CBSE - Pair of Linear Equations in Two Variables

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

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 = -1300
 

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

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,

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

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
x = 10 - 7 y /5
Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = - 4

x = 10 – 7 × 15/5 = 10 – 21 = - 11

Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get
y = 46/5 - 7x/5
y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6
 

Graphical representation

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.

(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

Comparing these equation with
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂= 0

We get
a₁ = 5, b₁ = -4, and c₁ = 8
a₂ =7, b₂ = 6 and c₂ = -9
a₁/a₂ = 5/7,
b₁/b₂ = -4/6 and
c₁/c₂ = 8/-9
Hence, a₁/a₂ ≠ b₁/b₂

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
a₁/a₂ = 9/18 = 1/2
b₁/b₂ = 3/6 = 1/2 and
c₁/c₂ = 12/24 = 1/2
Hence, a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, both lines are coincident

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂= 0

We get
a₁ = 6, b₁ = -3, and c₁ = 10
a₂ = 2, b₂ = -1 and c₂ = 9
a₁/a₂ = 6/2 = 3/1
b₁/b₂ = -3/-1 = 3/1 and
c₁/c₂ = 12/24 = 1/2
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, both lines are parallel

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 

Answer

(i) 3x + 2y = 5 ; 2x – 3y = 7
a₁/a₂ = 3/2
b₁/b₂ = -2/3 and
c₁/c₂ = 5/7
Hence, a₁/a₂ ≠ b₁/b₂
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -3/-6 = 1/2 and
c₁/c₂ = 8/9
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

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

(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
a₁/a₂ = 3/2/9 = 1/6
b₁/b₂ = 5/3/-10 = -1/6 and
c₁/c₂ = 7/14 = 1/2
Hence, a₁/a₂ ≠ b₁/b₂

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a₁/a₂ = 5/-10 = -1/2
b₁/b₂ = -3/6 = -1/2 and
c₁/c₂ = 11/-22 = -1/2
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂

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.

(v) 4/3x + 2y =8 ; 2x + 3y = 12
a₁/a₂ = 4/3/2 = 2/3
b₁/b₂ = /3 and
c₁/c₂ = 8/12 = 2/3
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂

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.

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

Answer

(i) x + y = 5; 2x + 2y = 10
a₁/a₂ = 1/2
b₁/b₂ = 1/2 and
c₁/c₂ = 5/10 = 1/2
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂
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.

x + y = 5

x = 5 - y 

And, 2x + 2y = 10
x = 10-2y/2
 

Graphical representation

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii) x – y = 8, 3x – 3y = 16
a₁/a₂ = 1/3
b₁/b₂ = -1/-3 = 1/3 and
c₁/c₂ = 8/16 = 1/2
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

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

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -1/2 and
c₁/c₂ = -6/-4 = 3/2
Hence, a₁/a₂ ≠ b₁/b₂

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y - 6 = 0
y = 6 - 2x

 

And, 4x - 2y -4 = 0
y = 4x - 4/2
 

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -2/-4 = 1/2 and
c₁/c₂ = 2/5
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂ 

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

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 

y + x = 36
 

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.

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

(i) Intersecting lines:
For this condition,
a₁/a₂ ≠ b₁/b₂
The second line such that it is intersecting the given line is
2x + 4y - 6 = 0 as
a₁/a₂ = 2/2 = 1
b₁/b₂ = 3/4 and
a₁/a₂ ≠ b₁/b₂

(ii) Parallel lines

For this condition,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Hence, the second line can be
4x + 6y - 8 = 0 as
a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 3/6 = 1/2 and
c₁/c₂ = -8/-8 = 1
and a₁/a₂ = b₁/b₂ ≠ c₁/c₂

(iii) Coincident lines
For coincident lines,
a₁/a₂ = b₁/b_2 = c₁/c₂
Hence, the second line can be
6x + 9y - 24 = 0 as
a₁/a₂ = 2/6 = 1/3
b₁/b₂ = 3/9 = 1/3 and
c₁/c₂ = -8/-24 = 1/3
and a₁/a₂ = b₁/b_2 = c₁/c₂

x - y + 1 = 0
x = y - 1
 

3x + 2y - 12 = 0

x = 12 - 2y/3
 

Graphical representation 

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).

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

Answer

(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 gets = 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 

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

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 = - 2 6
y = 13
So value of x = 3y
Putting value of y, we get
x = 3 × 13 = 39
Hence the numbers are 13 and 39.

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 degree 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 = 99Hence, the angles are 99º and 81º.

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.

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.

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.

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 + 5year
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/- 4
y = 10 year
Putting the value of y in equation first we get
x = 3 × 10 + 10
x = 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

(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)
2x –3y = 4 ... (ii)
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 methodx + 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)
3x + 4y = 10 ... (i)
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
9x – 2y = 7 ... (ii)
Multiplying equation (i) by 3, we get
9 x – 15 y = 11 ... (iii)
9x – 2y = 7 ... (ii)
Subtracting equation (ii) from equation (iii), we get
-13y = 5
y = -5/13
Putting value in equation (i), we get
3x – 5y = 4 ... (i)
3x - 5(-5/13) = 4
Multiplying by 13 we get
39x + 25 = 52
39x = 27
x =27/39 = 9/13
Hence our answer is x = 9/13 and y = - 5/13

By substitution method
3x – 5y = 4 ... (i)
Adding 5y both side we get
3x = 4 + 5y
Dividing by 3 we get
x = (4 + 5y )/3 ... (iv)
Putting this value in equation (ii) we get
9x – 2y = 7 ... (ii)
9 ((4 + 5y )/3) – 2y = 7
Solve it we get
3(4 + 5y ) – 2y = 7
12 + 15y – 2y = 7
13y = - 5
y = -5/13

Hence we get x = 9/13 and y = - 5/13 again.

(iv) x/2 + 2y/3 = - 1 and x – y/3 = 3
By elimination method
x/2 + 2y/3 = -1 ... (i)
x – y/3 = 3 ... (ii)
Multiplying equation (i) by 2, we get
x + 4y/3 = - 2 ... (iii)
x – y/3 = 3 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y/3 = -5
Dividing by 5 and multiplying by 3, we get
y = -15/5
y = - 3
Putting this value in equation (ii), we get
x – y/3 = 3 ... (ii)
x – (-3)/3 = 3
x + 1 = 3
x = 2

Hence our answer is x = 2 and y = −3.

By substitution method
x – y/3 = 3 ... (ii)
Add y/3 both side, we get
x = 3 + y/3 ... (iv)
Putting this value in equation (i) we get
x/2 + 2y/3 = - 1 ... (i)
(3+ y/3)/2 + 2y/3 = -1
3/2 + y/6 + 2y/3 = - 1
Multiplying by 6, we get
9 + y + 4y = - 6
5y = -15
y = - 3

Hence our answer is x = 2 and y = −3

(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

(ii) 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

(iii) 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.

(iv) 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.

(v) 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.

(i) x – 3y – 3 = 0
3x – 9y – 2 =0
a₁/a₂ = 1/3
b₁/b₂ = -3/-9 = 1/3 and
c₁/c₂ = -3/-2 = 3/2
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5
3x +2y = 8
a₁/a₂ = 2/3
b₁/b₂ = 1/2 and
c₁/c₂ = -5/-8 = 5/8
a₁/a₂ ≠ b₁/b₂

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,
x/b₁c₂-b₂c₁ = y/c₁a₂-c₂a₁ = 1/a₁b₂-a₂b₁
x/-8-(-10) = y/-15+16 = 1/4-3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
∴ x = 2, y = 1.

(iii) 3x – 5y = 20
6x – 10y = 40
a₁/a₂ = 3/6 = 1/2
b₁/b₂ = -5/-10 = 1/2 and
c₁/c₂ = -20/-40 = 1/2
a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a₁/a₂ = 1/3
b₁/b₂ = -3/-3 = 1 and
c₁/c₂ = -7/-15 = 7/15
a₁/a₂ ≠ b₁/b₂

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

x/45-(21) = y/-21-(-15) = 1/-3-(-9)
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
x = 4 and y = -1
∴ x = 4, y = -1.

2x + 3y -7 = 0

(a – b)x + (a + b)y - (3a +b –2) = 0

a₁/a₂ = 2/a-b = 1/2
b₁/b₂ = -7/a+b and
c₁/c₂ = -7/-(3a+b-2) = 7/(3a+b-2)
For infinitely many solutions,a₁/a₂ = b₁/b_{2 =} c₁/c₂

2/a-b = 7/3a+b-26a + 2b - 4 = 7a - 7b
a - 9b = -4 ... (i)

2/a-b = 3/a+b
2a + 2b = 3a - 3b
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.

3x + y -1 = 0

(2k –1)x + (k –1)y - (2k + 1) = 0

a₁/a₂ = 3/2k-1
b₁/b₂ = 1/k-1 and
c₁/c₂ = -1/-2k-1 = 1/2k+1
For no solutions,
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
3/2k-1 = 1/k-1 ≠ 1/2k+1

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

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

x/-20-(-18) = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1 and y/5 = 1
x = -2 and y = 5

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

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

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

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

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.

(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
Let 1/x = p and 1/y = q, then the equations changes as below:
p/2 + q/3 = 2
⇒ 3p + 2q -12 = 0 ... (i)
p/3 + q/2 = 13/6
⇒ 2p + 3q -13 = 0 ... (ii)

By cross-multiplication method, we get

p/-26-(-36) = q/-24-(-39) = 1/9-4
p/10 = q/15 = 1/5
p/10 = 1/5 and q/15 = 1/5
p = 2 and q = 3
1/x = 2 and 1/y = 3
Hence, x = 1/2 and y = 1/3

(ii) 2/√x +3/√y = 2
4/√x - 9/√y = -1
Let 1/√x = p and 1/√y = q, then the equations changes as below:
2p + 3q = 2 ... (i)
4p - 9q = -1 ... (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 ... (iii)
Adding equation (ii) and (iii), we get

10p = 5

p = 1/2 ... (iv)

Putting in equation (i), we get

2 × 1/2 + 3q = 2
3q = 1
q = 1/3

p = 1/√x = 1/2
√x = 2
x = 4
and
q = 1/√y = 1/3
√y = 3
y = 9
Hence, x = 4, y = 9

(iii) 4/x + 3y = 14
3/x - 4y = 23
Putting 1/x = p in the given equations, we get
4p + 3y = 14
⇒ 4p + 3y - 14 = 0
3p - 4y = 23
⇒ 3p - 4y -23 = 0
By cross-multiplication, we get
p/-69-56 = y/-42-(-92) = 1/-16-9
⇒ -p/125 = y/50 = -1/25
Now,
-p/125 = -1/25 and y/50 = -1/25
⇒ p = 5 and y = -2
Also, p = 1/x = 5
⇒ x = 1/5
So, x = 1/5 and y = -2 is the solution.

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 - 3/y-2 = 1
Putting 1/x-1 = p and 1/y-2 = q in the given equations, we obtain
5p + q = 2 ... (i)
6p - 3q = 1 ... (ii)
Now, by multiplying equation (i) by 3 we get
15p + 3q = 6 ... (iii)
Now, adding equation (ii) and (iii)
21p = 7
⇒ p = 1/3
Putting this value in equation (ii) we get,
 6×1/3 - 3q =1
 ⇒ 2-3q = 1
 ⇒ -3q = 1-2
 ⇒ -3q = -1
 ⇒ q = 1/3
Now,
p = 1/x-1 = 1/3
 ⇒1/x-1 = 1/3
 ⇒ 3 = x - 1
 ⇒ x = 4
Also,
q = 1/y-2 = 1/3
 ⇒ 1/y-2 = 1/3
 ⇒ 3 = y-2
 ⇒ y = 5
Hence, x = 4 and y = 5 is the solution.

(v) 7x-2y/xy = 5
 ⇒ 7x/xy - 2y/xy = 5
 ⇒ 7/y - 2/x = 5 ... (i)
8x+7y/xy = 15
 ⇒ 8x/xy + 7y/xy = 15
 ⇒ 8/y + 7/x = 15 ... (ii)
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
Now,
p = 1/x = 1
⇒ 1/x = 1
⇒ x = 1
also, q = 1 = 1/y
⇒ 1/y = 1
⇒ y = 1
Hence, x =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy
⇒ 6x/xy + 3y/xy = 6
⇒ 6/y + 3/x = 6 ... (i)
2x + 4y = 5xy
⇒ 2x/xy + 4y/xy = 5
⇒ 2/y + 4/x = 5 ... (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
6q + 3p - 6 = 0
2q + 4p - 5 = 0
By cross multiplication method, we get
p/-30-(-12) = q/-24-(-15) = 1/6-24
p/-18 = q/-9 = 1/-18
p/-18 = 1/-18 and q/-9 = 1/-18
p = 1 and q = 1/2
p = 1/x = 1 and q = 1/y = 1/2
x = 1, y = 2
Hence, x = 1 and y = 2

(vii) 10/x+y + 2/x-y = 4
15/x+y - 5/x-y = -2
Putting 1/x+y = p and 1/x-y = q in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q - 4 = 0 ... (i)
15p - 5q = -2
⇒ 15p - 5q + 2 = 0 ... (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/x-y = 1
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 .... (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

(viii) 1/3x+y + 1/3x-y = 3/4
1/2(3x-y) - 1/2(3x-y) = -1/8
Putting 1/3x+y = p and 1/3x-y = q in the given equations, we get
p + q = 3/4 ... (i)
p/2 - q/2 = -1/8
p - q = -1/4 ... (ii)
Adding (i) and (ii), we get
2p = 3/4 - 1/4
2p = 1/2
p = 1/4
Putting the value in equation (ii), we get
1/4 - q = -1/4
q = 1/4 + 1/4 = 1/2
p = 1/3x+y = 1/4
3x + y = 4 ... (iii)
q = 1/3x-y = 1/2
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

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.

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,

4(2/x + 5/y) = 1

2/x + 5/y = 1/4

3(3/x + 6/y) = 1

3/x + 6/y = 1/3

Putting 1/x = p and 1/y = q in these equations, we get

2p + 5q = 1/4

By cross multiplication, we get

p/-20-(-18) = q/-9-(-18) = 1/144-180

p/-2 = q/-1 = 1/-36

p/-2 = -1/36 and q/-1 = 1/-36

p = 1/18 and q = 1/36

p = 1/x = 1/18 and q = 1/y = 1/36

x = 18 and y = 36

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

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 get1200q = 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.

Let the age of Ani and Biju be x and y years respectively.

Therefore, age of Ani’s father, Dharam = 2 × x = 2x years

And age of Biju’s sister Cathy  = y / 2 years

Case (i) :When Ani is older than Biju by 3 years,

x − y = 3 …...............(i)

2x – y / 2 = 30

4x − y = 60 …...............(ii)

Subtracting (i) from (ii), we obtain

3x = 60 − 3 = 57

x = 57 / 3 = 19

Therefore, age of Ani = 19 years

And age of Biju = 19 − 3 = 16 years

Case (ii) :When Biju is older than Ani,

y − x = 3 …........ (I)

2x – y / 2 = 30

4x − y = 60 …...............(ii)

Adding (i) and (ii), we obtain

3x = 63

x = 21

Therefore, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years.

Let those friends were having Rs x and y with them.

Using the information given in the question, we obtain

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300 (i)

And, 6(x − 10) = (y + 10)

6x − 60 = y + 10

6x − y = 70 (ii)

Multiplying equation (ii) by 2, we obtain

12x − 2y = 140 (iii)

Subtracting equation (i) from equation (iii), we obtain

11x = 140 + 300

11x = 440

x = 40

Using this in equation (i), we obtain

40 − 2y = −300

40 + 300 = 2y

2y = 340

y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.

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 was d km. We know that,

x =d  / t

Or, d = xt (i)

Using the information given in the question, we obtain

By using equation (i), we obtain

3x − 10t = 30 (iii)

Adding equations (ii) and (iii), we obtain

x = 50

Using equation (ii), we obtain

(−2) × (50) + 10t = 20

−100 + 10t = 20

10t = 120

t = 12 hours

From equation (i), we obtain

Distance to travel = d = xt

= 50 × 12

= 600 km

Hence, the distance covered by the train is 600 km.

Let the number of rows be x and number of students in a row be y.

Total students of the class

= Number of rows × Number of students in a row

= xy

Using the information given in the question,

Condition 1

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3 (i)

Condition 2

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6 (ii)

Subtracting equation (ii) from (i),

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 3 + 6

y = 9

By using equation (i), we obtain

3x − 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = xy = 4 × 9 = 36

Let the number of rows be x and number of students in a row be y.

Total students of the class

= Number of rows × Number of students in a row

= xy

Using the information given in the question,

Condition 1

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3 (i)

Condition 2

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6 (ii)

Subtracting equation (ii) from (i),

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 3 + 6

y = 9

By using equation (i), we obtain

3x − 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = xy = 4 × 9 = 36

Given that,

∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠B

∠B = 2∠A

2 ∠A − ∠B = 0 … (i)

We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

∠A + ∠B + ∠C = 180°

∠A + ∠B + 3 ∠B = 180°

∠A + 4 ∠B = 180° … (ii)

Multiplying equation (i) by 4, we obtain

8 ∠A − 4 ∠B = 0 … (iii)

Adding equations (ii) and (iii), we obtain

9 ∠A = 180°

∠A = 20°

From equation (ii), we obtain

20° + 4 ∠B = 180°

4 ∠B = 160°

∠B = 40°

∠C = 3 ∠B

= 3 × 40° = 120°

Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

5x − y = 5

Or, y = 5x − 5

The solution table will be as follows.

3x − y = 3

Or, y = 3x − 3

The solution table will be as follows.

The graphical representation of these lines will be as follows.

It can be observed that the required triangle is ΔABC formed by these lines and y-axis.

The coordinates of vertices are A (1, 0), B (0, − 3), C (0, − 5).

(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 + q² x = p² + q²

(p² + q²) x = p² + q²

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

From equation (1), we obtain

ax + by = c

Or, bx − ay = 0 … (1)

ax + by = a² + b² … (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain

b²x − aby = 0 … (3)

a²x + aby = a³ + 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

(a − b) x − (a + b) x = (a² − 2ab − b²) − (a² + b²)

(a − b − a − b) x = − 2ab − 2b²

− 2bx = − 2b (a + b)

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

(v) 152x − 378y = − 74

76x − 189y = − 37

 … (1)

− 378x + 152y = − 604

− 189x + 76y = − 302 … (2)

Substituting the value of x in equation (2), we obtain

− (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

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

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°