Gujarat Board Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

 Gujarat Board Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Gujarat Board Textbook Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Ex 3.1

Question 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.
Solution:
Let the present age of Aftab be x years and the present age of his daughter be y years. Algebraic representation of the given situation is
x – 7 = 7(y – 7)
and x + 3 = 3(y + 3)
⇒ x – 7y + 42 = 0    …(1)
x – 3y – 6 = 0 …(2)
Graphical representation, we find two solutions for each equation.
For equation (1)
x – 7y + 42 = 0
x = 7y – 42
Table 1

For equation (2)
x – 3y – 6 = 0
x = 3y + 6
Table 2

We plot the points A(0, 6) and B(7, 7) corresponding to the solutions in table 1 on a graph paper to get the line AB representing the equation (1) and the points C(6, 0) and D(0, -2) on the same graph paper to get the line CD representing the equation (2).

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of one bat be ₹ x
and the cost of one ball be ₹ y.
Then, the algebraic representation of the given situation is
3x + 6y = 3900
and x + 3y = 1300
⇒ x + 2y = 1300 …(1)
x + 3y = 1300 …(2)
For graphical representation, we find two solutions for each equation.
For equation (1)
x + 2y = 1300
x = 1300 – 2y
Table 1

For equation (2)
x + 3y = 1300
x = 1300 – 3y
Table 2

We plot the points A(100, 600) and B(0, 650) corresponding to the solutions in table 1 on graph paper to get the line AB representing the equation (1) and the points C(400, 300) and D(100, 400) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing the equation (2).

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let the cost of 1 kg of apples be ₹ x
and the cost of 1 kg of grapes be ₹ y.
Algebraic representation of the given situation is
2x + y = 160 …(1)
4x + 2y = 300
⇒ 2x + y = 150 …(2)
For graphical representation we find two solutions for each equation.
For equation (1)
2x + y = 160
y = 160 – 2x
Table 1

For equation (2)
2x + y = 150
y = 150 – 2x
Table 2

We plot the points A(30, 100) and B(50, 60) corresponding to the solutions in table 1 on a graph paper to get the line AB representing the equation (1) and the points C(50, 50) and D(60, 30) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing the equation (2).

Ex 3.2

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

1. 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.
2. 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:
1. Let the number of boys and girls who took part in the quiz be x and y respectively. Then, the pair of linear equations formed is x + y – 10          …(1)
and y – x + 4             …(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of the equations. These two solutions of the equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1)
x + y – 10
⇒ y = 10 – x
Table 1

For equation (2)
y = x + 4
Table 2

We plot the points A(6, 4) and B(4, 6) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the paper. Also, we plot the points C(0, 4) and D(l, 5) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.

In the figure, we observe that the two lines intersect at the point P(3, 7). So x = 3 and y = 7 is the required solution of the pair of linear equations formed, i.e. the number of boys and girls who took part in the quiz is 3 and 7 respectively.

Verification:
Substitution x = 3 and y = 7 in (1) and (2), we find that both the equa-tions are satisfied as shown below:
x + y = 3 + 7 = 10
x + 4 = 3 + 4 = 7 = y
This verifies the solutions.

2. Let the cost of one pencil and a pen be ₹ x and ₹ y respectively. Then the pair of linear equations formed is
5x + 7y = 50        …(1)
and 7x + 5y = 46  …(2)
Let us draw the graphs of the equations (1) and (2) by finding two solutions for each of the equation. These two solutions of the equations (1) and (2) are given below in Table 1 and Table 2 respectively.
For equation (1)
5x + 7y = 50
⇒ 7y = 50 – 5x
⇒ y = 50−5x/7
Table 1

For equation (2)
7x + 5y = 46
⇒ 5y = 46 – 7x
46-7x
⇒ y = 46−7x/5
Table 2

We plot the points A(3, 5) and B(-4, 10) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure. Also, we plot the points C(-2, 12) and D(8, -2) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.

In the figure, we observe that the two lines intersect at the point A(3, 5). So x = 3 and y = 5 is the required solution of the pair of linear equations formed, i.e., the cost of one pencil is ₹ 3 and that of a pen is ₹ 5.
Verification:
Substituting x = 3 and y = 5 in (1) and (2), we find that both the equations are satisfied as shown below:
5x + 7y = 5(3) + 7(5) = 15 + 35 = 50
7x + 5y = 7(3) + 5(5) = 21 + 25 = 46
This verifies the solution.

 

Solution:
1. 3x + 2y = 5; 2x – 3y = 7
Here, a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7

Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one (unique) solution and therefore it is consistent.

2. 2x – 3y = 8; 4x – 6y = 9
Here, a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9

Hence, the given lines are parallel. So, the given pair of linear equations has no solution and therefore it is inconsistent.

Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one (unique) solution and therefore it is consistent.

4. 5x – 3y = 11; -10x + 6y = -22
Here, a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = 22


Hence, the given lines are coincident. So, the given pair of linear equations has infinitely many solutions and therefore it is consistent (dependent).

Hence, the given lines are coincident. So, the given pair of linear equations has infinitely many solutions and therefore it is consistent (dependent).

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

1. x + y = 5, 2x + 2y = 10
2. x – y – 8, 3x – 3y = 16
3. 2x + y – 6 = 0, 4x – 2y – 4 = 0
4. 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Hence, the lines represented by the equations (1) and (2) are coincident. Therefore, equations (1) and (2) have infinitely many common solutions, i.e., the given pair of linear equations is consistent.

For graphical representation, we draw the graphs of the equations (1) and (2) by finding two solutions for each of the equations. These two solutions of the equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1)
⇒ x + y = -5
y = 5 – x
Table 1

For equation (2)
2x + 2y = 10
⇒ 2y = 10 – 2x
⇒ y = 10−2x/2
⇒ y = 5 – x
Table 2

We plot the points A(0, 5) and B(5, 0) on a graph paper and join these points to form the line AB representing the equation (1) and points C(l, 4) and D(2, 3) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.

In the figure we observe that the two lines AB and CD are coincident.

Hence, the lines represented by the equations (1) and (2) are parallel. Therefore, equations (1) and (2) have no solution, i.e., the given pair of linear equation is inconsistent.

Hence, the lines represented by the equations (1) and (2) are intersecting. Therefore, equations (1) and (2) have exactly one (unique) solution i.e., the given pair of linear equation is consistent.

Graphical representation:
We draw the graphs of the equations (1) and (2) by finding two solutions for each of the equations. These two solutions of the equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1)
2x + y – 6 = 0
⇒ y = -2x + 6
Table 1

For equation (2)
4x – 2y – 4 = 0
⇒ 2y = 4x – 4
⇒ y = 4x−4/2
⇒ y = 2x – 2
Table 2

We plot the points A(0, -6) and B(3, 0) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure. Also we plot the points C(0, -2) and D(l, 0) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.

In the figure we observe that the two lines intersect at the point P(2, 2). So x – 2 and y – 2 is the required unique solution of the pair of linear equations formed.

Verification:
Substituting x = 2 and y = 2 in (1) and (2), we find that both the equations are satisfied as shown below:
2x – 2y – 6 = 2(2) + 2 – 6 = 0
4x – 2y – 4 = 4(2) – 2(2) – 4 = 0
This verifies the solution.

Hence, the lines represented by the equations (1) and (2) are parallel. Therefore, equations (1) and (2) have no solution, i.e., the given pair of linear equation is inconsistent.

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 the dimensions (i.e., length and width) of the garden be x m and y m respectively. Then,
x = y + 4
⇒ x – y = 4 …(1)
and 1/2 (2x + 2y) = 36
⇒ x +y = 36 …(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of the equations.
These two solutions of the equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1)
x – y = 4
⇒ y = x – 4
Table 1

For equation (2)
x + y = 36
y = 36 – x
Table 2

We plot the points A(4, 0) and B(2, -2) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure. Also, we plot the points C(20, 16), D(16, 20) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.

In the figure we observe that the two lines intersect at the point C(20, 16). So x = 20, y = 16 is the required solution of the pair of linear equations formed, i.e., the dimensions of the garden are 20 m and 16 m.

Verification:
Substituting x = 20 and y = 16 in (1) and (2), we find that both the equations are satisfied as shown below:
20 – 16 = 4
20 + 16 = 36
This verifies the solution.

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

1. Intersecting lines
2. Parallel lines
3. Coincident lines

Solution:
1. 2x + 3y – 8 = 0
3x + 2y – 7 = 0

2. 2x + 3y – 8 = 0
2x + 3y – 12 = 0

3. 2x + 3y – 8 = 0
4x + 6y – 16 = 0

Question 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 x-axis, and shade the triangular region.
Solution:
The given equations are
x – y + 1 = 0 …(1)
3x + 2y – 12 = 0 …(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of the equation. These two solutions of the equation (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1)
x – y + 1 = 0
⇒ y = x + 1
Table 1

For equation (2)
3x + 2y – 12 = 0
⇒ 2y = 12 – 3x
y =12−3x/2
Table 2


We plot the points A(0, 1) and B(-1, 0) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure. Also, we plot the points C(4, 0) and D(0, 6) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.

In the figure we observe that the coordinates of the vertices of the triangle formed by these given lines the x-axis are E(2, 3), B(-1, 0) and C(4, 0).

Ex 3.3

Question 1.
Solve the following pair of equations by substitution method.

Solution:
1. x + y = 14
x – y = 4
The given pair of linear equation is
x + y = 14 …..(1)
x – y = 4 …..(2)
From equation (1),
y = 14 – x …(3)
Substituting this value of y in equation (2), we get
x – (14 – x) = 4
⇒ x – 14 + x = 4
⇒ 2x – 14 = 4
⇒ 2x = 4 + 14
⇒ 2x = 18
⇒ x = 18/2 = 9
Substituting this value of x in equation (3), we get
y = 14 – 9 = 5
Therefore the solution is
x = 9, y = 5
Verification:
Substituting x = 9 and y = 5, we find that both the equations (1) and (2) are satisfied as shown below:
x + y = 9 + 5 = 14
x – y = 9 – 5 = 4
This verifies the Solution.

 

3. 3x – y = 3
9x – 3y = 9
The given pair of linear equations is
3x – y = 3 …(1)
9x – 3y = 9 …(2)
From equation (1),
y = 3x – 3 …(3)
Substituting this value of y in equation (2), we get
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 0
⇒ 9 = 9
Which is true. Therefore, equations (1) and (2) have infinitely many solutions.

4. 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
The given system of linear equation is
0.2x + 0.3y = 1.3 …(1)
0.4x + 0.5y = 2.3 …(2)
From equation (1),
0.3y = 1.3 – 0.2x
⇒ y = 1.3−0.2x/0.3 …(3)
Substituting this value of y in equation (2), we get
0.4x + 0.5(1.3−0.2x/0.3) = 2.3
⇒ 0.12x + 0.65 – 0.1x = 0.69
⇒ 0.12x – 0.1x = 0.69 – 0.65
⇒ 0.02x = 0.04
⇒ x = 0.04/0.02 = 2
Substituting this value of x in equation (3), we get

Therefore, the solution is x = 2, y = 3.
Verification:
Substituting x = 2 and y = 3, we find that both the equations (1) and (2) are satisfied as shown below:
0.2x + 0.3y = (0.2) (2) + (0.3) (3)
= 0.4 + 0.9= 1.3
0.4x + 0.5y = (0.4) (2) + (0.5) (3)
= 0.8 + 1.5 = 2.23
This verifies the solution.

   

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
The given pair of linear equations is
2x + 3y = 11 …….(1)
2x – 4y = -24 …….(2)
From equation (1),

Verification:
Substituting x = -2 and y = 5, we find that both the equations (1) and (2) are satisfied as shown below:
2x + 3y = 2(-2) + 3(5)
= -4 + 15 = 11
2x – 4y = 2(-2) – 4(5)
= -4 – 20 = -24
This verifies the solution.
Now, y = mx + 3
⇒ 5 = m(-2) + 3
⇒ -2m = 5 – 3
⇒ -2m = 2
⇒ m = 2/−2 = -1

Question 3.
Form the pair of linear equations in the following problems and find their solution by substitution method:

1. The difference between two numbers is 26 and one number is three times the other. Find them.
2. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
3. 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.
4. 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?
5. 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.
6. 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 age?

Solution:
1. Let the two numbers be x and y(x > y), then according to the question, the pair of linear equations formed is
x – y = 26 …(1)
x = 3y …(2)
Substituting the value of x from equation (2) in equation (1), we get
3y – y = 26
⇒ 2y = 26
⇒ y = 26/2
⇒ y = 13
Substituting this value ofy in equation (2), we get
x = 3(13) = 39
Hence, the required number are 39 and 13.
Verification:
Substituting x – 39 and y = 13, we find that both the equations (1) and (2) are satisfied as shown below:
x – y = 39 – 13 = 26
3y = 3(13) = 39 = x
This verifies the solution.

2. Let the larger and the smaller of two supplementary angles be x° and y° respectively.
Then, according to the question.
The pair of linear equations formed is
x° = y° + 18° …(1)
x° + y° = 180° …(2)
| ∵ The two angles are supplementary
Substituting the value of x° from equation (1) in equation (2), we get
y° + 18°+ y° = 180°
⇒ 2y° + 18° = 180°
⇒ 2y° = 180° – 18°
⇒ 2y° = 162°
⇒ y° = 162°/2 = 81°
Substituting this value ofy° in equation (1), we get
x° = 81° + 18° = 99°
Hence, the larger and the smaller of the two supplementary angles are 99° and 81° respectively.
Verification:
Substituting x° = 99° and y° = 81°, we find that both the equations (1) and (2) are satisfied as shown below:
y° + 18° = 81°+ 18° = 99° = x°
x° + y° = 99°+ 81° = 180°
This verifies the solution.

Hence, the cost of each bat is ₹ 500 and each that of ball is ₹ 50 respectively.
Verification:
Substituting x = 500 and y = 50, we find that both the equations (1) and (2) are satisfied as shown below:
7x + 6y = 7(500) + 6(50)
= 3500 + 300 = 3800
3x + 5y = 3(500) + 5(50)
= 1500 + 250 = 1750
This verifies the solution.

(iv) Let the fixed charges be ₹ x and the charge per kilometre be ₹ y.
Then, according to the question, the pair of
linear equations formed is
x + 10y = 105 …(1)
x + 15y = 155 …(2)
From equation (1)
x = 105 – 10y …(3)
Substituting this value of x in equation (2), we get
105 – 10y + 15y = 155
⇒ 5 + 5y = 155
⇒ 5y = 155 – 105
⇒ 5y = 50
⇒ y = 50/5 = 10
Substituting this value of y in equation (3), we get
x = 105 – 10(10)
= 105 – 100 = 5
Hence, the fixed charges are ₹ 5 and the charge per kilometre is ₹ 10.
Verification:
Substituting x = 5 and y = 10, we find that both the equations (1) and (2) are satisfied as shown below:
x + 10y = 5 + 10(10)
= 5 + 100 = 105
x + 15y = 5 + 15(10)
= 5 + 150 = 155
This verifies the solution.
Again, for travelling a distance of 25 km, a person will have to pay
= 5 + 10(25)
= 5 + 250 = ₹ 255

Substituting this value of y in equation (3), we get

Hence, the required fraction is 7/9.
Verification:
Substituting x = 7 and y = 9, we find that both the equations (1) and (2) are satisfied as shown below:

This verified the solution.

6. Let the present ages of Jacob and his son be x years and y years respectively.
Then, according to the question, the pair of linear equations formed is
x + 5 = 3(y + 5)
x – 5 = 7(y – 5)
⇒ x – 3y = 10 …(1)
x – 7y = -30 …(2)
From equation (1),
x = 3y + 10 …(3)
Substituting this value of x in equation (3), we get
3y + 10 – 7y = -30
-4y = -40
y = 10
Substituting y = 10 in equation (3), we get
x = 3(10) + 10
= 30 + 10 = 40
Hence, the present age of Jacob and his son are 40 years and 10 years respectively.
Verification:
Substituting x = 40 and y = 10, we find that both the equations (1) and (2) are satisfied as shown below:
x – 3y = 40 – 3(10)
= 40 – 30 = 10
x – 7y = 40 – 7(10)
= 40 – 70 = -30
This verifies the solution.

Ex 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:

II. By Substitution Method:
The given systems of equation is
x + y = 5 …(1)
2x – 3y = 4 …(2)
From equation (1),
y = 5 – x …(3)
Substitute this value ofy in equation (2), we get
2x – 3(5 – x) = 4
⇒ 2x – 15 + 3x = 4
⇒ 5x – 15 = 4
⇒ 5x = 15 + 4
⇒ 5x = 19
⇒ x = 19/5
Substituting this value of* in equation (3),
we get

So, the solution of the given system of equations is

Verification:
Substituting x = 19/5 and y = 6/5, we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.

2. 3x + 4y = 10 and 2x – 2y = 2
I. By Elimination method:
The given system of equation is
3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
Multiplying equation (2), by 2, we get
4x – 4y = 4 …(3)
Adding equation (1) and equation (3), we get

Substituting this value of x in equation (2), we get
2(2) – 2y = 2
⇒ 4 – 2y = 2
⇒ 2y = 4 – 2
⇒ 2y = 2
⇒ y = 2/2 = 1
So, the solution of the given system by equations is x = 2, y = 1.

II. By substitution method:
The given system of equation is
3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
From equation (2),
2y = 2x – 2
⇒ y = 2x−2/2
⇒ y = x – 1 …(3)
Substituting this value of y in equation (1),
we get
3x + 4(x – 1) = 10
⇒ 3x + 4x – 4 = 10
⇒ 7x – 4 = 10
⇒ 7x = 10 + 4
⇒ 7x = 14
⇒ x = 14/7 = 2
Substituting this value of x in equation (3), we get
y = 2 – 1
⇒ y = 1
So, the solution of the given system of equations is x = 2, y = 1.
Verification:
Substituting x = 2, y = 1, we find that both the equations (1) and (2) are satisfied as shown below:
3x + 4y = 3(2) + 4(1)
= 6 + 4 = 10
2x + 2y = 2(2) – 2(1)
= 4 – 2 = 2
Hence, the solution is correct.

3. 3x – 5y – 4 = 0 and 9x = 2y + 7
I. By Elimination method:
The given system of equation is
3x – 5y – 4 = 0 …(1)
9x = 2y + 7
⇒ 9x – 2y – 7 = 0 …(2)
Multiplying equation (1) by 3, we get
9x – 15y – 12 = 0 …(3)
Subtracting equation (3) from equation (2), we get
13y + 5 = 0
⇒ 13y = -5
⇒ y = −5/13
Substituting this value of y in equation (1), we get

So, the solution of the given system of equations is

  

II. By substitution method:
The given system of equation is

From equation (2),

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

1. 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?

2. 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?

3. 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 number. Find the number.

4. 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.

5. 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 five days. Find the fixed charge and the charge for each extra day.

Solution:
1. Let the fraction be x/y.
Then, according to the question,

⇒ x + 1 = y – 1
2x = y + 1
⇒ x – y = -2 …..(1)
⇒ 2x – y = 1 …..(2)
Subtracting equation (1) from equation (2), we get
x = 3
Substituting this value of x in equation (1), we get
3 – y = -2
⇒ y = 3 + 2
⇒ y = 5
Hence, the required fraction is 3/5.

Verification:
Substituting the values of x = 3 and y = 5, we find that both the equations cases are satisfied as shown below:

Hence, the solution is correct.

2. Let Nuri and Sonu be x years and y years old respectively at present.
Then, according to the question,
x – 5 = 3(y – 5)
x + 10 = 2(y + 10)
⇒ x – 5 = 3y – 15
x + 10 = 2y + 20
⇒ x – 3y = -10 …(1)
x – 2y = 10 …(2)
Subtracting equation (2) from equation (1), we get
-y = -20
⇒ y = 20
Substituting this value ofy in equation (2), we get
x – 2(20) = 10
⇒ x – 40 = 10
⇒ x = 40 + 10
⇒ x = 50
Hence, Nuri and Sonu are 50 years and 20 years old respectively at present.

Verification:
Substituting the values of x = 50 and y = 20, we find that both the equations (1) and (2) are satisfied as shown below:
x – 3y = 50 – 3(20)
= 50 – 60 = -10
x – 2y = 50 – 2(20)
= 50 – 40 = 10
Hence, the solution is correct.

3. Let the unit’s digit and the ten’s digit in the two-digit number be x and y respectively.
Then, the number = 10y + x
Also, the number obtained by reversing the
order of the digits = 10x + y
According to the question,
x + y = 9 …(1)
9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x + 2y
⇒ 11x – 88y = 0
⇒ x – 8y = 0 …(2)
Subtracting equation (2), from equation (1), we get
9y = 9
⇒ y = 9/9 = 1
Substituting this value of y in equation (1), we get
x + 1 = 9
⇒ x = 9 – 1 = 8
Hence, the required number is 18.
Verification:
Substituting x = 8 and y = 1, we find that both the equations (1) and (2) are satisfied as shown below:
x + y = 8 + 1 = 9
x – 8y = 8 – 8(1) = 0
Hence, the solution is correct.

4. Suppose that Meena received x notes of ₹ 50 and y notes of ₹ 100.
Then, according to the question,
x + y = 25 …(1)
50x + 100y = 2000
⇒ x + 2y = 40 …(2)
Subtracting equation (1) from equation (2), we get
y = 15
Substituting this value of y in equation (1), we get
x + 15 = 25
⇒ x = 25 – 15 = 10
Hence, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
Verification:
Substituting x = 10 and y = 15, we find that both the equations (1) and (2) are satisfied as shown below:
x + y = 10 + 15 = 25
x + 2y = 10 + 2(15)
= 10 + 30 = 40
Hence, the solution is correct.

5. Let the fixed charge be ₹ a and the charge for each extra day be ₹ b.
Then, according to the question,
a + 46 = 27 …(1)
(∵ Extra days = 7 – 3 = 4)
a + 2b = 21 …(2)
(∵ Extra days = 5 – 3 = 2)
Subtracting equation (2) from equation (1), we get
2b = 6
⇒ b = 6/2 = 3
Substituting this value of 6 in equation (2), we get
a + 2(3) = 21
⇒ a + 6 = 21
⇒ a = 21 – 6 = 15
Hence, the fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.

Verification:
Substituting a = 15 and 6 = 3, we find that both the equations (1) and (2) are satisfied as shown below:
a + 46 = 15 + 4(3)
= 15 + 12 = 27
a + 26 = 15 + 2(3)
= 15 + 6 = 21
Hence, the solution is correct.

Ex 3.5

Question 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.

1. x – 3y – 3 = 0, 3x – 9y – 2 = 0
2. 2x + y = 5, 3x + 2y = 8
3. 3x – 5y = 20, 6x – 10y = 40
4. x – 3y – 7 = 0, 3x – 3y – 15 = 0

Solution:
1. The given pair of linear equations is
x – 3y – 3 = 0 …(1)
3x – 9y – 2 = 0 …(2)
Here, a1 = 1, b1 = -3, c1 = -3
a2 = 3, b2 = -9, c2 = -2
We see that

Hence, the given pair of linear equations has no solution.

2. The given pair of linear equations is
2x + y = 5
3x + 2y = 8
⇒ 2x +y – 5 = 0
3x + 2y – 8 = 0
Here, a1 = 2, b1 = 1, c1 = -5
a2 = – 3, b2 = 2, c2 = -8
We see that

Hence, the required solution of the given pair of linear equations is x = 2, y = 1.
Verification:
Substituting x = 2, y = 1, we find that both the equations (1) and (2) are satisfied as shown below:
2x + y = 2(2) + 1
= 4 + 1 = 5
3x + 2y = 3(2) + 2(1)
= 6 + 2 = 8
Hence, the solution is correct.

3. The given pair of linear equations is
3x – 5y = 20 …(1)
6x – 10y = 40 …(2)
⇒ 3x – 5y – 20 = 0
6x – 10y – 40 = 0
Here, a1 = 3, b1 = -5, c1 = -20
a2 = 6, b2 = -10, c2 = -40
We see that

Hence, the given pair of linear equations has infinitely many solutions.

4. The given pair of linear equations is
x – 3y – 7 = 0 …(1)
3x – 3y – 15 = 0 …(2)
Here, a1 = 1 , b1 = -3, c1 = -7
a2 = 3, b2 = 3, c2 = -15
We see that

Hence, the required solution of the given pair of linear equations is x = 4, y = -1.
Verification:
Substituting x – 4, y = 1, we find that both the equations (1) and (2) are satisfied as shown below:
x – 3y – 7 = 4 – 3(-1) – 7 = 0
3x – 3y – 15 = 3(4) – 3(-1) – 15 = 0
Hence, the solution is correct.

Question 2.
1. 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
2. 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
Solution:
1. The given pair of linear equations is 2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
Here, a1 = 2, b1 = 3, c1 = 7
a2 = a – b, b2 = a + b, c2 = 3a + b – 2
Since, the given pair of linear equations has infinite solution,

 

2. The given pair of linear equations is:
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
⇒ 3x + y – 1 = 0
(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)
For having no solution, we must have

⇒ 3(k – 1) = 2k – 1
⇒ 3k – 3 = 2k – 1
⇒ 3k – 2k = 3 – 1
⇒ k = 2
Hence, the required value of k is 2.

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods.
8x + 5y = 9
3x + 2y = 4
Solution:

II. By Cross-multiplication method:
Let us write the given pair of linear equations is
8x + 5y – 9 = 0 …(3)
3x + 2y – 4 = 0 …(4)
To solve the equations (3) and (4) by cross-multiplication method, we draw the diagram below:

Hence, the required solution of the given pair of linear equations is
x = -2, y = 5.
Verification:
Substituting x = -2, y = 5, we find that both the equations (1) and (2) are satisfied as shown below:
8x + 5y = 8(-2) + 5(5)
= -16 + 25 = 9
3x + 2y = 3(-2) + 2(5)
= -6 + 10 = 4
Hence, the solution is correct.

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 charge and the cost of food per day.

(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.

(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 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?

(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.

Solution:
(i) Let the fixed charge be ₹ x and the cost of food per day be ₹ y.
Then according to the question,
x + 20y = 1000 …(1)
x + 26y = 1180 …(2)
⇒ x + 20y – 1000 = 0 …(3)
x + 26y – 1180 = 0 …(4)
To solve the equations (3) and (4) by cross-multiplication method, we draw the diagonal below:


Hence, the fixed charge is ₹ 400 and the cost of food per day is ₹ 30.
Verification:
Substituting x = 400, y = 30, we find out both the equations (1) and (2) are satisfied as shown below:
x + 20y = 400 + 20(30)
= 400 + 600 = 1000
x + 26y = 400 + 26(30)
= 400 + 780
= 1180
Hence, the solution is correct.

(ii) Let the fraction be x/y.
Then, according to the question,

3(x – 1) = y
4x = y + 8 …….(3)
⇒ 3x – y – 3 = 0
4x – y – 8 = 0
To solve the equations (3) and (4) by cross-multiplication method, we draw the diagram below:

Verification:
Substituting x = 5, y = 12, we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.

(iii) Suppose that Yash gave right answers to x questions and wrong answers toy questions. Then, the total number of questions in the test = x + y.
Also, according to the question,
3x – y = 40 ……(1)
4x – 2y = 50 …….(2)
⇒ 3x – y – 40 = 0 ……(3)
4x – 2y – 50 = 0 ……(4)
To solve the equations (3) and (4) by cross-multiplication method, we draw the diagram below:

Hence, Yash gave right answers to 15 questions and wrong answers to 5 questions. Therefore, total number of questions in the test = x + y = 15 + 5 = 20.
Verification:
Substituting x = 15, y = 5, we find that both the equations (1) and (2) are satisfied as shown below:
3x – y = 3(15) – 5
= 45 – 5 = 40
4x – 2y = 4(15) – 2 × (15)
= 60 – 10 = 50
Hence, the solution is correct.

(iv) Let the speeds of two cars be x km/hour and y km/hour respectively.
Case I:
When the cars travel in the same direction.
Let they meet at P.

Distance travelled by the car starting from A in 5 hours = AP = 5x km

Distance travelled by the car starting from B in 5 hours = BP = 5y km

Now,
AP – BP = 100
5x – 5y = 100
x – y = 20 …(1)

Case II:
When the cars travel towards each other.
Let they meet at Q.

Distance travelled by the car starting from A in 1 hour = AQ = x km
Distance travelled by the car starting from B in 1 hour = BQ = y km
Now, AQ + BQ = 100
x + y = 100 ……(2)
Equation (1) and (2) can be re-written as
x – y – 20 = 0 …(3)
x + y – 100 = 0 …(4)
To solve the equations (3) and (4) by cross-multiplication method, we draw the diagram below:


Hence, the speeds of the two cars are 60 km/hour and 40 km/hour respectively.
Verification:
Substituting x = 60, y = 40, we find that both the equations (1) and (2) are satisfied as shown below
x – y – 60 – 40 = 20
x + y = 60 + 40 = 100
Hence, the solution is correct.

(v) Let the dimensions (i.e., the length and the breadth) of the rectangle be x units and y units respectively.
Then, area of the rectangle
= Length × Breadth
= xy square units
According to the question,
xy – 9 = (x – 5)(y + 3)
⇒ xy – 9 = xy + 3x – 5y – 15
⇒ 3x – 5y – 6 = 0 …(1)
and xy + 67 = (x + 3)(y + 2)
⇒ xy + 67 = xy + 2x + 3y + 6
⇒ 2x + 3y – 61 = 0 …(2)
To solve the equations (1) and (2) by cross-multiplication method, we draw the diagram below:

Hence, the dimensions (i.e., the length and the breadth) of the rectangle are 17 units and 9 units respectively.
Verification:
Substituting x = 17, y = 9, we find that both the equations (1) and (2) are satisfied as shown below:
3x-5y – 6 = 3(17) – 5(9) – 6
= 51 – 45 – 6 = 0
2x + 3y – 61 = 2(17) + 3(9) – 61
= 34 + 27 – 61 = 0
Hence, the solution is correct.

Ex 3.6

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

       

Hence, the solution of the given pair of
equations is x = 4, y = 5.
Verification: Substituting x = 4, y = 5, we
find that both the equations (1) and (2) are
satisfied as shown below:

Hence, the solution is correct.

(v) The given pair of equations is

Then the equations (3) and (4) can be rewritten as
7υ – 2u = 5
8υ + 7u = 15
7υ – 2u – 5 = 0
8υ + 7u – 15 = 0
To solve the equations by the cross multiplication method, we draw the diagram below:

(vi) The given pair of equations is
6x + 3y = 6xy
2x + 4y = 5xy
(Dividing throughout by xy)

Then equations (1) and (2) can be rewritten as
6υ + 3u = 6
2υ + 4u = 5
Multiplying equation (6) by 3, we get
6υ + 12u = 15
Subtracting equation (5) from equation (7), we get

     

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 partly 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 her speed of rowing in still water be x kim/hour and the speed of the current be y km/hour.
Then, her speed of rowing downstream
= (x + y) km/hour
and, her speed of rowing upstream
= (x – y) km/hour

Substituting this value of x in equation (1), we get
6 + y = 10
y = 10 – 6 = 4
Hence, the speed of her rowing in still water
is 6 km/hour and the speed of the current is 4 km/hour.
Verification: Substituting x = 6, y = 4, we
find that both the equations (1) and (2) are
satisfied as shown below:
x + y = 6 + 4 = 10
x – y = 6 – 4 = 2
Hence, the solution is correct.

(ii) Let the time taken by 1 woman alone to finish the embroidery work be x days and the time taken by 1 man alone to finish the embroidery be y days.

 

Hence, the time taken by 1 woman alone to finish the embroidery work is 18 days and the time taken by 1 man alone to finish the embroidery work is 36 days.
Verification : Substituting x = 18, y = 36,
we find that both the equations (1) and (2) are satisfied as shown below:

(iii) Let the speed of the train and the bus be x km/hour and y km/hour respectively.
When she travels 60 km by train and the remaining (300 – 60) km, i.e., 240 km by bus, the time taken is 4 hours.

So, the solution of the equations (1) and (2)
is x = 60 and y = 80.
Hence, the speed of the train is 60 km/hour and the speed of the bus is 80 km/hour.
Verification: Substituting x= 60, y = 80,
we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.

Ex 3.7

Question 1.
The ages of two friends Aol and Rijo differ by 3 years, Mi’s father Dharam is twice as old as Axil and Riju is twice as old as his aislar Cathy. The ages of Cathy and Pharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the age-s of Ani and Biju be x years and yea respectively- Then, according to the question,
x – y = ±3 ………………….(1)
Age of Ani’s father Dharam = ax years
Age of Biju’a sister = y/2 years
According to the question,
2x – y/2
4x – y = 60 ……………… (2)

Case I: When x – y = 3
Then, we have
x – y = 3 ……………… (1)
4x – y = 60 ……………. (2)
Subtracting equation tI) from equation (2),
3x = 57
x = 57/3 = 19 years
Substituting the value of x in equation (1),
19 – y = 3
Ani’s age = 19 years
Biju’s age = 16 years
Verification:
x – y = 19 – 16 – 3
4x – y = 4 x 19 – 16
= 76 – 16 = 60
This verifies the solution.

Case II: When x – y = – 3
Then, we have
x – y = – 3 ……………… (3)
4x – y = 60 ………… (2)
Subtracting equation (3) from equation (2), we get
3x = 63
x = 63/3 = 21
Subetituting the value of x in the equation (3). we get
21 – y = – 3
y = 24
Ani’s age = 21 years
Bijus age = 24 years
verification:
x – y = 21 – 24 = – 3
4x – y = 4(21) – 24
= 84 – 24 = 60
This verifies the solution,

Question 2.
One says, “Give me a hundred, friend! I shall then becomes twice us nch as you”, The othcr replies. “1f you give me ten, I shall be six times as rich as you”. Tell mu what. is the amount of their (respcctive) capital.
[From the Bijaganita of Bhasknrn II]
[Hint : x+ 100 = 2(y – 100), y + 10 = 6 (x – 10)]
Solution:
Let the amounts of their respective capitals be
₹ x and ₹ y respectively.
Then, according to the question,
x + 100 = – 100
x – 2y = – 300 …………….. (1)
and 6(x – 10) = y + 10
6x – y = 70
From equatton (1),
x = 2y – 300
Substituting the value of x in equation (2), we get
6(2y – 300) – y = 70
12y – l800 – y = 70
11 y = 1870
y = 1870/11 = 170
Substituting this value of y in equation (3), we get
x = 2 (170) – 300
= 340 – 300 = 40
So, the solution of the equations (1) and (2)
is x= 40 and y 170. Hence, the amount.s of
their respective capitals are ₹ 40 and ₹ 170 reapcetivcly.
Verification : Substituting z = 40, y = 170,
we find thai both the equations (1) and (2) are satisfied as shown below:
x – 2y = 40 – 2 (170)
= 40 – 340 = – 300
6x – y = 6(40) – 170
=240 – 170 = 70
Hence, the solution we have got is correct.

Question 3.
A tain covered a certain diietaswe at a uniform speed. 1f the tain would have been 10 km/h faster, it would have taken 2 hours lete than the scheduled tiene. And, if the tain were slower by 10 km/h; it would have taken 3 hours more than the schedulod tiene. Find the distance covered by the train.
Solution:
Let the artual speed 0f the train 6ex km/hour
and the actual time taken byy hours Then.
Distance = Speed x Tinte
Distante = (xy) km
According to the question.
xy = (x + 10) (y – 2)
xy = xy – 2x + 10y – 20
2x – 10 y + 20 = 0
x – 5y + 10 = 0 ……………… (1)
[Dividing throughout by 2]
and, xy = (x + 10) (y – 2)
xy = xy – 2x + 10y – 20
3x – 10y – 30 = 0
To solve the equations (1) and (2) by the cross
multiplication method, we draw the diagram below:

So, the solution of the equations (1) and (2) is x = 50 andy = 12.
Hence, the distance covered by the train is 50 x 12 = 600 km.
Verification: Substituting x = 50, y = 12,
we find that both the equations (1) and (2) are
satisfied as shown below.
x – 5y + 10 = 50 – 5(12) + 10
= 50 – 60+ 10 = 0
3x – 10y – 30 = 3(50) – 10(12) – 30
= 150 – 120 – 30
= 0
Hence, the solution is correct.

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 the number of students in the class be x and the number of rows bey. Then, number of
 

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2 (∠A+∠B)Find the
three angles.
Solution.
We have
∠C = 3 ∠B = 2 (∠A+∠B) ……… (1)
We know that the sum of the measurea of three
angles of a triangle is 180°.
∴∠A + ∠B + ∠C = 180° ……………. (2)
(1) and (2) give
∠A + ∠B + 2 (∠A+∠B) = 180° ……… (2)
3∠A + 3∠B = 180°
∠A + ∠B = 60° ……………..(3)
(Dividing theought by 3)
and, ∠A + ∠B + 3∠B = 180°
∠A + 4 ∠B = 180°
Subtracting equation (3) from equation (4), we get
3∠B = 120°
∠B = 120°/3 = 40°
Substituting this value ∠B = 40 equation (3), we get
∠A + 40° = 60°
∠A = 60° – 40°
Again from(1)
∠C = 3 ∠B = 3(40°) = 120°
Hence, the three angles of the ∆ABC are given by ∠A = 20°,∠B = 40° and ∠C = 120°.
Verification: Substituting ∠A = 20°,∠B = 40° and ∠C = 120°.
we find that (1) is satisfied as shown below:
∠C = 3 ∠B = 2 (∠A+∠B)
120° = 3(40°) = 2( 20° + 40°)
Hence, the solution is correct

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3, Determine lo co-ordinate of the vertices of the triangle formed by these lines end they-axis.
Solution:
The given equations are
5x – y = 5
3x – y = 3
Let us draw the graphs of equations (1) and (2) by finding two solutions of each of the equations.
These two solutions of each of the equation (1) and (2) are given beiow in taHe 1 and table 2 respectively,
For equation (1)
y = 5x – 5

We plot the points A(1,0) and B(2, 5) on a graph pnpr and join these paints t0.

Form the line AB repreeenting the equation (1) aa ahown in the figure. Also, we plot the pointa C(2,3) and D(3, 6) on the same graph paper and join theae pointa to form the line CD representing the equation (2) ea ahown in the same figure.

In the figure, we aheerve that the coordinatea of the verticee of the triangle AEF are A(1,0), E(0,-5) and F(0,-3).

Question 7.
Solve the following pair of hnear equationa:

(ii) The given pair of linear equations is
ax + by = c ……………… (1)
bx + ay = 1 + c …………. (2)
ax + by – c = 0 ……………….(3)
bx + ay – (1 + c) = 0 ……………. (4)
To solve the equations by the cross multiplication method, we draw the diagram below:


Hence, the solution of the given pair of linear equations is

Verification: Substatituting

we find that both the equations (1) and (2) are satisfied as shown below:

This verifies the solution.

(iv) The given pair of linear equations is
(a – b) x +(a + b) y = a2 – 2ab -b2
(a + b)(x + y) = a2 + b2
(a + b) x + (a + b) y = a2 + b2 ………….(2)
Subtracting equation (2) from equation (1), we get

Substituting this value of x in equation (1), we get

(v) The given pair of linear equations is
152x – 378y = – 74 ……………. (1)
– 378x + 152y = – 604 ……………. (2)
Adding equation (1) and equation (2), we get
– 226x – 226y = – 678
x + y = 3 …………….(3)
[Dividing throughout by -226]
Subtracting equation (2) from equation (1), we get
530x – 530y = 530
x – y = 1
[Dividing throughout by 530]
Adding equation (3) and equation (4), we get
2x = 4
x = 4/2 = 2
Subtracting equation (4) from equation (3), we get
2y = 2
y = 2/2 = 1
Hence, the solution of the given pair of linear equations is x = 2, y = 1.
Verification: Substituting z = 2, y = 1, we
find that both the equations (1) and (2) are satisfied as shown below:
152x – 378y = (152) (2) – (378)(1)
= 304 – 378 = – 74
– 378x + 152y
= ( – 378)(2) + (152)(1)
= – 736 + 152 = – 604
This verifies the solution.

Question 8
ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

Solution:
We know that the opposite angles of a cyclic quadriLateral are supplementary, therefore,
∠A + ∠C = 180°
4y + 20° + (-4x) = 180°
4y – 4x = 160°
y – x = 40° ………….. (1)
[Dividing theought by 4]
and, ∠B + ∠D = 180°
3y – 5° + (-7x + 5)° = 180°
3y – 5° – 7x + 5° = 180°
3y – 7x = 180° ……………. (2)
From equation (1),
y = 40° + x
Substituting this value of y In equation (2), we get
3(40° + x) – 7x = 180°
120°+ 3x – 7x = 180°
-4x = 60°
x = 60/−4 = – 15°
Subet.tutLng x = – 15° in equation (3), we get
y = 40° – 15° = 25°
∠A 4(25) + 20 = 120°
∠B = 3(25) – 5 = 70°
∠C = – 4 x (-15) = 60°
∠D = – 7 x (-15) + 5
= 105 + 5 = 110°
Hence, the anglee of the cyclic quadrilateral are
∠A = 120°, ∠B = 70°, ∠C = 60°, and ∠D = 110°
Verification: Substituting x = – 15°, y = 25°,
we find that both the equations (I) and (2) are satisfied aa shown below.
y – x = 15 + 25 = 40°
3y – 7x = 7(15) + 3 (25)
= 105° + 75° = 180°
This verifies the solution.