 + 5𝑦 = 9
3𝑥 + 2𝑦 = 4
Sol.3) 8𝑥 + 5𝑦 = 9        ... (i)
3𝑥 + 2𝑦 = 4        ... (ii)
From equation (ii), we get
𝑥 = 4 − 2𝑦/3        ... (iii)
Putting this value in equation (i), we get
8 (4 − 2𝑦/3 ) + 5𝑦 = 9
32 − 16𝑦 + 15𝑦 = 27 
−𝑦 = −5
𝑦 = 5        ... (iv)
Putting this value in equation (ii), we get
3𝑥 + 10 = 4
𝑥 = −2
Hence, 𝑥 = −2, 𝑦 = 5
By cross multiplication again, we get
8𝑥 + 5𝑦 − 9 = 0
3𝑥 + 2𝑦 − 4 = 0

Q.4) i) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
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 Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.
Sol.4) i) Let 𝑥 be the fixed charge of the food and y be the charge for food per day. According to
the question,
𝑥 + 20𝑦 = 1000 ... (i)
𝑥 + 26𝑦 = 1180 ... (ii)
Subtracting equation (i) from equation (ii), we get
6𝑦 = 180
𝑦 = 180/6 = 30
Putting this value in equation (i), we get
𝑥 + 20 × 30 = 1000
𝑥 = 1000 − 600
𝑥 = 400
Hence, fixed charge = Rs.400 and charge per day = Rs.30

Q.4) ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
Sol.4) ii) Let the fraction be 𝑥/𝑦
According to the question,
𝑥 − 1/𝑦 = 1/3
⇒ 3𝑥 − 𝑦 = 3... (i)
𝑥/𝑦 + 8 = 1/4
⇒ 4𝑥 − 𝑦 = 8 ... (ii)
Subtracting equation (i) from equation (ii), we get
x = 5 ... (iii)
Putting this value in equation (i), we get
15 − 𝑦 = 3
𝑦 = 12
Hence, the fraction is 5/12.
Q.4) 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?
Sol.4) iii) Let the number of right answers and wrong answers be 𝑥 and 𝑦 respectively.
According to the question,
3𝑥 − 𝑦 = 40 ... (i)
4𝑥 − 2𝑦 = 50
⇒ 2𝑥 − 𝑦 = 25 ... (ii)
Subtracting equation (ii) from equation (i), we get
𝑥 = 15 ... (iii)
Putting this value in equation (ii), we get
30 − 𝑦 = 25
𝑦 = 5
Therefore, number of right answers = 15 And number of wrong answers = 5
Total number of questions = 20

Q.4) 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 different 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?
Sol.4) iv) Let the speed of 1st car and 2nd car be 𝑢 𝑘𝑚/ℎ and 𝑣 𝑘𝑚/ℎ.
Respective speed of both cars while they are travelling in same direction = (𝑢 − 𝑣)𝑘𝑚/ℎ
Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (𝑢 + 𝑣) 𝑘𝑚/ℎ
According to the question,
5(𝑢 − 𝑣) = 100
⇒ 𝑢 − 𝑣 = 20 ... (i)
1(𝑢 + 𝑣) = 100 ... (ii)
Adding both the equations, we get
2𝑢 = 120
𝑢 = 60 𝑘𝑚/ℎ ... (iii)
Putting this value in equation (ii), we obtain 𝑣 = 40 𝑘𝑚/ℎ
Hence, speed of one car = 60 𝑘𝑚/ℎ and speed of other car = 40 𝑘𝑚/ℎ

Q.4) 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 square units. Find the dimensions of the rectangle.
Sol.4) v) Let length and breadth of rectangle be 𝑥 unit and 𝑦 unit respectively.
Area = xy
According to the question,
(𝑥 − 5)(𝑦 + 3) = 𝑥𝑦 − 9
⇒ 3𝑥 − 5𝑦 − 6 = 0 ... (i)
(𝑥 + 3)(𝑦 + 2) = 𝑥𝑦 + 67
⇒ 2𝑥 − 3𝑦 − 61 = 0 ... (ii)
By cross multiplication, we get
𝑥/305 − (−18) = 𝑦/−12 − (−183) = 1/9 − (−10) 
𝑥/323 = 𝑦/171 = 1/19
𝑥 = 17, 𝑦 = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

Exercise 3.6

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

Putting 1/𝑥 = 𝑝 and 1/𝑦 = 𝑞 in (i) and (ii) we get,
7𝑞 − 2𝑝 = 5 ... (iii)
8𝑞 + 7𝑝 = 15 ... (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49𝑞 − 14𝑝 = 35 ... (v)
16𝑞 + 14𝑝 = 30 ... (vi)
Now, adding equation (v) and (vi) we get,
49𝑞 − 14𝑝 + 16𝑞 + 14𝑝 = 35 + 30
⇒ 65𝑞 = 65 ⇒ 𝑞 = 1
Putting the value of 𝑞 in equation (iv)
8 + 7𝑝 = 15
⇒ 7𝑝 = 7
⇒ 𝑝 = 1

3𝑥 − 𝑦 = 2 ... (iv)
Adding equations (iii) and (iv), we get
6𝑥 = 6
𝑥 = 1 ... (v)
Putting the value in equation (iii), we get
3(1) + 𝑦 = 4
𝑦 = 1
Hence, 𝑥 = 1 and 𝑦 = 1

Q.2) (i) Formulate the following problems as a pair of equations, and hence find their solutions:
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.
Sol.2) i) Let the speed of Ritu in still water and the speed of stream be 𝑥 𝑘𝑚/ℎ and 𝑦 𝑘𝑚/ℎ respectively.
Speed of Ritu while rowing Upstream = (𝑥 − 𝑦) 𝑘𝑚/ℎ
Downstream = (𝑥 + 𝑦) 𝑘𝑚/ℎ
According to question,
2(𝑥 + 𝑦) = 20 ⇒ 𝑥 + 𝑦 = 10 ... (i)
2(𝑥 − 𝑦) = 4 ⇒ 𝑥 − 𝑦 = 2 ... (ii)
Adding equation (i) and (ii), we get
Putting this equation in (i), we get 𝑦 = 4
Hence, Ritu's speed in still water is 6 𝑘𝑚/ℎ and the speed of the current is 4 𝑘𝑚/ℎ.

Q.2) 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.
Sol. Let the number of days taken by a woman and a man be 𝑥 and 𝑦 respectively.
Therefore, work done by a woman in 1 𝑑𝑎𝑦 = 1/𝑥
According to the question,

Hence, number of days taken by a woman = 18 and
number of days taken by a man = 36

Q.2) iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and 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.
Sol.2 Let the speed of train and bus be 𝑢 𝑘𝑚/ℎ and 𝑣 𝑘𝑚/ℎ respectively.
According to the given information,
60/𝑢 + 240/𝑣 = 4 ... (i)
100/𝑢 + 200/𝑣 = 25/6 ... (ii)
Putting 1/𝑢 = 𝑝 and 1/𝑣 = 𝑞 in the equations, we get
60𝑝 + 240𝑞 = 4 ... (iii)
100𝑝 + 200𝑞 = 25/6 
600𝑝 + 1200𝑞 = 25 ... (iv)
Multiplying equation (iii) by 10, we get
600𝑝 + 2400𝑞 = 40 .... (v)
Subtracting equation (iv) from (v), we get
1200𝑞 = 15
𝑞 = 15/200 = 1/80 ... (vi)
Putting equation (iii), we get
60𝑝 + 3 = 4
60𝑝 = 1
𝑝 = 1/60
𝑝 = 1/𝑢  = 1/60 and 𝑞 = 1/𝑣 = 1/80
𝑢 = 60 and 𝑣 = 80
Hence, speed of train = 60 𝑘𝑚/ℎ and speed of bus = 80 𝑘𝑚/ℎ.

Exercise 3.7

Q.1) The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani 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.
Sol.1) Let the age of Ani and Biju be 𝑥 𝑦𝑒𝑎𝑟𝑠 and 𝑦 𝑦𝑒𝑎𝑟𝑠 respectively.
Age of Dharam = 2𝑥 𝑦𝑒𝑎𝑟𝑠 and Age of Cathy = (𝑦/2) 𝑦𝑒𝑎𝑟𝑠
According to question, 𝑥 – 𝑦 = 3             … (1)
And 2𝑥 − (𝑦/2) = 30
⇒ 4𝑥 – 𝑦 = 60                                        … (2)
Subtracting (1) from (2), we obtain:
3𝑥 = 60 – 3 = 57
⇒ 𝑥 = Age of Ani = 19 𝑦𝑒𝑎𝑟𝑠
Age of Biju = 19 – 3 = 16 𝑦𝑒𝑎𝑟𝑠
Again, According to question, 𝑦 – 𝑥 = 3       … (3)
And 2𝑥 − (𝑦/2) = 30
⇒ 4𝑥 – 𝑦 = 60… (4)
Adding (3) and (4), we obtain:
3𝑥 = 63 ⇒ 𝑥 = 21
Age of Ani = 21 𝑦𝑒𝑎𝑟𝑠
Age of Biju = 21 + 3 = 24 𝑦𝑒𝑎𝑟𝑠

Q.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]
Sol.2) Let the money with the first person and second person be 𝑅𝑠. 𝑥 and 𝑅𝑠. 𝑦 respectively.
According to the question,
𝑥 + 100 = 2(𝑦 – 100)
⇒ 𝑥 + 100 = 2𝑦 – 200
⇒ 𝑥 – 2𝑦 = – 300                 … (1)
Again, 6(𝑥 – 10) = (𝑦 + 10)
⇒ 6𝑥 – 60 = 𝑦 + 10
⇒ 6𝑥 – 𝑦 = 70                      … (2)
Multiplying equation (2) by 2, we obtain:
12𝑥 – 2𝑦 = 140                    … (3)
Subtracting equation (1) from equation (3), we obtain:
11𝑥 = 140 + 300
⇒ 11𝑥 = 440
⇒ 𝑥 = 40
Putting the value of x in equation (1), we obtain:
40 – 2𝑦 = – 300
⇒ 40 + 300 = 2𝑦
⇒ 2𝑦 = 340
⇒ 𝑦 = 170
Thus, the two friends had Rs. 40 and Rs. 170 with them.

Q.3) A train covered a certain distance at a uniform speed. If the train would have been 10 km/h 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.
Sol.3) Let the speed of the train be 𝑥 𝑘𝑚/ℎ and the time taken by train to travel the given distance be 𝑡 ℎ𝑜𝑢𝑟𝑠 and the distance to travel be 𝑑 𝑘𝑚.
Since Speed = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑/𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡ℎ𝑎𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
⇒ 𝑥 = 𝑑/𝑡
⇒ 𝑑 = 𝑥𝑡        ……….. (i)
According to the question
𝑥 + 10 = 𝑑/𝑡−2
⇒ (𝑥 + 10)(𝑡 − 2) = 𝑑
⇒ 𝑥𝑡 + 10𝑡 − 2𝑥 − 20 = 𝑑
⇒ − 2𝑥 + 10𝑡 = 20      ………… (ii) [Using eq. (1)]
Again, 𝑥 − 10 = 𝑑/𝑡+3
⇒ (𝑥 − 10)(𝑡 + 3) = 𝑑
⇒ 𝑥𝑡 − 10𝑡 + 3𝑥 − 30 = 𝑑
⇒ 3𝑥 − 10𝑡 = 30       …….. (iii) [Using eq. (1)]
Adding equations (2) and (3), we obtain: 𝑥 = 50
Substituting the value of 𝑥 in equation (2), we obtain:
(−20) × (50) + 10𝑡 = 20
⇒ – 100 + 10𝑡 = 20
⇒ 10𝑡 = 120𝑡 = 12
From equation (1), we obtain:
⇒ 𝑑 = 𝑥𝑡 = 50 × 12 = 600
Thus, the distance covered by the train is 600 km.

Q.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.
Sol.4) Let the number of rows be 𝑥 and number of students in a row be 𝑦.
Total number of students in the class = Number of rows 𝑥
Number of students in a row = 𝑥𝑦
According to the question,
Total number of students = (𝑥 – 1) (𝑦 + 3)
⇒ 𝑥𝑦 = (𝑥 – 1) (𝑦 + 3)
⇒ 𝑥𝑦 = 𝑥𝑦 – 𝑦 + 3𝑥 – 3
⇒ 3𝑥 – 𝑦 – 3 = 0
⇒ 3𝑥 – 𝑦 = 3… (1)
Total number of students = (𝑥 + 2) (𝑦 – 3)
⇒ 𝑥𝑦 = 𝑥𝑦 + 2𝑦 – 3𝑥 – 6
⇒ 3𝑥 – 2𝑦 = – 6… (2)
Subtracting equation (2) from (1), we obtain:       𝑦 = 9
Substituting the value of y in equation (1), we obtain:
3𝑥 – 9 = 3
⇒ 3𝑥 = 9 + 3 = 12
⇒ 𝑥 = 4
Number of rows = 𝑥 = 4
Number of students in a row = 𝑦 = 9
Hence, Total number of students in a class = 𝑥𝑦 = 4 × 9 = 36

Q.5) In a Δ𝐴𝐵𝐶, ∠𝐶 = 3 ∠𝐵 = 2 (∠𝐴 + ∠𝐵). Find the three angles.
Sol.5) Let ∠𝐶 = 3, ∠𝐵 = 2∠(𝐴 + 𝐵) = 𝑥
∠𝐶 = 𝑥, -----(1)
3∠𝐵 = 𝑥

Hence the measures of ∠𝐴 = 20°, ∠𝐵 = 40°and ∠𝐶 = 120°respectively.

Q.6) Draw the graphs of the equations 5𝑥 − 𝑦 = 5 and 3𝑥 − 𝑦 = 3 Determine the coordinate of the vertices of the triangle formed by these lines and the y-axis.
Sol.6) 5𝑥 − 𝑦 = 5
⇒ 𝑦 = 5𝑥 − 5
Three solutions of this equation can be written in a table as follows:

It can be observed that the required triangle is Δ𝐴𝐵𝐶.
The coordinates of its vertices are A (1, 0), B (0, –3), C (0, –5).

Q.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) 𝑥/𝑎 − 𝑦/𝑏 = 0
𝑎𝑥 + 𝑏𝑦 = 𝑎² + 𝑏²
(iv) (𝑎 − 𝑏) 𝑥 + (𝑎 + 𝑏) 𝑦 = 𝑎² − 2𝑎𝑏 − 𝑏²
(𝑎 + 𝑏) (𝑥 + 𝑦) = 𝑎² + 𝑏_²
(v) 152𝑥 − 378𝑦 = − 74
− 378𝑥 + 152𝑦 = − 604
Sol.7) (i)𝑝𝑥 + 𝑞𝑦 = 𝑝 − 𝑞 … (1)
𝑞𝑥 − 𝑝𝑦 = 𝑝 + 𝑞 … (2)
Multiplying equation (1) by p and equation (2) by q, we obtain
𝑝²𝑥 + 𝑝𝑞𝑦 = 𝑝² − 𝑝𝑞 … (3)
𝑞²𝑥 − 𝑝𝑞𝑦 = 𝑝𝑞 + 𝑞² … (4)
Adding equations (3) and (4), we obtain
𝑝²𝑥 +𝑞²𝑥 = 𝑝² + 𝑞²
(𝑝² + 𝑞²) 𝑥 = 𝑝² + 𝑞²
𝑥 = 𝑝²+ 𝑞²/𝑝²+𝑞² = 1
From equation (1), we obtain
𝑝 (1) + 𝑞𝑦 = 𝑝 − 𝑞
𝑞𝑦 = − 𝑞
𝑦 = − 1
(ii) 𝑎𝑥 + 𝑏𝑦 = 𝑐               … (1)
𝑏𝑥 + 𝑎𝑦 = 1 + 𝑐                … (2)
Multiplying equation (1) by a and equation (2) by 𝑏, we obtain
𝑎²𝑥 + 𝑎𝑏𝑦 = 𝑎𝑐                    … (3)
𝑏₂𝑥 + 𝑎𝑏𝑦 = 𝑏 + 𝑏𝑐                 … (4)
Subtracting equation (4) from equation (3),
(𝑎² − 𝑏²) 𝑥 = 𝑎𝑐 − 𝑏𝑐 – 𝑏

(iii) 𝑥/𝑎 − 𝑦/𝑏 = 0
Or, 𝑏𝑥 − 𝑎𝑦 = 0 … (1)
𝑎𝑥 + 𝑏𝑦 = 𝑎² + 𝑏² … (2)
Multiplying equation (1) and (2) by b and a respectively, we obtain
𝑏² − 𝑎𝑏𝑦 = 0 … (3)
𝑎²𝑥 + 𝑎𝑏𝑦 = 𝑎² + 𝑎𝑏² … (4)
Adding equations (3) and (4), we obtain
𝑏²𝑥 + 𝑎²𝑥 = 𝑎3 + 𝑎𝑏²
𝑥 (𝑏²+ 𝑎²) = 𝑎 (𝑎² + 𝑏²)
𝑥 = 𝑎
By using (1), we obtain
𝑏 (𝑎) − 𝑎𝑦 = 0
𝑎𝑏 − 𝑎𝑦 = 0
𝑎𝑦 = 𝑎𝑏
𝑦 = 𝑏

(iv) (𝑎 − 𝑏) 𝑥 + (𝑎 + 𝑏) 𝑦 = 𝑎² − 2𝑎𝑏 − 𝑏² … (1)
(𝑎 + 𝑏) (𝑥 + 𝑦) = 𝑎² + 𝑏²
(𝑎 + 𝑏) 𝑥 + (𝑎 + 𝑏) 𝑦 = 𝑎² + 𝑏² … (2)
Subtracting equation (2) from (1), we obtain

𝑥 = 𝑎 + 𝑏
Using equation (1), we obtain
(𝑎 − 𝑏) (𝑎 + 𝑏) + (𝑎 + 𝑏) 𝑦 = 𝑎² − 2𝑎𝑏 − 𝑏²
𝑎² − 𝑏² + (𝑎 + 𝑏) 𝑦 = 𝑎² − 2𝑎𝑏 − 𝑏²
(𝑎 + 𝑏) 𝑦 = − 2𝑎𝑏
𝑦 = −2𝑎𝑏/𝑎+𝑏
(v) 152𝑥 − 378𝑦 = − 74
76𝑥 − 189𝑦 = − 37
𝑥 = 189𝑦−37/76 … (1)
− 378𝑥 + 152𝑦 = − 604
− 189𝑥 + 76𝑦 = − 302 … (2)
Substituting the value of 𝑥 in equation (2), we obtain
−189 ( 189𝑦−37/76) + 76𝑦 = −302
− (189)² 𝑦 + 189 × 37 + (76)2 𝑦 = − 302 × 76
189 × 37 + 302 × 76 = (189)2 𝑦 − (76)2 𝑦
6993 + 22952 = (189 − 76) (189 + 76) 𝑦
29945 = (113) (265) 𝑦
𝑦 = 1
From equation (1), we obtain
𝑥 = 189(1)−37/76
𝑥 = (189−37)/76 = 152/76
𝑥 = 2

Q.8) ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

Sol.8) Therefore, ∠𝐴 + ∠𝐶 = 180
4𝑦 + 20 − 4𝑥 = 180
− 4𝑥 + 4𝑦 = 160
𝑥 − 𝑦 = − 40 ……(i)
Also, ∠𝐵 + ∠𝐷 = 180
3𝑦 − 5 − 7𝑥 + 5 = 180
− 7𝑥 + 3𝑦 = 180 ………(ii)
Multiplying equation (i) by 3, we obtain
3𝑥 − 3𝑦 = − 120 ……..(iii)
Adding equations (ii) and (iii), we obtain
− 7𝑥 + 3𝑥 = 180 − 120
− 4𝑥 = 60
𝑥 = −15
By using equation (i), we obtain
𝑥 − 𝑦 = − 40
−15 − 𝑦 = − 40
𝑦 = −15 + 40 = 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°