SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in two Variables

SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in two Variables Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in two Variables Notes and select need one. SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in two Variables Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in two Variables

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per State Council of Educational Research and Training (SCERT) Book guidelines. SEBA Class 10 Maths Chapter 3 Pair of Linear Equations in two Variables are part of All Subject Solutions. Here we have given SEBA Solutions For Class 10 Maths for All Chapters, You can practice these here.

NEW VIDEO CLASS

Class 10 Mathematics Video Classes

Step-by-step easy concept explanation aur full questions Answer ke liye watch karein ya app download karein:

Watch on YouTube

Get Full Class on App

Pair of Linear Equations in two Variables

Chapter: 3

Exercise 3.1

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

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

Ans: Let the number of boys be x and the number of girls be y.

According to question,

x + y = 10

⇒ y = 10 − x …(i)

and

y = x + 4 …(ii)

Graphical representation:

Two points whose coordinates satisfy equation (i) are tabulated as:

	A	B
x	5	3
y = 10 – x	5	7

Two points whose coordinates satisfy equation (ii) are tabulated as:

	C	D
x	0	2
y = x + 4	4	6

The points A(5, 5) and B(3, 7) are plotted on the graph paper. On joining these points, we get which is the graph of equation (i).

Similarly, the points C(0, 4) and D(2, 6) are plotted on the same graph paper. On joining these points, we get which is the graph of equation (ii).

It is seen that the graphs of the two equations intersect at the point B(3, 7).

Thus, the two equations have a unique solution (3,7) or x = 3, y = 7.

Therefore, the number of boys is 3 and that of girls is 7.

(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.

Ans: Let the cost of one pencil be ₹ x and that of one pen be ₹ y.

According to question,

5x + 7y = 50…(i)

and 7x + 5y = 46…(ii)

Graphical representation:

From equation (i), we have

5x + 7y = 50

⇒ 7y = 50 − 5x

⇒ y = 50−5x/7

Two points whose coordinates satisfy equation (i) are tabulated as:

	A	B
x	10	3
y = 50−5x/7	0	5

From equation (ii), we have

7x + 5y = 46

⇒ 5y = 46 − 7x

⇒ Y = 46−7x/5

Two points whose coordinates satisfy equation (ii) are tabulated as:

	C	D
x	0	-2
y = 46 – 7x-7/5	46/5 = 9.2	12

The points A (10, 0) and B (3, 5) are plotted on the graph paper. On joining these points, we get vec which is the graph of equation (i).

Similarly, the points C (0, 9.2) and D (-2, 12) are plotted on the same graph paper. On joining these points, we get which is the graph of equation (ii).

It is seen that the graphs of the two equations intersect at the point B (3, 5). Thus, the two equations have a unique solution (3, 5) or x = 3 y = 5 .

Therefore, the cost of one pencil is 3 and that of one pen is 5.

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 coincident:

(i) 5x − 4y + 8 = 0

7x + 6y − 9 = 0

Ans: The given pair of linear equations is:

5x − 4y + 8 = 0 …(i)

and 7x + 6y − 9 = 0 …(ii)

Here, a₁ = 5, b₁ = −4, c₁ = 8

and a₂ = 7, b₂ = 6, c₂ = −9

Now, a₁/a₂ = 5/7

b₁/b₂ = -4/6 = -2/3

Since a₁/a₂ ≠ b₁/b₂, therefore the lines representing the given pair of linear equations intersect at a point.

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

Ans: The given pair of linear equations is:

9x + 3y + 12 = 0…(i)

and 18x + 6y + 24 = 0…(ii)

Here, a₁ = 9, b₁ = 3, c₁ = 12

and a₂ = 18, b₂ = 6, c₂ = 24

Now, a₁/a₂ = 9/18 = ½

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = 12/24 = 1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂ therefore the lines representing the given pair of linear equations are coincident lines.

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Ans: The given pair of linear equations is:

6x – 3y + 10 = 0 …. (i)

and 2x – y + 9 = 0 …. (ii)

Here, a₁ = 6, b₁ = -3, c₁= 10

and a₂ = 2, b₂ = -1, c₂ = 9

Now,

a₁/a₂ = 6/2 = 3

b₁/b₂ = (-3)/(-1) = 3

c₁/c₂ = 10/9

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, therefore the lines representing the given pair of linear equations are parallel lines.

3. On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5; 2x – 3y = 7

Ans: The given pair of linear equations is:

3x + 2y = 5

⇒ 3x + 2y – 5 = 0 …(i)

and 2x – 3y = 7

⇒ 2x – 3y – 7 = 0 …(ii)

here, a₁ = 3, b₁ = 2, c₁ = -5

and, a₂ = 2, b₂ = -3, c₂ = -7

Now, a₁/a₂ = 3/2

b₁/b₂ = 2/-3 = -2/3

Since a₁/a₂ ≠ b₁/b₂ Therefore the given pair of linear equations is consistent and has a unique solution.

(ii) 2x – 3y = 8; 4x – 6y = 9

Ans: The given pair of linear equations is:

2x – 3y = 8

⇒ 2x – 3y – 8 = 0 …(i)

and, 4x – 6y = 9

⇒ 4x – 6y – 9 = 0 …(ii)

Here, a₁ = 2, b₁ = -3, c₁ = -8

and, a₂ = 4, b₂ = -6, c₂ = -9

Now, a₁/a₂ = 2/4 = 1/2

b₁/b₂ = -3/-6 = 1/2

c₁/c₂ = -8/-9 = 8/9

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ therefore the given pair of linear equations is inconsistent and has no solution.

(iii) 3/2 x + 5/3 y = 7; 9x – 10y = 14

Ans: The given pair of linear equations is:

3/2 x + 5/3 y = 7

⇒ 3/2 x + 5/3 y – 7 = 0 …(i)

and 9x – 10y = 14

⇒ 9x – 10y – 14 = 0 …(ii)

Here, a₁ = 3/2, b₁ = 5/3, c₁ = -7

and a₂ = 9, b₂ = -10, c₂ = -14

Now,

Since a₁/a₂ ≠ b₁/b₂ therefore the given pair of linear equations is consistent and has a unique solution.

(iv) 5x – 3y = 11; -10x + 6y = -22

Ans: The given pair of linear equations is:

5x – 3y = 11

⇒ 5x – 3y – 11 = 0 …(i)

and -10x + 6y = 22

⇒ -10x + 6y + 22 = 0 …(ii)

Here, a₁ = 5, b₁ = -3, c₁ = -11

and a₂ = -10, b₂ = 6, c₂ = 22

Now, a₁/a₂ = 5/-10 = -1/2

b₁/b₂ = -3/6 = -1/2

c₁/c₂ = -11/22 = -1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂ therefore the given pair of linear equations is consistent and has infinitely many solutions.

(v) 4/3 x + 2y = 8 ; 2x + 3y = 12

Ans: The given pair of linear equations is:

4/3 x + 2y – 8

⇒ 4/3 x + 2y – 8 = 0 …(i)

and 2x + 3y = 12

⇒ 2x + 3y – 12 = 0 …(ii)

Here, a₁ = 4/3, b₁ = 2, c₁ = -8

and a₂ = 2, b₂ = 3, c₂ = -12

Now,

Since a₁/a₂ = b₁/b₂ = c₁/c₂, therefore the given pair of linear equations is consistent and has infinitely many solutions.

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

(i) x + y = 5, 2x + 2y = 10

Ans: The given pair of linear equations is:

x + y = 5

⇒ x + y – 5 = 0 …. (i)

and 2x + 2y = 10

⇒ 2x + 2y – 10 = 0 …. (ii)

Here, a₁ = 1, b₁ = 1, c₁ = -5

and a₂ = 2, b₂ = 2, c₂ = -10

Now,

a₁/a₂ = 1/2

b₁/b₂ = 1/2

c₁/c₂ = -5/-10 = 1/2

Since

a₁/a₂ = b₁/b₂ = c₁/c₂ therefore the given pair of linear equations is consistent.

Graphical representation:

From equation (i), we have

y = 5 – x

Two points whose coordinates satisfy equation (i) are tabulated as:

	A	B
x	0	5
y = 5 – x	5	0

From equation (ii), we have

2y = 10 – 2x

⇒ y = (10 – 2x)/2

Two points whose coordinates satisfy equation (ii) are tabulated as:

	C	D
x	2	4
y = (10 – 2x)/2	3	1

The points A (0, 5) and B (5, 0) are plotted on the graph paper. On joining these points, we get which is the graph of equation (i).

Similarly, the points C (2, 3) and D (4, 1) are plotted on the same graph paper. On joining these points, we get which is the graph of equation (ii).

It is seen from the graph that both the lines and coincide i.e. both the equations are equivalent and one can be derived from the other. So, the given pair of linear equations has infinitely many solutions.