Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Solution

 

BSEB Bihar Board Class 10th Maths Solutions 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 us denote the present age of daughter and her father Aftab as x years and y years respectively. Then, algebraic representation is given by the following equations:
7(x – 7) = y – 7 or 7x – 49 = y – 7
or 7x – y = 42 ………………. (1)
and 3(x + 3) = y + 3 or 3x + 9 = y + 3
or 3x – y = – 6 ……………….. (2)
To obtain the equivalent graphical representation, we find two points on the line representing each equation. That is, we find two solutions of each equation. These solutions are given below in the tables:
For 7x – y = 42

For 3x – y = – 6

To represent these equations graphically, we plot the points A(6, 0) and B(5, – 7) to get the graph of (1) and the points C(0, 6) and D(- 2, 0) give the graph of (2).

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Solution:
Let us denote the cost of one bat by Rs x and one ball by Rs y. Then, the algebraic representation is given by the following equations:
3x + 6y = 3900 ……………… (1)
and x + 2y = 1300 …………….. (2)
To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation. These solutions are given below in the table:

We plot the points A(0, 650) and B(1300, 0) to obtain the geometric representation of 3x + 6y = 3900 and C(500, 400) and D(100, 600) to obtain the geometric representation of x + 2y = 1300.

We observe that these lines are coincident.

Question 3.
The cost of 2 kg of apples and 1 kg 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.
Solution:
Let us denote the cost of 1 kg of apples by Rs x and cost of 1 kg of grapes by Rs y.
Then the algebraic representation is given by the following equations:
2x + y = 160 …………….. (1)
4x + 2y = 300 ……………….. (2)
To find the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation.

Geometric representation is shown on the previous pages which is a pair of parallel lines.

BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solution 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, 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 Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.
Solution:
1. Let us denote the number of boys by x and the number of girls by y. Then, the equations formed are
x + y = 10 ………………. (1)
and y = x + 4 i.e., – x + y = 4 ……………… (2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of these equations. The solutions of the equations are given in the tables:

Plot the points and draw the lines passing through them to represent the equations as shown.
These two lines intersect at (3, 7). So, x = 3 and y = 7 is the required solution.

Hence, the number of boys and girls are 3 and 7 respectively.

Verification:
Put x = 3 and y = 7 in (1) and (2), we find that both the equations are satisfied.

2. Let us denote the cost of one pencil by Rs x and one pen by Rs y. Then, the equations, formed are
5x + 7y = 50 ……………….. (1)
and 7x + 5y = 46 ………………… (2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of these equations. The solutions of the equations are given in tables:

Plot the points and draw the lines passing through them to represent the equations as shown.

These two lines intersect at (3, 5). So, x = 3 and y = 5 is the required solution.
Hence, the cost of one pencil is Rs 3 and that of one pen is Rs 5.

Verification:
Put x = 3 and y = 5 in (1) and (2), we find that both the equations are satisfied.

Question 2.
On comparing the ratios a1a2, b1b2 and c1c2, find out whether the lines representing the following pairs
of linear equations intersect at a point, are parallel or coincide.

1. 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
3. 6x – 3y + 10 = 0; 2x – y + 9 = 0

Solution:
1. The given pair of linear equations is
5x – 4y + 8 = 0 ……………… (1)
and 7x + 6y – 9 = 0 …………….. (2)
Here, 918 = 36 = 1224 [∵ Each = 12]
∴ lines (1) and (2) are coincident lines.

2. The given pair of linear equations is
9x + 3y + 12 = 0 ………………… (1)
and 18x + 6y + 24 = 0 ……………………… (2)
Here, 918 = 36 = 1224 [∵ Each = 12]
∴ lines (1) and (2) are coincident lines.

3. The given pair of linear equations is
6x – 3y + 10 = 0 ……………….. (1)
and 2x – y + 9 = 0 ……………… (2)
Here, 62 = −3−1 ≠ 109
∴ lines (1) and (2) are parallel lines.

Question 3.
On comparing the ratios a1a2, b1b2 and c1c2, find out whether the following pairs of linear equations are consistent, or inconsistent.

1. 3x + 2y = 5; 2x – 3y = 7
2. 2x – 3y = 8; 4x – 6y = 9
3. 32x + 53y = 7; 9x – 10y = 14
4. 5x – 3y = 11; – 10x + 6y = – 22
5. 43x + 2y = 8; 2x + 3y = 12

Solution:
1. 3x + 2y = 5; 2x – 3y = 7. That is, they are:
3x + 2y – 5 = 0; 2x – 3y – 7 = 0
a1 = 3, b1 = 2, c1 = – 5; a2 = 2, b2 = – 3, c2 = – 7

∴ The pair of linear equations is consistent.

2. 2x – 3y = 8; 4x – 6y = 9. That is they are 2x – 3y – 8 = 0; 4x – 6y – 9 = 0
a1 = 2, b1 = – 3, c1 = – 8; a2 = 4, b2 = – 6, c2 = – 9

∴ The pair of linear equations is consistent.

3. 32x + 53y = 7; 9x – 10y = 14. That is, they are 32x + 53y – 7 = 0 9x – 10y – 14 = 0
a1 = 32, b1 = 53, c1 = – 7; a2 = 9, b2 = – 10, c2 = – 14

∴ The pair of linear equations is consistent.

4. 5x – 3y = 11; – 10x + 6y = – 22. That is, they are:
5x – 3y – 11 = 0; – 10x + 6y + 22 = 0
a1 = 5, b1 = – 3, c1 = – 11; a2 = – 10, b2 = 6, c2 = 22

∴ The pair of linear equations is consistent (and dependent).

5. 43x + 2y = 8; 2x + 3y = 12. That is, they are:
43x + 2y – 8 = 0; 2x + 3y – 12 = 0
a1 = 43, b1 = 2, c1 = – 8; a2 = 2, b2 = 3, c2 = – 12

∴ The pair of linear equations is consistent (and dependent).

Question 4.
Which of the following pairs of linear equations are consistent? Obtain the solution in such cases graphically.

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

Solution:
1. Graph of x + y = 5 :
We have: x + y = 5 or y = 5 – x
When x = 0, y = 5; when x = 5, y = 0.
Thus, we have the following table:

Plot the points A(0, 5) and B(5, 0) on the graph paper, Join A and B and extend it on both sides to obtain the graph of x + y = 5.

Graph of 2x + 2y = 10:
We have: 2x + 2y = 10 or 2y = 10 – 2x
When x = 1, y = 5 – 1 = 4; when x = 2, y = 5 – 2 = 3
Thus, we have the following table:

Plotting the points C(1, 4) and D(2, 3) on the graph paper and drawing a line passing through these points on the same graph paper, we obtain the graph of 2x + 2y = 10. We find that C and D both lie on the graph of x + y = 5. Thus, the graphs of the two equations are coincident. Consequently, every solution of one equation is a solution of the other. Hence, the system of equations has infinitely many solutions, i.e., consistent and dependent.

2. Graph of x – y = 8:
We have: x – y = 8 or y = x – 8
When x = 0, y = – 8 when x = 8, y = 0
Thus, we have the following table:

Plot the points A(0, – 8) and B(8, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of x – y = 8 as shown.
Graph of 3x – 3y = 16:
We have: 3x – 3y = 16 or 3y = 3x – 16 or y = 3x−163
When x = 0, y = −163 = – 513;
When x = 163 = 513, y = 0
Thus, we have the following table:

Plot the points C(0, −163) and D(163, 0) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 3x – 3y = 16 as shown. We find the graphs of x – y = 8 and 3x – 3y = 16 are parallel. So, the two lines have no common point. Hence, the given equations has no solution, i.e., inconsistent.

3. Graph of 2x + y – 6 = 0:
We have: 2x + y – 6 = 0 or y = 6 – 2x
When x = 0, y = 6 – 0 = 6; when x = 3, y = 6 – 6 = 0
Thus, we have the following table:

Plot the points A(0, 6) and B(3, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 2x + y – 6 = 0 as shown.
Graph of 4x – 2y – 4 = 0:
We have: 4x – 2y – 4 = 0 or 2y = 4x – 4 or y = 2x – 2
When x = 0, y = – 2; when x – 1, y = 0
Thus, we have the following table:


Plotting the points C(0, – 2) and D(2, 0) on the same graph and drawing a line joining them as shown, we obtain the graph of 4x – 2y – 4 = 0.
Clearly, the two lines intersect at point P(2, 2).
Hence, x = 2, y = 2 is the solution of the given equations, i.e., consistent.

Verification:
Putting x = 2, y = 2 in the given equations, we find that both the equations are satisfied.

4. Graph of 2x – 2y – 2 = 0:
We have:
2x – 2y – 2 = 0 or 2y = 2x – 2 or y = x – 1
When x = 2, y = 0; when x = 0, y = – 2
Thus, we have the following table:

Plot the points A(1, 0) and B(0, – 1) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 2x – 2y – 2 = 0.
Graph of 4x – 4y – 5 = 0:
We have:
4x – 4y – 5 = 0 or 4y = 4x – 5 or y = 4x−54
When x = 0, y = – 54; when x = 54, y = 0.
Thus, we have the following table:

Plot the points C(0, −54) and D(54, 0) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 4x – 4y – 5 = 0 as shown. We find the graphs of these equations are parallel lines. So, the two lines have no common point. Hence, the given system of equations has no solution, i.e., 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 length of the garden be x m and its width be y m.
Then, perimeter = 2( Length + Width)
= 2(x + y)
Therefore, half perimeter = (x + y)
But it is given as 36
∴ (x + y) = 36 ………………. (1)
Also, x = y + 4 i.e; x – y = 4 ………………… (2)
For finding the solution of (1) and (2) graphically, we form the following tables:

Draw the graphs by joining points (20, 16) and (24, 12) and points (10, 6) and (16, 12). The two lines intersect at point (20, 16) as shown in the figure.

Hence, Length = 20 m and width = 16 m.

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:
Given linear equation is 2x + 3y – 8 = 0 ……………… (1)
1. For intersecting lines, we know that
a1a2≠b1b2
Any intersecting line may be taken as
5x + 2y – 9 – 0

2. For parallel lines, a1a2 = b1b2 ≠ c1c2
Any line parallel to (1) may be taken as
4x + 6y – 3 = 0

3. For coincident lines, a1a2 = b1b2 = c1c2
Any line coincident to (1) may be taken as
6x + 9x – 24 = 0
Note: Answers can differ from person to person.

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:
For the graph of x – y + 1 = 0:
We have:
x – y + 1 = 0 or y = x + 1
When x = 0, y = 1; when x = – 1, y’= 0.
Thus, we have the following table:

Plot the points A(0, 1) and B(- 1, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of x – y + 1 = 0.

For the graph of 3x + 2y – 12 = 0:
We have:
3x + 2y – 12 = 0 or 2y = 12 – 3x or y = 12−3x2
When x = 4, y = 12−122 = 0
When x = 0, y = 122 = 6
Thus, we have the following table:

Plot the points C(4, 0) and D(0, 6) on the same graph paper and draw a line passing through these two points to obtain the graph of equation 3x + 2y – 12 = 0. Clearly, we obtain a ∆ PBC formed by the given lines and the x-axis. The co-ordinates of the vertices are P(2, 3), B(- 1, 0) and C(4, 0). Triangular region is shaded as shown.

BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

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

2. s – t = 3
s3 + t2 = 6

3. 3x – y = 3
9x – 3y = 9

4. 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

5. 2x−−√ + 3y−−√ = 0
3–√x – 8–√y = 0

6.m 3x2 – 5y3 = – 2
x3 + y2 = 136
Solution:
1. The given system of equations is
x + y = 14 …………………. (1)
and x – y = 4 ……………… (2)
From (1), y = 14 – x
Substituting y = 14 – x in (2), we get
x – (14 – x) = 14 or x – 14 + x = 4 or 2x = 4 + 14 or 2x = 18
or x = 9
Putting x = 9 in (1), we get
9 + y = 14 or y = 14 – 9 or y = 5
Hence, the solution of the given system of equations is
x = 9, y = 5

2. The given system of equations is
s – t = 3 …………… (1)
s3 + t2 = 6
From (1), s = 3 + t
Substituting s = 3 + t in (2), we have
3+t3 + t2 = 6 or 2(3 + t) + 3t = 36
or 6 + 2t + 3t = 36 or 5t = 30 or t = 6
Putting t = 6 in (1), we get
s – 6 = 3 or s = 3 + 6 = 9
Hence, the solution of the given system of equations is
s = 9, t = 6

3. The given system of equations is
3x – y = 3 …………. (1)
and 9x – 3y = 9 …………… (2)
From (1), y = 3x – 3
Substituting y = 3x – 3 in (2), we get
9x – 3(3x – 3) = 9 or 9x – 9x + 9 = 9
or 9 = 9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y. This situation has arisen because both the given equations are the same.
∴ Equations (1) and (2) have infinitely many solutions.

4. The given system of equations is
0.2x + 0.3y = 1.3 ⇒ 2x + 3y = 13 …………….. (1)
and 0.4x + 0.5y = 2.3 ⇒ 4x + 5y = 23 ……………….. (2)
From (2), 5y = 23 – 4x or y = 23−4x5
Substituting y = 23−4x5 in (1), we get
2x + 3(23−4x5) = 13 or 10x + 69 – 12x = 65
or – 2x = – 4 or x = 2
Putting x = 2 in (1), we get
2 × 2 + 3y = 13 or 3y = 13 – 4 = 9
or y = 93 = 3
Hence, the solution of the given system of equations is
x = 2, y = 3

5. The given system of equations is
2x−−√ + 3y−−√ = 0 ………………. (1)
and 3x−−√ – 8y−−√ = 0 ………………… (2)
From (2), 8y−−√ = 3x−−√ or y = 3√x8√
Substituting y = 3√x8√ in (1), we get

Putting x = 0 in (1), we get
0 + 3y−−√ = 0 or y = 0

6. The given system of equations is

or 20x + 27x + 36 = 130
or 47x = 130 – 36 or 47x = 94
or x = 9447 = 2
Putting x = 2 in (1), we get
9 × 2 – 10y = – 12 or – 10y = – 12 – 18
or – 10y = – 30 or y = −30−10 = 3
Hence, the solution of the given system of equations is
x = 2, y = 3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
The given system of equations is
2x + 3y = 11 ……………… (1)
and 2x – 4y = – 24 ……………….. (2)
From (1), 3y = 11 – 2x or y = 11−2x3
Substituting y = 11−2x3 in (2), we get
2x – 4(11−2x3) = – 24 or 6x – 44 + 8x = – 72
or 14x = – 72 + 44 or 14x = – 28
or x = −2814 = – 2
Putting x = – 2 in (1), we get
2(- 2) + 3y = 11 or 3y = 11 + 4
or 3y = 15
or y = 153 = 5
putting x = – 2, y = 5 in y = mx + 3, we get
5 = m(- 2) + 3 or – 2m = 5 – 3
or – 2m = 2 or m = – 1

Question 3.
Form the pair of linear equations for the following problems and find their solutions 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 Rs 3800. Later, she buys 3 bats and 5 balls for Rs 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 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 travelling adistance of 25 km?

5. A fraction becomes 911, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes. 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 ages?
Solution:
1. Let the numbers be x and y (x > y). Then,
x – y = 26 ……………… (1)
and x = 3y ……………… (2)
From (2), y = x3
Substituting y = x3 in (1), we get
x – x3 = 26 or 3x – x = 78
or 2x = 78 or x = 39
Putting x = 39 in (2), we get
39 = 3y or y = 393 = 13
Hence, the required numbers are 39 and 13.

2. Let the angles be x° and y° (x > y). Then,
x + y = 180 ……………….. (1)
and x = y + 18 …………………. (2)
Substituting x = y + 18 in (1), we get
y + 18 + y = 180 or 2y = 180 – 18
or 2y = 162 or y = 1622 = 81
Putting y = 81 in (2), we get –
x = 81 + 18 = 99
Thus, the angles are 99° and 81°.

3. Let the cost of one bat and one ball be Rs x and Rs y respectively. Then,
7x + 6y = 3800 ………………… (1)
and 3x + 5y = 1750 ………………….. (2)
From (2), 5y = 1750 – 3x or y = 1750−3x5
Substituting y = 1750−3x5 in (1), we get
7x + 6(1750−3x5) = 3800 or 35x + 10500 – 18x = 19000
or 17x = 19000 – 10500 or 17x = 8500
or x = 850017 = 500
Putting x = 500 in (2), we get
3(500) + 5y = 1750 or 5y = 1750 – 1500
or 5y = 250 or y = 2505 = 50
Hence, the cost of one bat is Rs 500 and the cost of one ball is Rs 50.

4. Let the fixed charges of taxi be Rs x and the running charges be Rs y per km.
According to the given condition, we have:
x + 10y = 105 …………….. (1)
x + 15y = 155 ………………… (2)
From (1), x = 105 – 10y
Substituting x = 105 – 10y in (2), we get
105 – 10y + 15y = 155 or 105 + 5y = 155
or 5y = 155 – 105 or 5y = 50
or y = 10
Putting y = 10 in (1), we get
x + 10 × 10 = 105 or x = 105 – 100
or x = 5
Total charges for travelling a distance of 25 km
= x + 25y = Rs (5 + 25 × 10) = Rs 255
Hence, the fixed charge is Rs 5, the charge per km is Rs 10 and the total charges for travelling a distance of 25 km is Rs 255.

5. Let the fraction be xy
Then, according to the given conditions, we have:
x+2y+2 = 911 and x+3y+3 = 56
or 11x + 22 = 9y + 18 and 6x + 18 = 5y + 15
or 11x – 9y = – 4 ……………… (1)
and 6x – 5y = – 3 ……………… (2)
From (2), 5y = 6x + 3 or y = 6x+35
Substituting y = 6x+35 in (1), we get
11x – 96x+35 = – 4 or 55x – 54x – 27 = – 20
or x = – 20 + 27 or x = 7
Putting x = 7 in (1), we get
11(7) – 9y = – 4 or – 9y = – 4 – 77
or – 9y = – 81 or y = −81−9 = 9
Hence, the given fraction is 79.

6. Let the present age of Jacob be x years and the present age of his son be y years.
Five years hence,
Jacob’s age = (x + 5)
years Son’s age = (y + 5) years
Five years ago,
Jacob’s age = (x – 5) years
Son’s age = (y – 5) years
As per question, we get
(x + 5) = 3(y + 5) or x – 3y = 10 ……………. (1)
and (x – 5) = 7(y – 5) or x – 7y = – 30 …………….. (2)
From (1), x = 3y + 10
Substituting x = 3y + 10 in (2), we get
3y + 10 – 7y = – 30 or – 4y = – 30 – 10 or – 4y = 40 or y = 10
Putting y = 10 in (1), we get
x – 3 × 10 = 10 or x = 10 + 30 = 40
Hence, present age of Jacob is 40 years and that of his son is 10 years.

BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

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

1. x + y = 5 and 2x – 3y = 4
2. 3x + 4y = 10 and 2x – 2y = 2
3. 3x – 5y – 4 = 0 and 9x = 2y + 7
4. x2 + 2y3 = – 1 and x – y3 = 3

Solution:
1. By elimination method:
The given system of equations is
x +y = 5 ……………….. (1) and
2x – 3y = 4 ………………….. (2)
Multiplying (1) by 3, we get
3x + 3y = 15 ……………….. (3)
Adding (2) and (3), we get
5x = 19 or x = 195
Putting x = 195 in (1), we get
195 + y = 5 or y = 5 – 195 = 25−195 = 65
Hence, x = 195, y = 65.

By substitution method:
The given system of equations is
x + y = 5 ……………. (1) and
2x – 3y = 4 ……………….. (2)
From (1), y = 5 – x
Substituting y = 5 – x in (2), we get
2x – 15 + 3x = 4 or 5x = 4 + 15
or 5x = 19 or x = 195
Putting x = 195 in (1), we get
195 + y = 5 or y = 5 – 195 = 25−195 = 65
Hence, x = 195, y = 65.

2. By elimination method:
The given system of equations is
x2 + 2y3 = – 1 or 3x + 4y = – 6 ……………… (1)
and x – y3 = 3 or 3x – y = 9 ………………. (2)
Multiplying (2) by 4 and adding to (1), we get
15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3
Hence, x = 2, y = – 3
By substitution method:
The given system of equations is
x2 + 2y3 = – 1 or 3x + 4y = – 6 ……………….. (1)
and x – y3 = 3 or 3x – y = 9 …………….. (2)

From (2), Putting y = 3x – 9 in (1), we get y = 3x – 9
3x + 4(3x – 9) = – 6 or 3x + 12x – 36 = – 6
or 15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3
Hence, x = 2, y = – 3

3. By elimination method:
The given system of equations is
3x – 5y – 4 = 0 or 3x – 5y = 4 ……………….. (1)
and 9x = 2y + 7 or 9x – 2y = 7 ………………….. (2)
Multiplying (1) by 3 , we get
9x – 15y = 12 ………………. (3)
Substracting (1) by 3, we get
– 13y = 5 or y = – 513
Putting y = – 513 in (1), we get

By substitution method:
The given system of equations is

4. By elimination method:
The given system of equations is

Multiplying (2) by 4 and adding to (1), we get
15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3

By substitution method:
The given system of equations is

From (2), putting y = 3x – 9 in (1), we get y = 3x – 9
3x + 4(3x – 9) = – 6 or 3x + 12x – 36 = – 6
or 15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3
Hence, x = 2, y = – 3.

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12 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 digits. Find the number.

4. Meena went to a 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.

5. 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 she 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.
Solution:
1. Let x be the numerator and y be the denominator of the fraction. So, the fraction is xy.
By given conditions:
x+1y−1 = 1 or x + 1 = y – 1
or x – y = – 2 ……………….. (1)
and xy+1 = 12 or 2x = y + 1
or 2x – y = 1 ……………….. (2)
Subtracting (2) from (1), we get
(x – y) – (2x – y) = – 2 – 1
or x – y – 2x + y = – 3
or – x = – 3 or x = 3
Substituting x = 3 in (1), we get
3 – y = – 2 or y = 5
Hence, the required fraction is 35.

2. Let the present age of Nuri be x years
and the present age of Sonu be y years.
Five years ago,
Nuri’s age = (x – 5) years
Sonu’s age = (y – 5) years
As per conditions,
x – 5 = 3(y – 5) or x – 5 = 3y – 15
or x – 3y = – 15 + 5
or x – 3y = – 10 ………………… (1)
Ten years later,
Nuri’s age = (x + 10) years
As per condition’s,
x + 10 = 2 (y + 10) or x + 10 = 2y + 20
or x – 2y = 20 – 10
or x – 2y = 10 ………………… (2)
Putting y = 20 in (2), we get
x – 2(20) = 10 or x = 10 + 40 = 50
Nuri’s present age = 50 years
and Sonu’s present age = 20 years

3. Let the digits in the unit’s place and ten’s place be x and y respectively.
∴ Number = 10y + x
If the digits are reversed, the new number = 10y + x
As per conditions:
x + y = 9 ……………… (1)
and 9(10y + x) = 2(10x + y)
or 90y + 9x = 20x + 2y
or 20x – 9x + 2y – 90y = 0
From (1), y = 9 – x
Putting y = 9 – x in (2), we get
11x – 88(9 – x) = 0
or 11x – 88 × 9 + 88x = 0
or 99x = 88 × 9
Putting x = 8 in (1), we get
8 + y = 9 or y = 1
Hence, the number = 10y + x = 10 × 1 + 8 = 18

4. Let the number of Rs 50 notes be x and number of Rs 100 notes be y. Then,
x + y – 25 …………….. (1)
and 50x + 100y = 2000
or x + 2y = 40 …………….. (2)
On subtracting (1) from (2), we get
y = 15
Putting y = 15 in (1), we get
x + 15 = 25 or x = 10
Hence, number of Rs 50 notes = 10
and number of Rs 100 notes = 15

5. Let the fixed charges for 3 days be Rs x and charges per day be Rs y.
∴ By the given conditions,
x + 4y = 27 ………….. (1)
and x + 2y = 21 ……………… (2)
Subtracting (2) from (1), we get
2y = 6 or y = 3 Putting y = 3 in (1), we get
x + 4 × 3 = 27 or x = 27 – 12 = 15
∴ Fixed charges = Rs 15 and
charges per day = Rs 3

BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or 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 system of equations is
x – 3y – 3 = 0 and
3x – 9y – 2 = 0
These are of the form
a1x + b1y + c1 = 0
and a2x + b2y + c2 = 0
where a1 = 1, b1 = – 3, c1 = – 3 and a2 = 3, b2 = – 9, c2 = – 2

So, the given system of equations has no solution, i.e., it is inconsistent.

2. The given system of equations may be written as 2x + y – 5 = 0 and 3x + 2y – 8 = 0
These are of the form
a1x + b1y + c1 = 0 and x2x + b2y + c2 = 0
where a1 = 2, b1 = 1, c1 = – 5 and a2 = 3, b2 = 2 and c2 = – 8

So, the given system of equations has a unique solution.
To find the solution, we use the cross-multiplication method. By cross-multiplication, we have:

Hence, the given system of equations has a unique solution given by x = 2, y = 1.

3. The given system of equations may be written as
3x – 5y – 20 – 0 and 6x – 10y – 40 = 0
The given equations are of the form
a1x + b1y + c1 = – 20 and a2 = 6, b2 = – 10, c2 = – 40.

So, the given system of equations has infinitely many solutions.

4. The given system of equations is
x – 3y – 7 = 0 and 3x – 3y – 15 = 0
The given equations are of the form
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where a1 = 1, b1 = – 3, c1 = – 7 and a2 = 3, b2 = – 3, c2 = – 15

So, the given system of equations has a unique solution.
To find the solution, we use cross-multiplication method. By cross-multiplication, we have:

Hence, the given system of equations has a unique solution given by
x = 4, y = – 1

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. We know that the system of equations
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has infinite number of solutions if
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ The given system of equations can be written as follows:
2x + 3y – 7 = 0
(a – b)x + (a – t – b)y – 3a – b + 2 = 0
It will have infinite number of solutions, if

2a + 2b = 3a – 3b and 6a + 2b – 4 = 7a – 7b
a = 5b and a – 9b = – 4
Putting a = 5b in a = 5b = – 4, we get
56 – 96 = – 4 or – 46 = – 4 or 6 = 1
Putting b = 1 in a = 56, we get
a = 5(1) = 5
Hence, the given system of equations will have infinitely many solutions, if
a = 5 and b = 1

2. We know that the system of equations
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
has no solution if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
The given system of equations can be written as follows :
3x + y – 1 = 0
(2k – 1)x + (k – 1)y – 2k – 1 = 0
It will have no solution, if

Clearly, for k = 2, we have:
\(\frac{1}{k-1}\) ≠ \(\frac{1}{2k+1}\)
Hence, the given system of equations will have no solution, if k = 2.

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
The given system of equations is
8x + 5y = 9 …………….. (1)
and 3x + 2y = 4 ……………. (2)
By substitution method:
Substituting y = \(\frac{9-8x}{5}\) in (2), we get
or 15x + 18 – 16x = 20 or – x = 2 or x = – 2
Putting x = – 2 in (1), we get
8(- 2) + 5y = 9 or 5y = 9 + 16 = 25
or y = \(\frac{25}{5}\) = 5
Hence, x = – 2, y = 5 is the solution of the given system of equations.

By cross-multiplication method:
The given equations can be written as follows:
8x + 5y – 9 – 0
3x + 2y – 4 = 0
So, we have:

Hence, x = – 2, y = 5 is the solution of the given system of equations.

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
1. 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.

2. A fraction becomes \(\frac{1}{3}\) when 1 is subtracted from the numerator and it becomes \(\frac{1}{4}\) when 8 is added to its denominator. Find the fraction.

3. 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?

4. 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?

5. 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:
1. Let the fixed charges be Rs x and charges per day be Rs y.
∴ By the given conditions,
x + 20y = 1000 …………….. (1)
and x + 26y = 1180 ……………….. (2)
Subtracting (1) from (2), we get
6y = 180 or y = 30
From (1), x + 20 × 30 = 1000
x = 1000 – 600 = 400
∴ Fixed charges = Rs 400
and cost of food per day = Rs 30

2. Let the fraction be \(\frac{x}{y}\).
According to the given conditions:
\(\frac{x-1}{y}\) = \(\frac{1}{3}\) or 3x – 3 = y
or 3x – y – 3 = 0 …………….. (1)
and \(\frac{x}{y+8}\) = \(\frac{1}{4}\) or 4x = y + 8
or 4x – y – 8 = 0 ………………. (2)
Solving (1) and (2) by cross-multiplication method, we have:

Hence, the required fraction is \(\frac{5}{12}\).

3. Let Yash answer x questions correctly and y questions incorrectly. According to the given conditions:
3x – y = 40 ………………. (1)
and 4x – 2y = 50 ………………….. (2)
Multiplying (1) by 2 and subtracting (2) from the result so obtained, we get
2x = 30 or x = 15
From (1), 3 × 15 – y = 40 or – y = 40 – 45
or – y = – 5 or y = 5
∴ Total number of questions in the test are 15 + 5 = 20.

4. Let X and Y be two cars starting from places A and B respectively. Let the speed of car X be x km/h and that of Y be y km/h.
Case I: When two cars move in the same direction.
Let these cars meet at point Q. Then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 5 hours.
∴ Distance travelled by car X in 5 hours = 5x km
or AQ = 5x
Distance travelled by car Y in 5 hours = 5y km
or BQ = 5y
Clearly, AQ – BQ = AB
or 5x – 5y = 100
or x – y = 20 [∵ AB = 100 km]

Case II: When two cars move in opposite directions.
Let these cars meet at point P. Then,
Distance travelled by car X = AP
Distance travelled by car Y = BP
In this case, two cars meet in 1 hour.
∴ Distance travelled by car X in 1 hour = x km
or AP = x
Distance travelled by car Y in 1 hour = y km
or BP = y
Clearly, AP + PB = AB
or x + y = 100 ……………. (2) [∵ AB = 100 km]
Solving (1) and (2), we have:
x = 60, y = 40
Hence, speed of car X is 60 km/h and speed of car Y is 40 km/h.

5. Let the length and breadth of the rectangle be x and y units respectively. Then,
Area = xy sq. units
If length is reduced by 5 units and breadth is increased by 3 units, then area is reduced by 9 sq. units.
∴ xy – 9 = (x – 5)(y + 3)
or xy – 9 = xy + 3x – 5y – 15
or 3x – 5y – 6 = 0 …………….. (1)
When length is increased by 3 units and breadth by 2 units, the area is increased by 67 sq. units.
∴ xy + 67 = (x + 3 )(y + 2)
or xy + 67 = xy + 2x + 3y + 6
or 2x + 3y – 61 = 0 ………………… (2)
Solving (1) and (2), we get
\(\frac{x}{305+18}\) = \(\frac{-y}{-183+12}\) = \(\frac{1}{9+10}\)
or x = \(\frac{323}{19}\) = 17, y = \(\frac{171}{19}\) = 9
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.

BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:
1. 12x + 13y = 2
13x + 12y = 136

2. 2x√ + 3y√ = 2
4x√ – 9y√ = – 1

3. 4x + 3y = 14
3x – 4y = 23

4. 5x−1 + 1y−2 = 2
6x−1 – 3y−2 = 1

5. 7x−2yxy = 5
8x+7yxy = 15

6. 6x + 3y = 6xy
2x + 4y = 5xy

7. 10x+y + 2x−y = 4
15x+y – 5x−y = – 2

8. 13x+y + 13x−y = 34
12(3x+y) – 12(3x−y) = −18
Solution:
1. Taking 1x = u and 1y = v, the given system of equations becomes
12u + 13v = 2 or 3u + 2v = 12 ……………… (1)
13u + 12v = 136 or 2u + 3v = 13 ………………. (2)
Multliplying (1) by 3 and (2) by 2, we have:
9u + 6v = 36 ………………. (3)
and 4u + 6v = 26 ………………. (4)
Substracting (4) from (3), we get
5u = 10 or u = 2
Putting u = 2 in (3), we get
18 + 6v = 36 or 6v = 18 or v = 3
Now, u = 2 gives 1x = 2 or x = 12
and v = 3 gives 1y = 3 or y = 13
Hence, the solution is x = 12, y = 13.

2. The given system of equation is
2x√ + 3y√ = 2 and 4x√ – 9y√ = – 1
Putting u = 1x√ and v = 1y√, the given equations becomes
2u + 3v = 2 ………………. (1)
and 4u – 9v = – 1 ……………… (2)
6u + 9v = 6 …………………… (3)
Adding (2) and (3), we get
10u = 5 or u = 510 = 12
Putting u = 12 in (1), we get
2 × 12 + 3v = 2 or 3v = 1 or v = 13

Hence, the solution is x = 4, y = 9.

3. The given system of equation is
4x + 3y = 14 …………………. (1)
and 3x – 4y = 23 ………………. (2)
Multliplying (1) by 4 and (2) by 3, we get
16x + 12y = 56 ……………….. (3)
and 9x – 12y = 69 ……………….. (4)
Adding (3) and (4), we get
25x = 125 or x = 25125 = 15
Putting x = 15 in (1), we get
4 × 5 + 3y = 14 or 3y = 14 – 20
or 3y = – 6 or y = – 2
Hence, the solution is x = 15, y = – 2.

4. Let u = 1x−1 and v = 1y−2. Then, the given system of equations becomes
5u + v = 2 …………………. (1)
and 6u – 3v = 1 ……………….. (2)
Multliplying (1) by 3, we get
15u + 3v = 6 ………………… (3)
Adding (2) and (3), we get
21u = 7 or u = 13
Putting u = 13 in (1), we get
53 + v = 2 or v = 2 – 53 = 6−53 = 13
Now, u = 13 gives
1x−1 = 13 or x – 1 = 3 or x = 4
v = 13 gives
1y−2 = 13 or y – 2 = 3 or y = 5
Hence, the solution is x = 4, y = 5.

5. The given system of equation is

Let u = 1x and v = 1y. Then, the above equations become 7v – 2u = 5 …………… (1)
and 8v + 7u = 15 …………….. (2)
Multliplying (1) by 7 and (2) by 2, we get
49u – 14u = 35 ………………. (3)
and 16v + 14u = 30 ………………. (4)
Adding (3) and (4), we get
65v = 65 or v = 1
Putting v = 1 in (1), we get
7 – 2u = 5 or – 2u = – 2 or u = 1
Now, u = 1 gives 1x = 1 or x = 1
and v = 1 gives 1y = 1 or y = 1
Hence, the solution is x = 1, y = 1.

6. The given system of equations is 6x + 3y = 6xy and 2x + 4y = 5xy, where x and y are non-zero.
Since x ≠ 0, y ≠ 0, we have xy ≠ 0.
On dividing each one of the given equatìons by xy, we get
3x + 6y = 6 and 4x + 2y = 5
Taking 1x = u and 1y = v, the above equations become
3u + 6v = 6 …………….. (1)
and 4u + 2v = 5 ………………. (2)
Multiplying (2) by 3, we get
12u + 6v = 15
Subtracting (1) from (3), we get
9u = 15 – 6 = 9 or u = 1
Putting u = 1 in (1), we get
3 × 1 + 6v = 6 or 6v = 6 – 3 = 3
or v = 12
Now, u = 1 gives 1x = 1 or x = 1
and v = 12 gives 1y = 12 or y = 2
Hence, the given system of equations has one solution x = 1, y = 2.

7. The given system of equations is
10x+y + 2x−y = 4
15x+y – 5x−y = – 2
Putting u = 1x+y and v = 1x−y, the given equations become
10u + 2v = 4 or 5u + v = 2 ………………. (1)
and 15u – 5v = – 2
Putting u = 1x+y and v = 1x−y, the given equations become
10u + 2v = 4 or 5u + v = 2 ……………… (1)
and 15u – 5v = – 2 ………………… (2)
Multiplying (1) by 5, we get
25u + 5v = 10 ………………… (3)
Adding (2) and (3), we get
40u = 8 or u = 15
Putting u = 15 in (1), we get
5(15) + v = 2 or v = 2 – 1 = 1
Now, u = 15 gives x + y = 5 ……………….. (4)
v = 1 gives 1x−y = 1
or x – y = 1 …………………. (5)

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions:
1. 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.

2. 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.

3. 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:
1. Let the speed of the boat in still water be x km/h and the speed of the current be y km/h. Then,
Speed upstream = (x – y) km/h
Speed downstream = (x + y) km/h
Time taken to cover 20 km downstream
= 2 hours
or 20x+y = 2 or x + y = 10 ……………….. (1)
Time taken to cover 4 km upstream = 2 hours
or 4x−y = 2 or x – y = 2 ………………. (2)
Adding (1) and (2) we get,
2x = 12 or x = 6
Putting x = 6 in (1), we get
6 + y = 10 or y = 10 – 6 = 4
Hence, the speed of boat in still water = 6km/hr
and speed of current = 4km/hr

2. Let 1 woman can finish the embroidery in x days and 1 man can finish the embroidery in x days and 1 man can finish the embroidery in y days.
Then, 1 woman’s 1 day’s work = 1x
1 man’s 1 day’s work = 1y
∴ 2x + 5y = 14 and 3x + 6y = 13
Putting 1x = u and 1y = v, these equations become
2u + 5v = 14 …………….. (1)
and 3v + 6v = 13 ……………. (2)
Multiplying (1) by 3 and (2) by 2 and substracting, we get
3v = 112 or v = 136
Putting v = 136 in (1), we get

Thus, 1 woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.

3. Let the speed of train be x km/hr and speed of bus be y km/hr.

Multiplying eqn. (3) by 5 and eqn. (4) by 6, we get

∴ Speed of train is 60 km/hr and speed on bus is 80 km/hr.

BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as 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?
Solution:
Let the ages of Ani and Biju be x years and y years respectively. Then,
x – y = ±3 [Given]
Dharam’s age = 2x, and Cathy’s age = [Math Processing Error] Clearly, Dharam is older than Cathy.
Thus, we have the following two systems of linear equations:
x – y = ±3 ……………… (1)
and 4x – y = 60 ………………… (2)
x – y = – 3 ……………… (3)
and 4x – y = 60 ………………. (4)
Subtracting (1) from (2), we get
3x = 57 or x = 19
Putting x = 19 in (1), we get
19 – y = 3 or y = 19 – 3 = 16
Subtracting (4) from (3), we get
3x = 63 or x = [Math Processing Error] = 21
Putting x = 21 in (3), we get
21 – y = – 3 or y = 21 + 3 = 24
Hence, Ani’s age = 19 years and Biju’s age = 16 years
or Ani’s age = 21 years and Biju’s age = 24 years

Question 2.
One says, “Give me a hundred, friend ! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital?
[From the Bijaganita of Bhaskara II]
Solution:
Let the friends be named as A and B. Let A has Rs x and B has Rs y.
As per question, we have:
x + 100 = 2(y – 100)
or x – 2y + 300 = 0 ……………… (1)
and y + 10 = 6(x – 10)
or 6x – y – 70 = 0 …………… (2)
Multiplying (2) by 2, we get
12x – 2y – 140 = 0 …………… (3)
Subtracting (3) from (1), we get
– 11x + 440 = 0 or – 11x = – 440 or x = 40
Putting x = 40 in (1), we get
40 – 2y + 300 = 0 or – 2y = – 340 or y = 170

Question 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.
Solution:
Let the original speed of the train be x km/h and let the time taken to complete the journey be y hours.
Then, the distance covered = xy km

Case I:
When speed = (x + 10) km/h
and time taken = (y – 2) hours
In this case, distance = (x + 10)(y – 2)
or xy = (x + 10)(y – 2)
or xy = xy – 2x + 10y – 20
or 2x – 10y + 20 = 0 ………………… (1)

Case II:
When speed = (x – 10) km/h
and time taken = (y + 3) hours
In this case, distance = (x – 10)(y + 3)
or xy = (x – 10)(y + 3)
or xy = xy + 3x – 10y – 30
or 3x – 10y – 30 = 0 …………………… (2)
Subtracting (1) from (2), we get
x – 50 = 0 or x = 50
Putting x = 50 in (1), we get
100 – 10y + 20 = 0 or – 10y = – 120
or y = 12
∴ The original speed of the train = 50 km/h
The time taken to complete the journey = 12 hours
∴ The length of the journey = Speed × Time
= (50 × 12) km
= 600 km

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more.
Find the number of students in the class.
Solution:
Let originally there be x students in each row. Let there bey rows in total. Therefore, total number of students is xy.

Case I:
When 3 students are taken extra in each row, then the number of rows in this case becomes (y – 1).
∴ xy = (x + 3)(y – 1)
or xy = xy – x + 3y – 3
or x – 3y + 3 = 0 …………….. (1)

Case II:
When 3 students are taken less in each row, then the number of rows becomes (y + 2).
∴ xy = (x – 3)(y + 2)
or xy = xy + 2x – 3y – 6
or 2x – 3y – 6 = 0 ……………. (2)
Solving (1) and (2), we get
x = 9, y = 4
Hence, the number of students = xy = 9 × 4 = 36.

Question 5.
In a ∆ ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x°, ∠B = y°. Then,
∠C = 3∠B or ∠C = 3y°
or 3y = 2(x + y) or y = 2x
or 2x – y = 0 ………………… (1)
Since ∠A, ∠B and ∠C are angles of a triangle, therefore
∠A + ∠B + ∠C = 180°
or x + y + 3y = 180° or x + 4y = 180°…………………. (2)
Putting y = 2x in (2), we get
x + 8x = 180° or 9x = 180°
or x = 20°
Putting x = 20° in (1), we get
y = 2 × 20° = 40°
Hence, ∠A = 20°, ∠B = 40° and ∠C = 3y° = 3 × 40° = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.
Solution:
The given system of equations is
5x – y = 5 …………….. (1)
and 3x – y = 3 …………… (2)
Since both the equations are linear, so their graphs must be a pair of straight lines.
Consider the equation (1),
5x – y = 5 or y = 5x – 5
When x = 1, y = 5 – 5 = 0;
When x = 2, y = 10 – 5 = 5
Thus, we have the following table:

Plot the points A(1, 0) and B(2, 5) on the graph paper. Join AB and extend it on both sides to obtain the graph of 5x – y = 5.
Now, consider the equation (2),
3x – y = 3 or y = 3x – 3
When x = 0, y = 0 – 3 = – 3;
When x = 2, y = 3(2) – 3 = 3
Thus, we have the following table:

Plot the points C(0, – 3) and D(2, 3) on the same graph paper. Join CD and extend it on both sides to obtain the graph of 3x – y = 3.

We know that when a line meets y-axis, the value of x is 0. In the equation 5x – y = 5, if we put x – 0, we get y = – 5. Thus, the line 5x – y = 5 meets the y-axis at (0, – 5).
Similarly, the line 3x – y = 3 meets the y-axis at (0, – 3).
Thus, the co-ordinates of ∆ ACE formed by these two lines and the y-axis are A(1, 0), C(0, – 3) and E(0, – 5).

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

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

(iii) [Math Processing Error] – [Math Processing Error] = 0
ax + by = a2 + b2

(iv) Solve for x and y:
(a – b)x + (a + b)y = a2 – 2ab – b2
(a + b)(x + y) = a2 + b2

(v) 152x – 378y = – 74
– 378x + 152y = – 604
Solution:
(i) The given pair of equations is
px + qy = p – q or px + qy – (p – q) = 0
qx – py – p + q or qx – py – (p + q) = 0
By cross-multiplication, we get

Hence, the required solution is x = 1 and y = – 1.

(ii) The given system of equations may be written as
ax + by – c = 0
bx + ay – (1 + c) = 0
By cross-multiplication, we have:

Hence, the required solution is

(iii) The given system of equations is
[Math Processing Error] – [Math Processing Error] = 0 or bx – ay = 0 ……………. (1)
and ax + by – (a2 + b2) = 0 ………………. (2)
By cross-multiplication, we have:

Hence, the required solution is x = a and y = b.

(iv) The given pair of equations may be rewritten as
(a – b)x + (a + b)y – (a2 – 2ab – b2) = 0
and (a + b)x + (a + b)y – (a2 + b2) = 0
By cross-multiplication, we have:

Hence, the required solution is x = a + b and y = [Math Processing Error].

(v) We have:
152x – 378y = – 74 ……………… (1)
and – 378x + 152y = – 604 ……………….. (2)
Adding (1) and (2), we get
– 226x – 226y = – 678 or x + y = 3 ……………. (3)
Subtracting (1) from (2), we get
– 530x + 530y = – 530 or x – y = 1 …………….. (4)
Adding (3) and (4), we get
2x = 4 or x = 2
Putting x = 2 in (1), we get
2 + y = 3 or y = 1
Hence, the required solution is x = 2 and y = 1.

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

Solution:
We know that the sum of the opposite angles of a cyclic quadrilateral is 180°. In the cyclic quadrilateral ABCD, angles A and C, angles B and D form pairs of opposite angles.
∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°
or (4y + 20°) + 4x = 180°
and (3y – 5°) + (7x + 5°) = 180°
or 4x + 4y – 160° = 0 and 7x + 3y – 180° = 0
or x + y – 40° = 0 ………………… (1)
and 7x + 3y – 180° = 0 …………….. (2)
Multiplying (1) by 3, we get
3x + 3y – 120° = 0 …………… (3)
Subtracting (3) from (2), we get
4x – 60° = 0 or x = 15°
Putting x = 15° in (1), we get
15° + y – 40° = 0 or y – 25° = y = 25°
Hence, ∠A = 4y + 20° = 4 × 25° + 20°
= 100° + 20° = 120°
∠B = 3y – 5° = 3 × 25° – 5°
= 75° – 5°
= 70°
∠C = 4x = 4° × 15° = 60°
and ∠D = 7x + 5° = 7 × 15° + 5°
= 105° + 5°
= 110°.