** NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables- **This article curated by the GeeksforGeeks experts, contains free NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables to help students develop an easy approach to solving problems related to this chapter.

NCERT Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables, includes topics such as Introduction to linear equations, Pair of linear equations, Graphical representation of Linear equations, Equations reducible to linear form, and Application of linear equations in real-life problems.

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables**: Exercise 3.1

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables**

### 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:**

Present age of Aftab = x

And, the present age of his daughter = y

Seven years ago,Age of Aftab = x-7

Age of his daughter = y-7

Three years after,Age of Aftab = x+3

Age of his daughter = y+7

Here, According to the given condition,

x−7 = 7(y−7)

x−7 = 7y−49

x−7y = −42 ………………………(I)To draw the line for Eqn. (I), we need at least two solutions of the equation, So, we can use the following table to draw the graph:

x y 0 6 -42 0 And, According to the another given condition,

x+3 = 3(y+3)

x+3 = 3y+9

x−3y = 6 …………..…………………(II)To draw the line for Eqn. (II), we need at least two solutions of the equation, So, we can use the following table to draw the graph:

x y 0 -2 6 0 The graphical representation of Eqn. (I) and Eqn. (II) is:

### 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:**

The cost of a bat = ₹ x

And, the cost of a ball = ₹ y

Here, According to the Given condition,

3x+6y = 3900 …………………….(I)

x+3y = 1300 …………………….(II)To draw the line for Eqn. (I), we need at least two solutions of the equation, So, we can use the following table to draw the graph:

x y 0 650 1300 0 To draw the line for Eqn. (II), we need at least two solutions of the equation, So, we can use the following table to draw the graph:

x y 100 400 1300 0 The graphical representation of Eqn. (I) and Eqn. (II) is:

### Question 3. The cost of 2 kg of apples and 1kg 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:**

The cost of 1 kg of apples = ₹ x

And, cost of 1 kg of grapes = ₹ y

Here, According to the given conditions,

2x+y = 160 ………………………(I)

4x+2y = 300 ………………………(II)To draw the line for Eqn. (I), we need at least two solutions of the equation, So, we can use the following table to draw the graph:

x y 0 160 80 0 To draw the line for Eqn. (II), we need at least two solutions of the equation, So, we can use the following table to draw the graph:

x y 0 150 75 0 The graphical representation of Eqn. (I) and Eqn. (II) is:

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables**: Exercise 3.2

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables**

### Question 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.

**Solution:**

Let’s take,

Number of girls = x

Number of boys = y

According to the given conditions,

x + y = 10 -(1)

x – y = 4 -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.

For equation (1)

x + y = 10, So, we can use the following table to draw the graph:

x y 0 10 10 0 For equation (2)

x – y = 4, So, we can use the following table to draw the graph:

x y 0 -4 4 0 The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (7, 3).

Hence, the number of girls are 7 and number of boys are 3 in a class.

### (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.

**Solution:**

Let’s take,

Cost for one pencil = x

Cost for one pencil = y

According to the given conditions,

5x + 7y = 50

-(1)7x + 5y = 46

-(2)So, to construct a graph, we need to find at least two solutions of the given equation.

For equation (1)

5x + 7y = 50, So, we can use the following table to draw the graph:

x y 3 5 10 0 For equation (2)

7x + 5y = 46, So, we can use the following table to draw the graph:

x y 3 5 8 -2 The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (3, 5).

Hence, the cost of a pencil is ₹ 3 and cost of a pen is ₹ 5.

### Question 2. On comparing the ratios [Tex]\frac{a_1}{a_2}, \frac{b_1}{b_2}[/Tex], and [Tex]\frac{c_1}{c_2}[/Tex], 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

**Solution:**

In the given equations,

a1 = 5

a2 = 7

b1 = -4

b2 = 6

c1 = 8

c2 = -9

Now, here

a1/a2 = 5/7

b1/b2 = -4/6 = -2/3

c1/c2 = 8/-9

As, here

[Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}[/Tex]

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

### (ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

**Solution:**

In the given equations,

a

_{1}= 9a

_{2}= 18b

_{1}= 3b

_{2 }= 6c

_{1}= 12c

_{2}= 24Now, here

a

_{1}/a_{2 }= 9/18 = 1/2b

_{1}/b_{2}= 3/6 = 1/2c

_{1}/c_{2}= 12/24 = 1/2As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

### (iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0

**Solution:**

In the given equations,

a

_{1}= 6a

_{2}= 2b

_{1}= -3b

_{2}= -1c

_{1}= 10c

_{2}= 9Now, here

a

_{1}/a_{2}= 6/2 = 3b

_{1}/b_{2}= -3/-1 = 3c

_{1}/c_{2 }= 10/9As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

### Question 3. On comparing the ratios [Tex]\frac{a_1}{a_2}, \frac{b_1}{b_2}[/Tex], and [Tex]\frac{c_1}{c_2}[/Tex], find out whether the following pair of linear equations are consistent, or inconsistent.

### (i) 3x + 2y = 5; 2x – 3y = 7

**Solution:**

In the given equations,

a

_{1}= 3a

_{2}= 2b

_{1}= 2b

_{2}= -3c

_{1}= -5c

_{2}= -7Now, here

a

_{1}/a_{2}= 3/2b

_{1}/b_{2}= 2/-3c

_{1}/c_{2 }= -5/-7As, here

[Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}[/Tex]

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

Pair of linear equations are CONSISTENT.

### (ii) 2x – 3y = 8; 4x – 6y = 9

**Solution:**

In the given equations,

a

_{1}= 2a

_{2}= 4b

_{1}= -3b

_{2}= -6c

_{1}= -8c

_{2}= -9Now, here

a

_{1}/a_{2}= 2/4 = 1/2b

_{1}/b_{2}= -3/-6 = 1/2c

_{1}/c_{2}= -8/-9 = 8/9As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

Pair of linear equations are INCONSISTENT.

### (iii) [Tex]\frac{3}{2}x + \frac{5}{3}y[/Tex] = 7; 9x – 10y = 14

**Solution:**

In the given equations,

a

_{1}= 3/2a

_{2}= 9b

_{1}= 5/3b

_{2}= -10c

_{1 }= -7c

_{2}= -14Now, here

[Tex]\frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{1}{6}[/Tex]

[Tex]\frac{b_1}{b_2} = \frac{\frac{5}{3}}{-10} = \frac{-1}{6}[/Tex]

c

_{1}/c_{2}= -7/-14 = 1/2As, here

[Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}[/Tex]

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

Pair of linear equations are CONSISTENT.

### (iv) 5x – 3y = 11; – 10x + 6y = –22

**Solution:**

In the given equations,

a

_{1}= 5a

_{2 }= -10b

_{1 }= -3b

_{2}= 6c

_{1 }= -11c

_{2}= 22Now, here

a

_{1}/a_{2 }= 5/-10 = -1/2b

_{1}/b_{2}= -3/6 = -1/2c

_{1}/c_{2}= -11/22 = -1/2As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

Pair of linear equations are CONSISTENT.

### (v) [Tex]\frac{4}{3}x[/Tex] + 2y = 8; 2x + 3y = 12

**Solution:**

In the given equations,

a

_{1}= 4/3a

_{2 }= 2b

_{1}= 2b

_{2}= 3c

_{1}= -8c

_{2}= -12Now, here

[Tex]\frac{a_1}{a_2} = \frac{\frac{4}{3} }{2} = \frac{2}{3}[/Tex]

b

_{1}/b_{2}= 2/3c

_{1}/c_{2}= -8/-12 = 2/3As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

Pair of linear equations are CONSISTENT.

### Question 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

**Solution:**

In the given equations,

a1 = 1

a2 = 2

b1 = 1

b2 = 2

c1 = -5

c2 = -10

Now, here

a1/a2 = 1/2

b1/b2 = 1/2

c1/c2 = -5/-10 = 1/2

As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

Pair of linear equations are CONSISTENT.

x + y = 5-(1)

2x + 2y = 10-(2)

So, to construct a graph, we need to find at least two solutions of the given equation.

For equation (1)

x + y = 5, So, we can use the following table to draw the graph:

x y 0 5 5 0 For equation (2)

2x + 2y = 10, So, we can use the following table to draw the graph:

x y 0 5 5 0 The graph will be as follows for Equation (1) and (2):

### (ii) x – y = 8, 3x – 3y = 16

**Solution:**

In the given equations,

a

_{1 }= 1a

_{2}= 3b

_{1}= -1b

_{2}= -3c

_{1}= -8c

_{2}= -16Now, here

a

_{1}/a_{2}= 1/3b

_{1}/b_{2}= 1/3c

_{1}/c_{2}= -8/-16 = 1/2As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}[/Tex]

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

Pair of linear equations are INCONSISTENT.

### (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

**Solution:**

In the given equations,

a

_{1}= 2a

_{2}= 4b

_{1}= 1b

_{2}= -2c

_{1}= -6c

_{2 }= -4Now, here

a

_{1}/a_{2}= 2/4 = 1/2b

_{1}/b_{2}= 1/-2c

_{1}/c_{2 }= -6/-4 = 3/2As, here

[Tex]\frac{a1}{a2} ≠ \frac{b1}{b2} [/Tex]

Pair of linear equations are CONSISTENT.

2x + y – 6 = 0-(1)

4x – 2y – 4 = 0-(2)

So, to construct a graph, we need to find at least two solutions of the given equation.

For equation (1)

2x + y – 6 = 0, So, we can use the following table to draw the graph:

x y 0 6 3 0 For equation (2)

4x – 2y – 4 = 0, So, we can use the following table to draw the graph:

x y 0 -2 1 0 The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (2, 2).

### (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

**Solution:**

In the given equations,

a

_{1}= 2a

_{2}= 4b

_{1 }= -2b

_{2}= -4c

_{1 }= -2c

_{2}= -5Now, here

a

_{1}/a_{2}= 2/4 = 1/2b

_{1}/b_{2}= -2/-4 = 1/2c

_{1/}c_{2 }= -2/-5 = 2/5As, here

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}[/Tex]

Pair of linear equations are 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’s take,

length = x

breadth = y

Half the perimeter of a rectangular garden = [Tex]\frac{2(x+y)}{2}[/Tex] = x + y

According to the given conditions,

x = y + 4—(1)

x + y = 36—(2)

So, to construct a graph, we need to find at least two solutions of the given equation.

For equation (1)

x = y + 4, So, we can use the following table to draw the graph:

x y 0 -4 4 0 For equation (2)

x + y = 36, So, we can use the following table to draw the graph:

x y 0 36 36 0 The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (20, 16).

Hence, length is 20 m and breadth is 16 m of rectangle.

### 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:

### (i) intersecting lines

**Solution:**

Linear equation in two variables such that pair so formed is intersecting lines, so it should satisfy the given conditions

[Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}[/Tex]

By rearranging, we get

[Tex]\frac{a_1}{b_1} ≠ \frac{a_2}{b_2}[/Tex]

Hence, the required equation should not be in ratio of 2/3

Hence, another equation can be 2x – 9y + 9 = 0

where the ratio is 2/-9

and, [Tex]\frac{2}{3} ≠ \frac{2}{-9}[/Tex]

### (ii) parallel lines

**Solution:**

Linear equation in two variables such that pair so formed is parallel lines, so it should satisfy the given conditions

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}[/Tex]

By rearranging, we get

[Tex]\frac{a_1}{b_1} = \frac{a_2}{b_2}[/Tex]

[Tex]\frac{b_1}{c_1} ≠ \frac{b_2}{c_2}[/Tex]

Hence, the required equation a

_{2}/b_{2}should be in ratio of 2/3 and b_{2}/c_{2}should not be equal to 3/-8Hence, another equation can be 4x + 6y + 9 = 0

where the ratio a2/b2 is 2/3

and,[Tex] \frac{b_2}{c_2} ≠ \frac{3}{-8}[/Tex]

### (iii) coincident lines

**Solution:**

Linear equation in two variables such that pair so formed is parallel lines, so it should satisfy the given conditions

[Tex]\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} [/Tex]

By rearranging, we get

[Tex]\frac{a1}{b1} = \frac{a2}{b2} [/Tex]

[Tex]\frac{b1}{c1} = \frac{b2}{c2} [/Tex]

Hence, the required equation a2/b2 should be in ratio of 2/3 and b2/c2 should be equal to 3/-8

Hence, another equation can be 4x + 6y -16 = 0

where the ratio a2/b2 is 2/3

and, b2/c2 = 3/-8

### 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:**

x – y + 1 = 0

-(1)3x + 2y – 12 = 0

-(2)So, to construct a graph, we need to find at least two solutions of the given equation.

For equation (1)

x – y + 1 = 0, So, we can use the following table to draw the graph:

x y 0 -1 1 0 For equation (2)

3x + 2y – 12 = 0, So, we can use the following table to draw the graph:

x y 0 6 4 0 The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (2, 3), and x-axis at (−1, 0) and (4, 0).

Hence, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:** Exercise 3.3

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:**

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

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

**(i) x + y = 14 and x – y = 4**

**(i) x + y = 14 and x – y = 4**

**Solution:**

x + y = 14 ……….. (1)

x – y = 4 ………….. (2)

x = 14 – y

Substitute x in (2)

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Transposing 2

y = 10/2

y = 5

x = 14 – y

x = 9

Therefore, x = 9 and y = 5.

**(ii) s – t = 3 and (s/3) + (t/2) = 26**

**(ii) s – t = 3 and (s/3) + (t/2) = 26**

**Solution:**

s – t = 3 …….. (1)

(s/3) + (t/2) = 6 …………. (2)

s = 3 + t

Now, substitute the value of s in (2)

(3 + t) / 3 + (t/2) = 6

Taking 6 as LCM

(2(3 + t) + 3t) / 6 = 6

(6 + 2t + 3t) / 6 = 6

(6 + 5t) = 36

5t = 30

t = 6

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

**(iii) 3x – y = 3 and 9x – 3y = 9**

**(iii) 3x – y = 3 and 9x – 3y = 9**

**Solution:**

3x – y = 3 ……….. (1)

9x – 3y = 9 ……….(2)

From (1)

x = (3 + y) / 3

Substitute x in (2)

9(3 + y) / 3 – 3y = 9

9 + 3y – 3y = 9

0 = 0

Therefore, y has infinite values and x = (3 + y)/3 also has infinite values.

**(iv) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3**

**(iv) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3**

**Solution:**

0.2x + 0.3y = 1.3 ……… (1)

0.4x + 0.5y = 2.3 …….. (2)

From (1)

x = (1.3 – 0.3y) / 0.2

Putting x in (2)

0.4(1.3 – 0.3y) / 0.2 + 0.5y = 2.3

2(1.3 – 0.3y) + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

2.6 – 0.1 y = 2.3

0.1 y = 0.3

y = 3

Substitute y in (1)

x = (1.3 – 0.3(3)) / 0.2 = (1.3 – 0.9) / 0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

**(v) √2x + √3y = 0 and √3x – √8y = 0**

**(v) √2x + √3y = 0 and √3x – √8y = 0**

**Solution:**

√2 x + √3 y = 0 …………… (1)

√3 x – √8 y = 0 ………….. (2)

From (1)

x = – (√3/√2)y

Putting x in (2)

√3(-√3/√2)y – √8y = 0

(-3/√2)y – √8y = 0

-3y – 4y = 0

-7y = 0

y = 0

Therefore

x = 0

Therefore, x = 0 and y = 0.

**(vi) (3x/2) – (5y/3) = -2 and (x/3) + (y/2) = (13/6)**

**(vi) (3x/2) – (5y/3) = -2 and (x/3) + (y/2) = (13/6)**

**Solution:**

(3x/2) – (5y/3) = -2 ……………. (1)

(x/3) + (y/2) = 13/6 ………. (2)

From (1)

(3/2)x = -2 + (5y/3)

(3/2)x = (-6 + 5y) / 3

x = ((-6 + 5y) / 3) * 2/3

⇒ x = 2(-6 + 5y) / 9 = (-12 + 10y) / 9

Putting x in (2)

((-12 +10y)/9)/3 + y/2 = 13/6

(-12 + 10y)/27 + y/2 = 13/6

Taking 54 as LCM

-24 + 20y + 27y = 117

47y = 117 + 24

47y = 141

y = 3

x = (-12 + 30) / 9

x = 18/9

x = 2

Therefore, x = 2 and y = 3.

**Question 2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.**

**Question 2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.**

**Solution:**

2x + 3y = 11…………………………..(1)

2x – 4y = -24………………………… (2)

From (1)

x = (11 – 3y) / 2

Substituting x in equation (2)

2(11 – 3y) / 2 – 4y =- 24

11 – 3y – 4y = -24

-7y = -24 – 11

-7y = -35

y = 5

Putting y in (1)

x = (11 – 3 × 5) / 2 = -4/2 = -2

x = -2, y = 5

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore, the value of m is -1.

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

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

**(i) The difference between two numbers is 26 and one number is three times the other. Find them.**

**(i) The difference between two numbers is 26 and one number is three times the other. Find them.**

**Solution:**

Let the two numbers be x and y

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of y

3x – x = 26

2x = 26

x = 13

y = 39

Therefore, the numbers are 13 and 39.

**(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.**

**(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.**

**Solution:**

Let the larger angle by x

^{o}and smaller angle be y^{o}.Sum of two supplementary pair of angles is 180

^{o}.x + y = 180

^{o}……………. (1)x – y = 18

^{o }……………..(2)From (1)

x = 180 – y

Substituting in (2)

180

^{ }– y – y = 18-2y = -162

162 = 2y

y = 81

^{o}x = 180 – y

x = 180 – 81

= 99

Therefore, the angles are 99

^{o}and 81^{o}.

**(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.**

**(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.**

**Solution:**

Let the cost of a bat be Rs. x and cost of a ball be Rs. y.

7x + 6y = 3800 ………………. (i)

3x + 5y = 1750 ………………. (ii)

From (i)

y = (3800 – 7x) / 6………………..(iii)

Substituting (iii) in (ii)

3x + 5(3800 – 7x) / 6 =1750

Taking 6 as LCM

18x + 19000 – 7x = 10500

11x = 10500 – 19000

⇒ -17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800 – 7 × 500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

**(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?**

**(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?**

**Solution:**

Let the fixed charge be Rs x and per km charge be Rs y.

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1)

x = 105 – 10y ………………. (3)

Substituting the value of x in (2)

105 – 10y + 15y = 155

5y = 50

y = 10

Putting the value of y in (3)

x = 105 – 10 × 10

x = 105 – 100

x = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

**(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.**

**(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.**

**Solution:**

Let the fraction be x/y.

(x + 2)/(y + 2) = 9/11

By cross multiplication

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

By cross multiplication

(x + 3)/(y + 3) = 5/6

6x + 18 = 5y + 15

6x – 5y = -3 ………………. (2)

From (1)

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2)

6(-4 + 9y)/11 -5y = -3

Taking 11 as the LCM

-24 + 54y – 55y = -33

-24 – y = -33

-y = -33 + 24

-y = -9

y = 9

Substituting the value of y in (3)

x = (-4 + 9 × 9)/11

x = (-4 + 81)/11

x = 77/11

x = 7

Hence, the fraction is 7/9.

**(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?**

**(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?**

**Solutions:**

Let the age of Jacob = x years and that of son be y years.

(x + 5) = 3(y + 5) ………………… (i)

(x – 5) = 7(y – 5) ………………….. (ii)

From (i)

x + 5 = 3y + 15

x – 3y = 10……………. (iii)

From (ii)

x – 5 = 7y – 35

x – 7y = -30……………..(iv)

Subtracting (iv) from (iii)

-3y + 7y = 40

4y = 40

Transposing 4

y = 40/4

y = 10

Putting y = 10 in (iii)

x – 30 = 10

x = 40

Therefore, present age of Jacob is 40 years and that of son is 10 years

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:** Exercise 3.4

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:**

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

### (i) x + y = 5 and 2x – 3y = 4

**Solution:**

Here, the two given eqn. are as follows:

x + y = 5 ……….(I)

2x – 3y = 4 ………..(II)

ELIMINATION METHOD:Multiply equation (I) by 2, and then subtract (II) from it, we get

5y = 6

y = 6/5

Now putting y=6/5 in eqn. (I), we get

x + 6/5 = 5

x = (5–(6/5))

x = 19/5

SUBSTITUTION METHOD:From (I), we get

y=5–x…….(III)

Now substituting the value of y in eqn. (II), we get

2x – 3(5–x) = 4

2x – 15+3x = 4

5x = 4+15

x = 19/5

As, putting x = 19/5, in eqn. (III), we get

y = 5 – 19/5

y = 6/5

Hence, by elimination method and substitution method we get,

x = 19/5 and y = 6/5.

### (ii) 3x + 4y = 10 and 2x – 2y = 2

**Solution:**

Here, the two given eqn. are as follows:

3x + 4y = 10 ……….(I)

2x – 2y = 2 ………..(II)

ELIMINATION METHOD:Multiply equation (II) by 2, and then add it to (I), we get

7x = 14

x = 14/7

x = 2

Now putting x = 2 in eqn. (I), we get

3(2) + 4y = 10

4y = 10 – 6

y = 4/4

y = 1

SUBSTITUTION METHOD:From (II), we get

x = (2+2y)/2

x = y+1 …….(III)

Now substituting the value of x in eqn. (I), we get

3(y+1) + 4y = 10

3y + 3 + 4y = 10

7y = 10 – 3

y = 7/7

y = 1

As, putting y = 1, in eqn. (III), we get

x = 1+1

x = 2

Hence, by elimination method and substitution method we get,

x = 2 and y = 1.

### (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

**Solution:**

Here, the two given eqn. are as follows:

3x – 5y – 4 = 0

9x = 2y + 7

By rearranging we get,

3x – 5y = 4 ……….(I)

9x – 2y = 7 ………..(II)

ELIMINATION METHOD:Multiply equation (I) by 3, and then subtract (II) from it, we get

–13y = 5

y = -5/13

Now putting y = – 5/13 in eqn. (I), we get

3x – 5(– 5/13) = 4

3x = 4 – (25/13)

3x = 27/13

x = 9/13

SUBSTITUTION METHOD:From (I), we get

3x – 5y = 4

x = (4+5y)/3 …….(III)

Now substituting the value of x in eqn. (II), we get

9((4+5y)/3) – 2y = 7

3(4+5y) – 2y = 7

12+15y – 2y = 7

13y = – 5

y = – 5/13

As, putting y = – 5/13, in eqn. (III), we get

x=(4+5(– 5/13))/3

x = 9/13

Hence, by elimination method and substitution method we get,

–x=9/13 and y=5/13.

### (iv) x/2 + 2y/3 = -1 and x – y/3 = 3

**Solution:**

Here, the two given eqn. are as follows:

x/2 + 2y/3 = –1 ………..(A)

x – y/3 = 3 ……………….(B)

by rearranging (multiply (A) by 6 and multiply (B) by 3) we get,

3x + 4y = – 6 ……….(I)

3x – y = 9 ………..(II)

ELIMINATION METHOD:Subtract (II) from (I), we get

5y = – 15

y = – 3

Now putting y = -3 in eqn. (II), we get

3x – (– 3) = 9

3x = 9 – 3

x = 6/3

x = 2

SUBSTITUTION METHOD:From (II), we get

3x – y = 9

y = 3x – 9 …….(III)

Now substituting the value of y in eqn. (I), we get

3x + 4(3x – 9) = – 6

3x + 12x – 36 = – 6

15x = – 6 + 36

x = 30/15

x = 2

As, putting x = 2, in eqn. (III), we get

y = 3(2) – 9

y = – 3

Hence, by elimination method and substitution method we get,

–x=2 and 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:

### (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?

**Solution:**

Let the fraction be p/q, where p is numerator and q is denominator.

Here, According to the given condition,

(p+1)/(q – 1) = 1 ………………..(A)

and,

p/(q+1) = 1/2 …………………..(B)

Solving (A), we get

(p+1) = q – 1

–p–q =2 ……………………….(I)Now, solving (B), we get

2p = (q+1)

–2pq = 1……………………….(II)When equation (I) is subtracted from equation (II) we get,

p = 3

Now putting p = 3 in eqn. (I), we get

3 – q = – 2

q = 3+2

q = 5

So, p = 3 and q = 5.

Hence, the fraction p/q is 3/5.

### (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

**Solution:**

Let us assume, present age of Nuri is x

And present age of Sonu is y.

Here, According to the given condition, the equation formed will be as follows :

x – 5 = 3(y – 5)

–x – 3y =10 …………………………………..(I)Now,

x + 10 = 2(y +10)

x – 2y = 10 …………………………………….(II)Subtract eqn. (I) from (II), we get

y = 20

Now putting y = 20 in eqn. (II), we get

x – 2(20) = 10

x = 10+40

x = 50

Hence,

Age of Nuri is 50 years

Age of Sonu is 20 years.

### (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

**Solution:**

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number = 10y + x

And, reverse number = 10x + y

eg: 23

x = 3 and y = 2

So, 23 can be represented as = 10(2) + 3 = 23Here, According to the given condition

x + y = 9 …………………….(I)and,

9(10y + x) = 2(10x + y)

90y + 9x = 20x + 2y

88y = 11x

x = 8y

–x8y = 0 ………………………………………………………….. (II)Subtract eqn. (II) from (I) we get,

9y = 9

y = 1

Now putting y = 1 in eqn. (II), we get

x – 8(1) = 0

x = 8

Hence, the number is 10y + x

=10 × 1 + 8

Number = 18

### (iv) 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.

**Solution:**

Let the number of ₹ 50 notes be x and the number of ₹100 notes be y

Here, According to the given condition

x + y = 25 ……………………………….. (I)

50x + 100y = 2000 ………………………………(II)Divide (II) by 50 and then subtract (I) from it.

y = 15

Now putting y = 15 in eqn. (I), we get

x + 15 = 25

x = 10

Hence, Manna has 10 notes of ₹ 50 and 15 notes of ₹ 100.

### (v) 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 for five days. Find the fixed charge and the charge for each extra day.

**Solution:**

Let the fixed charge for the first three days be ₹ x and,

The charge for each day extra be ₹ y.

Here, According to the given condition,

x + 4y = 27 …….…………………………. (I)

x + 2y = 21 ………………………………………….. (II)Subtract (II) from (I), we get

2y = 6

y = 3

Now putting y = 3 in eqn. (II), we get

x + 4(3) = 27

x = 27 – 12

x = 15

Hence, the fixed charge is ₹15

And the Charge per day is ₹ 3

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables **: Exercise 3.5

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables**

### 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.

### (i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

**Solution:**

Here,

a

_{1}= 1, b_{1}= -3, c_{1}= -3a

_{2}= 3, b_{2}= -9, c_{2}= -2So,

[Tex]\frac{a_1}{a_2} = \frac{1}{3} [/Tex]

[Tex]\frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3} [/Tex]

[Tex]\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2} [/Tex]

As, [Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2} [/Tex]

Hence, the given pairs of equations have

.no solution

### (ii) 2x + y = 5, 3x + 2y = 8

**Solution:**

Rearranging equations, we get

2x + y -5 = 0

3x + 2y -8 = 0

Here,

a

_{1}= 2, b_{1}= 1, c_{1}= -5a

_{2}= 3, b_{2}= 2, c_{2}= -8So,

[Tex]\frac{a_1}{a_2} = \frac{2}{3} [/Tex]

[Tex]\frac{b_1}{b_2} = \frac{1}{2} [/Tex]

As, [Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2} [/Tex]

Hence, the given pairs of equations have

.unique solutionFor cross multiplication,

[Tex]\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}} [/Tex]

[Tex]\frac{x}{(1)(-8)-(2)(-5)} = \frac{y}{(-5)(3)-(-8)(2)} = \frac{1}{(2)(2)-(3)(1)} [/Tex]

[Tex]\frac{x}{(-8)+10} = \frac{y}{(-15)+16} = \frac{1}{4-3} [/Tex]

[Tex]\frac{x}{2} [/Tex] = y = 1

Hence,

[Tex]\frac{x}{2} [/Tex] = 1

x = 2and,

y = 1Hence, the

.required solution is x = 2 and y = 1

### (iii) 3x – 5y = 20, 6x – 10y = 40

**Solution:**

Rearranging equations, we get

3x – 5y – 20 = 0

6x – 10y – 40 = 0

Here,

a

_{1}= 3, b1 = -5, c1 = -20a

_{2}= 6, b_{2}= -10, c_{2}= -40So,

[Tex]\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} [/Tex]

[Tex]\frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} [/Tex]

[Tex]\frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2} [/Tex]

As,

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} [/Tex]

Hence, the given pairs of equations have

.infinitely many solutions

### (iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

**Solution:**

Here,

a

_{1}= 1, b1 = -3, c1 = -7a

_{2}= 3, b_{2}= -3, c_{2}= -15So,

[Tex]\frac{a_1}{a_2} = \frac{1}{3} [/Tex]

[Tex]\frac{b_1}{b_2} = \frac{-3}{-3} = 1 [/Tex]

As,

[Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2} [/Tex]

Hence, the given pairs of equations have

.unique solutionFor cross multiplication,

[Tex]\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}} [/Tex]

[Tex]\frac{x}{(-3)(-15)-(-3)(-7)} = \frac{y}{(-7)(3)-(-15)(1)} = \frac{1}{(1)(-3)-(3)(-3)} [/Tex]

[Tex]\frac{x}{45-21} = \frac{y}{(-21)+15} = \frac{1}{-3+9} [/Tex]

[Tex]\frac{x}{24} = \frac{y}{-6} = \frac{1}{6} [/Tex]

Hence,

x = [Tex]\frac{24}{6} [/Tex]

x = 4and, y = [Tex]\frac{-6}{6} [/Tex]

y = -1Hence, the

.required solution is x = 4 and y = -1

### Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

### 2x + 3y = 7

### (a – b) x + (a + b) y = 3a + b – 2

**Solution:**

Here,

a

_{1}= 2, b1 = 3, c1 = -7a

_{2}= a-b, b_{2}= a+b, c_{2}= -(3a+b-2)For having infinite number of solutions, it has to satisfy below conditions:

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} [/Tex]

[Tex]\frac{2}{a-b} = \frac{3}{a+b} = \frac{7}{(3a+b-2)} [/Tex]

Now, on comparing

[Tex]\frac{2}{a-b} = \frac{3}{a+b} [/Tex]

2(a+b) = 3(a-b)

2a+2b = 3a – 3b

a – 5b = 0

…………………(1)And, now on comparing

[Tex]\frac{3}{a+b} = \frac{7}{(3a+b-2)} [/Tex]

3(3a+b-2) = 7(a+b)

9a+3b-6 = 7a+7b

2a-4b-6=0

Reducing form, we get

a-2b-3=0

…………………(2)Now, new values for

a

_{1}= 1, b_{1}= -5, c_{1}= 0a

_{2}= 1, b_{2}= -2, c_{2}= -3

Solving Eq(1) and Eq(2), by cross multiplication,[Tex]\mathbf{\frac{a}{b_1c_2-b_2c_1} = \frac{b}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}} [/Tex]

[Tex]\frac{a}{(-5)(-3)-(-2)(0)} = \frac{b}{(0)(1)-(-3)(1)} = \frac{1}{(1)(-2)-(1)(-5)} [/Tex]

[Tex]\frac{a}{15-0} = \frac{b}{0+3} = \frac{1}{-2+5} [/Tex]

[Tex]\frac{a}{15} = \frac{b}{3} = \frac{1}{3} [/Tex]

Hence,

a = [Tex]\frac{15}{3} [/Tex]

a = 5and, b = [Tex]\frac{3}{3} [/Tex]

b = 1Hence, For values

pair of linear equations have an infinite number of solutionsa = 5 and b = 1

### (ii) For which value of k will the following pair of linear equations have no solution?

### 3x + y = 1

### (2k – 1) x + (k – 1) y = 2k + 1

**Solution:**

Here,

a

_{1}= 3, b_{1}= 1, c_{1}= -1a

_{2}= (2k-1), b_{2}= k-1, c_{2}= -(2k+1)For having no solution, it has to satisfy below conditions:

[Tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2} [/Tex]

[Tex]\frac{3}{2k-1} = \frac{1}{k-1} ≠ \frac{1}{2k+1} [/Tex]

Now, on comparing

[Tex]\frac{3}{2k-1} = \frac{1}{k-1} [/Tex]

3(k-1) = 2k-1

3k-3 = 2k-1

k = 2And, now on comparing

[Tex]\frac{1}{k-1} ≠ \frac{1}{2k+1} [/Tex]

2k+1 ≠ k-1

k ≠ -2Hence, for

the pair of linear equations have no solutionk = 2 and 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:**

8x + 5y – 9 = 0 …………………(1)

3x + 2y – 4 = 0 …………………(2)

Substitution MethodFrom Eq(2), we get

x = 4-2y/3

Now, substituting it in Eq(1), we get

8(4-2y/3) + 5y – 9 = 0

32-16y/3 + 5y – 9 = 0

32 – 16y + 15y – 27 = 0

y = 5Now, substituting y = 5 in Eq(2), we get

3x + 2(5) – 4 = 0

3x = -6

x = -2

Cross Multiplication MethodHere,

a

_{1}= 8, b_{1}= 5, c_{1}= -9a

_{2}= 3, b_{2}= 2, c_{2}= -4So,

[Tex]\frac{a_1}{a_2} = \frac{8}{3} [/Tex]

[Tex]\frac{b_1}{b_2} = \frac{5}{2} [/Tex]

As,

[Tex]\frac{a_1}{a_2} ≠ \frac{b_1}{b_2} [/Tex]

Hence, the given pairs of equations have unique solution.

For cross multiplication,

[Tex]\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}} [/Tex]

[Tex]\frac{x}{(5)(-4)-(2)(-9)} = \frac{y}{(-9)(3)-(-4)(8)} = \frac{1}{(8)(2)-(3)(5)} [/Tex]

[Tex]\frac{x}{(-20)+18} = \frac{y}{(-27)+32} = \frac{1}{16-15} [/Tex]

[Tex]\frac{x}{-2} = \frac{y}{5} [/Tex] = 1

Hence,

[Tex]\frac{x}{-2} [/Tex] = 1

x = -2and, [Tex]\frac{y}{-5} [/Tex] = 1

y = 5Hence,

.the required solution is x = -2 and y = 5

### 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 charges and the cost of food per day.

**Solution:**

Let’s take,

Fixed charge = x

Charge of food per day = y

According to the given question,

x + 20y = 1000

……………….. (1)x + 26y = 1180

………………..(2)Subtracting Eq(1) from Eq(2) we get

6y = 180

y = 30Now, substituting y = 30 in Eq(2), we get

x + 20(30) = 1000

x = 1000 – 600

x= 400.Hence,

.fixed charges is ₹ 400 and charge per day is ₹ 30

### (ii) A fraction become [Tex]\frac{1}{3} [/Tex] when 1 is subtracted from the numerator, and it becomes [Tex]\frac{1}{4} [/Tex] when 8 is added to its denominator. Find the fraction.

**Solution:**

Let the fraction be [Tex]\frac{x}{y} [/Tex].

So, as per the question given,

[Tex]\frac{x-1}{y} = \frac{1}{3} [/Tex]

3x – y = 3

…………………(1)[Tex]\frac{x}{y+8} = \frac{1}{4} [/Tex]

4x –y =8

………………..(2)Subtracting Eq(1) from Eq(2) , we get

x = 5Now, substituting x = 5 in Eq(2), we get

4(5)– y = 8

y = 12Hence, the fraction is [Tex]\mathbf{\frac{5}{12}} [/Tex].

### (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?

**Solution:**

Let’s take

Number of right answers = x

Number of wrong answers = y

According to the given question;

3x−y=40

……….……..(1)4x−2y=50

2x−y=25

……………….(2)Subtracting Eq(2) from Eq(1), we get

x = 15Now, substituting x = 15 in Eq(2), we get

2(15) – y = 25

y = 30-25

y = 5Hence, number of right answers = 15 and number of wrong answers = 5

Hence,

total number of questions = 20

### (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?

**Solution:**

Let’s take

Speed of car from point A = x km/he

Speed of car from point B = y km/h

When car travels in the same direction,

5x – 5y = 100

x – y = 20

………………(1)When car travels in the opposite direction,

x + y = 100

………………..(2)Subtracting Eq(1) from Eq(2), we get

2y = 80

y = 40Now, substituting y = 40 in Eq(1), we get

x – 40 = 20

x = 60Hence, the speed of car

from point A = 60 km/hSpeed of car

from point B = 40 km/h.

### (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:**

Let’s take

Length of rectangle = x unit

Breadth of the rectangle = y unit

Area of rectangle will be = xy sq. units

According to the given conditions,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0

……………………(1)(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0

……………………..(2)Using cross multiplication method, we get,

[Tex]\frac{x}{(-5)(-61)-(3)(-6)} = \frac{y}{(-6)(2)-(-61)(3)} = \frac{1}{(3)(3)-(2)(-5)} [/Tex]

[Tex]\frac{x}{305 +18} = \frac{y}{-12+183} = \frac{1}{9+10} [/Tex]

[Tex]\frac{x}{323} = \frac{y}{171} = \frac{1}{19} [/Tex]

Hence,

[Tex]\frac{x}{323} = \frac{1}{19} [/Tex]

x = 17and, [Tex]\frac{y}{171} = \frac{1}{19} [/Tex]

y = 9Hence, the required solution is

x = 17 and y = 9.

Length of rectangle = 17 units

Breadth of the rectangle = 9 units

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:** Exercise 3.6

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:**

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

### (i) [Tex]\frac{1}{2x} + \frac{1}{3y} = 2 [/Tex]

### [Tex]\frac{1}{3x} + \frac{1}{2y}= \frac{13}{6} [/Tex]

**Solution:**

Lets, take 1/x = a and 1/y = b

Here, the two given equation will be as follows:

[Tex]\frac{a}{2} [/Tex] + [Tex]\frac{b}{3} [/Tex] = 2

Multiply it by 6, we get

3a + 2b = 12 -(1)

and,

[Tex]\frac{a}{3} [/Tex] + [Tex]\frac{b}{2} [/Tex] = [Tex]\frac{13}{6} [/Tex]

Multiply it by 6, we get

2a + 3b = 13 -(2)

Now, by using Elimination method,

Multiply eq(1) by 2 and multiply eq(2) by 3, and then subtract them

5b = 15

b = 3

Now putting b = 3 in eq(1), we get

3a + 2(3) = 12

a = 6/3

a = 2

So, Now As

a = 1/x = 2

x = 1/2

b = 1/y = 3

y = 1/3

### (ii) [Tex]\frac{2}{√x} + \frac{3}{√y} = 2 [/Tex]

### [Tex]\frac{4}{√x} – \frac{9}{√y} = -1 [/Tex]

**Solution:**

Lets, take 2/√x = a and 3/√y = b

Here, the two given equation will be as follows:

a + b = 2 -(1)

and,

2a – 3b =-1 -(2)

Now, by using Elimination method,

Multiply eq(1) by 3, and then add them

5a = 5

a = 1

Now putting a = 1 in eq(1), we get

1 + b = 2

b = 1

So, Now As

a = 2/√x = 1

√x = 2

x = 4

b = 3/√y = 1

√x = 3

y = 9

### (iii) [Tex]\frac{4}{x} [/Tex] + 3y = 14

### [Tex]\frac{3}{x} [/Tex] – 4y = 23

**Solution:**

Lets, take 1/x = a

Here, the two given equation will be as follows:

4a + 3y = 14 -(1)

and,

3a – 4y = 23 -(2)

Now, by using Elimination method,

Multiply eq(1) by 3 and multiply eq(2) by 4, and then subtract them

-25y = 50

y = -2

Now putting y = -2 in eq(1), we get

4a + 3(-2) = 14

4a = 20

a = 5

So, Now As

a = 1/x = 5

x = 1/5

y = -2

### (iv) [Tex]\frac{5}{x-1} + \frac{1}{y-2} = 2 [/Tex]

### [Tex]\frac{6}{x-1} – \frac{3}{y-2} = 1 [/Tex]

**Solution:**

Lets, take [Tex]\frac{1}{x-1} [/Tex] = a and, [Tex]\frac{1}{y-2} [/Tex] = b

Here, the two given equation will be as follows:

5a + b = 2 -(1)

and,

6a – 3b = 1 -(2)

Now, by using Elimination method,

Multiply eq(1) by 3, and then add them

21a = 7

a = 1/3

Now putting a = 1/3 in eq(1), we get

5(1/3) + b = 2

b = 2 – 5/3

b = 1/3

So, Now As

a = [Tex]\frac{1}{x-1} = \frac{1}{3} [/Tex]

x – 1 = 3

x = 4

b = [Tex]\frac{1}{y-2} = \frac{1}{3} [/Tex]

y – 2 = 3

y = 5

### (v) [Tex]\frac{7x-2y}{xy} = 5 [/Tex]

### [Tex]\frac{8x+7y}{xy}=15 [/Tex]

**Solution:**

[Tex]\frac{7}{y} – \frac{2}{x} [/Tex] = 5

[Tex]\frac{8}{y} + \frac{7}{x} [/Tex] = 15

Lets, take 1/x = a and 1/y = b

Here, the two given equation will be as follows:

7b – 2a = 5 -(1)

and,

8b + 7a = 15 -(2)

Now, by using Elimination method,

Multiply eq(1) by 7, multiply eq(2) by 2 and then add them

65b = 65

b = 1

Now putting b = 1 in eq(1), we get

7(1) – 2a = 5

2a = 7 – 5

a = 1

So, Now As

a = 1/x = 1

x = 1

b = 1/y = 1

y = 1

### (vi) 6x + 3y = 6xy

### 2x + 4y = 5xy

**Solution:**

Divide both the equations by xy, we get

[Tex]\frac{6}{y} + \frac{3}{x} [/Tex] = 6

[Tex]\frac{2}{y} + \frac{4}{x} [/Tex] = 5

Lets, take 1/x = a and, 1/y = b

Here, the two given equation will be as follows:

6b + 3a = 6

Divide the above equation by 2,

2b + a = 2 -(1)

and,

2b + 4a = 5 -(2)

Now, by using Elimination method,

Subtract eq(1) from eq(2), we get

3a = 3

a = 1

Now putting a = 1 in eq(1), we get

2b + 1 = 2

b = 1/2

So, Now As

a = 1/x = 1

x = 1

b = 1/y = 1/2

y = 2

### (vii) [Tex]\frac{10}{x+y} + \frac{2}{x-y}=4 [/Tex]

### [Tex]\frac{15}{x+y} – \frac{5}{x-y}=-2 [/Tex]

**Solution:**

Lets, take [Tex]\frac{1}{x+y} [/Tex] = a and [Tex]\frac{1}{x-y} [/Tex] = b

Here, the two given equation will be as follows:

10a + 2b = 4

Divide the above equation by 2,

5a + b = 2 -(1)

and,

15a – 5b = -2 -(2)

Now, by using Elimination method,

Multiply eq(1) by 3 and subtract them,

8b = 8

b = 1

Now putting b = 1 in eq(1), we get

5a + 1 = 2

a = 1/5

So, Now As

a = [Tex]\frac{1}{x+y} [/Tex] = [Tex]\frac{1}{5} [/Tex]

x + y = 5 -(3)

b = [Tex]\frac{1}{x-y} [/Tex] = 1

x – y = 1 -(4)

By adding eq(3) and (4), we get

2x = 6

x = 3 and y = 2

### (viii) [Tex]\frac{1}{3x+y} + \frac{1}{3x-y}= \frac{3}{4} [/Tex]

### [Tex]\frac{1}{2(3x+y)} – \frac{1}{2(3x-y)} = \frac{-1}{8} [/Tex]

**Solution:**

Lets, take [Tex]\frac{1}{3x+y} [/Tex] = a

and, [Tex]\frac{1}{3x-y} [/Tex] = b

Here, the two given equation will be as follows:

a + b = 3/4 -(1)

and,

[Tex]\frac{a}{2} – \frac{b}{2} = \frac{-1}{8} [/Tex]

Multiply it by 2, we get

a – b = -1/4 -(2)

Now, by using Elimination method,

Add eq(1) and eq(1), we get

2a = 1/2

a = 1/4

Now putting a = 1/4 in eq(1), we get

[Tex]\frac{1}{4} [/Tex] + b = [Tex]\frac{3}{4} [/Tex]

b = 1/2

So, Now As

a =[Tex] \frac{1}{3x+y} = \frac{1}{4} [/Tex]

3x + y = 4 -(3)

b = [Tex]\frac{1}{3x-y} = \frac{1}{2} [/Tex]

3x – y = 2 -(4)

By adding eq(3) and eq(4), we get

6x = 6

x = 1 and y = 1

### 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.

**Solution:**

Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = (x + y) km/h

Upstream = (x – y) km/h

As Speed = [Tex]\frac{Distance}{Time} [/Tex]

According to the given question,

x + y = 20/2

x + y = 10 -(1)

and,

x – y = 4/2

x – y = 2 -(2)

Add eq(1) and eq(2), we get

2x = 12

x = 6 and y = 4

Hence, speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

### (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.

**Solution:**

Let’s take,

The total number of days taken by women to finish the work = x

The total number of days taken by men to finish the work = y

Work done by women in one day will be = 1/x

Work done by women in one day will be = 1/y

So, according to the question

4([Tex]\frac{2}{x} + \frac{5}{y} [/Tex]) = 1

And, 3([Tex]\frac{3}{x} + \frac{6}{y} [/Tex]) = 1

Lets, take 1/x = a and, 1/y = b

Here, the two given equation will be as follows:

4(2a + 5b) = 1

8a + 20b = 1 -(1)

and,

3(3a + 6b) = 1

9a + 18b = 1 -(2)

Now, by using Cross multiplication method,

[Tex]\frac{a}{20-18} = \frac{b}{9-8} = \frac{1}{180-144} [/Tex]

[Tex]\frac{a}{2} = \frac{b}{1} = \frac{1}{36} [/Tex]

a = [Tex]\frac{2}{36} = \frac{1}{18} [/Tex]

b = 1/36

So, Now As

a =[Tex] \frac{1}{x} = \frac{1}{18} [/Tex]

x = 18

b = [Tex]\frac{1}{y} = \frac{1}{36} [/Tex]

y = 36

Hence, number of days taken by women to finish the work = 18 days

Number of days taken by men to finish the work = 36 days.

### (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:**

Lets, take

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

[Tex]\frac{60}{x} + \frac{240}{y} [/Tex] = 4

[Tex]\frac{100}{x} + \frac{200}{y} = \frac{25}{6} [/Tex]

Lets, take 1/x = a and 1/y = b

Here, the two given equation will be as follows:

60a + 240b = 4

Divide it by 4, we get

15a + 60b = 1 -(1)

and,

100a + 200b = 25/6

Divide it by 25/6, we get

24a + 48b = 1 -(2)

Now, by using Cross multiplication method,

[Tex]\frac{a}{60-48} = \frac{b}{24-15} = \frac{1}{1440-720} [/Tex]

[Tex]\frac{a}{12} = \frac{b}{9} = \frac{1}{720} [/Tex]

a = [Tex]\frac{12}{720} [/Tex] = [Tex]\frac{1}{60} [/Tex]

b = [Tex]\frac{9}{720} [/Tex] = [Tex]\frac{1}{80} [/Tex]

So, Now As

a = [Tex]\frac{1}{x} [/Tex] = [Tex]\frac{1}{60} [/Tex]

x = 60

b = [Tex]\frac{1}{y} [/Tex] = [Tex]\frac{1}{80} [/Tex]

y = 80

Hence, speed of the train = 60 km/h

Speed of the bus = 80 km/h

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:** Exercise 3.7

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:**

### 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:**

The age difference between Ani and Biju = 3 yrs.

Either Biju is 3 years older than that of Ani,Case 1:y – x = 3

or Ani is 3 years older than Biju.Case 2:x – y = 3

Given, Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let’s take,

Ani’s age = x

and Biju’s age = y

So, Dharam’s age will be = 2x

And the age of Biju sister (Cathy) will be = [Tex]\frac{y}{2} [/Tex]

CASE 1 (y > x)According to the given condition,

y – x = 3 -(1)

and, 2x−y/2 = 30

Multiply it by 2, we get

4x – y = 60 -(2)

Add eq(1) and (2)

3x = 63

x = 63/3

x = 21

Now putting x = 21 in eq(1), we get

y – 21 = 3

y = 24

Therefore, the age of Ani = 21 years

And the age of Biju is = 24 years.

CASE 2: (x > y)According to the given condition,

x – y = 3 -(1)

and, 2x−y/2 = 30

Multiply it by 2, we get

4x – y = 60 -(2)

Now, by using Elimination method,

Subtract eq(1) from eq(2)

3x = 57

x = 57/3

x = 19

Now putting x = 19 in eq(1), we get

19 – y = 3

y = 19 – 3

y = 16

Therefore, the age of Ani = 19 years

And the age of Biju is = 16 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]

### [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

**Solution:**

Let there be two persons as A and B.

Let’s take,

Money person A has = ₹ x

Money person B has = ₹ y

So, According to the given conditions, we have

x + 100 = 2(y – 100)

x – 2y = – 300 -(1)

And

6(x – 10) = (y + 10)

6x -y = 70 -(2)

Now, by using Elimination method,

Multiply eq(2) by 2 and subtract eq(1) from it

11x = 440

x = 40

Now putting x = 40 in eq(1), we get

40 – 2y = – 300

2y = 300 + 40

y = 340/2

y = 170

Hence,

Person A had Rs 40 and person B had Rs 170 with them.

### 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 speed of the train = x km/hr

The time taken by the train to travel a distance = t hours

The distance to travel = d km.

As we know,

Distance traveled by train = Speed of the train × Time taken to travel that distance

d = xt -(1)

According to the given equation,

d = (x + 10) × (t – 2)

d = xt + 10t – 2x – 20

as d = xt, we get

10t – 2x = 20 -(2)

and 2

^{nd}condition,d = (x – 10) × (t + 3)

d = xt – 10t + 3x – 30

as d = xt, we get

10t – 3x = -30 -(3)

Now, by using Elimination method,

Subtract eq(3) from eq(2), we get

x = 50 km/h

Now putting x = 50 in eq(1), we get

10t – 2(50) = 20

10t = 120

t = 12 hours

Now, distance will be = 50 × 12 = 600 km

Hence, the distance covered by the train is 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 the number of rows = x

and, the number of students in a row = y.

Total number of students = Number of rows x Number of students in a row = xy -(1)

Here, total no. of student will be same always

According to the given condition,

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

xy = (x – 1)(y + 3)

xy = xy – y + 3x – 3

3x – y = 3 -(2)

and,

Total Number of students = (x + 2 ) ( y – 3 )

xy = xy + 2y – 3x – 6

3x – 2y = -6 -(3)

Now, by using Elimination method,

Subtract eq(3) from eq(2), we get

y = 9

Now putting y = 9 in eq(1), we get

3x – 9 = 3

3x = 12

x = 4

Number of total students in a class = xy

= 4 x 9

= 36

Hence, No. of students = 36

### Question 5. In a ∆ABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

**Solution:**

Here, according to the given conditions

3 ∠B = 2 (∠A + ∠B)

3 ∠B = 2∠A + 2∠B

∠B = 2∠A -(1)

and, ∠C = 3∠B

∠C = 3 (2∠A)

∠C = 6∠A -(2)

As we know,

∠A + ∠B + ∠C = 180° (Angle sum property)

∠A + (2∠A) + (6∠A) = 180° -(From eq(1) and eq(2))

9∠A = 180

∠A = 20°

∠B = 2∠A = 2(20°) = 40°

∠C = 6∠A = 6(20°) = 120°

Hence, the three angles are 20°, 40° and 120°.

### Question 6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y axis.

**Solution:**

5x – y = 5 -(1)

3x – y = 3 -(2)

y-axis, x = 0 -(3)

To get the co-ordinates of the vertices of the triangle formed, lets get there intersection point,

Intersection of eq(1) and (2)

Now, by using Elimination method,

Subtract eq(2) from (1), we get

2x = 2

x = 1

Now putting x = 1 in eq(1), we get

5(1) – y = 5

y = 0

So, Intersection of eq(1) and (2) vertices are (1, 0)

Now, Intersection of eq(1) and (3)

As x =0 substitutes it in eq(1), we get

5(0) – y = 5

y = -5

So, Intersection of (I) and (III) vertices are (0, -5)

Now, Intersection of eq(2) and (3)

As x =0 substitutes it in eq(2), we get

3(0) – y = 3

y = -3

So, Intersection of eq(2) and (3) vertices are (0,-3)

Hence, the co-ordinates of the vertices of the triangle formed are (1, 0), (0, -5) and (0, -3)

### Question 7. Solve the following pair of linear equations:

### (i) px + qy = p – q

### qx – py = p + q

**Solution:**

px + qy = p – q -(1)

qx – py = p + q -(2)

Multiplying p to equation (1) and q to equation (2), we get

p

^{2}x + pqy = p^{2}− pq -(3)q

^{2}x − pqy = pq + q^{2}-(4)Add eq(3) and (4), we get

p

^{2}x + q^{2}x = p^{2}+ q^{2}(p

^{2}+ q^{2}) x = p^{2}+ q^{2}x = [Tex]\frac{(p^2 + q^2)}{p^2 + q^2} [/Tex]

x = 1

Now putting x = 1 in eq(1), we get

p(1) + qy = p – q

qy = p-q-p

qy = -q

y = -1

### (ii) ax + by = c

### bx + ay = 1 + c

**Solution:**

ax + by = c -(1)

bx + ay = 1 + c -(2)

Multiplying a to equation (1) and b to equation (2), we get

a

^{2}x + aby = ac -(3)b

^{2}x + aby = b + bc -(4)Subtract equation eq(4) from equation (3),

(a

^{2}– b^{2}) x = ac − bc– bx = [Tex]\frac{(ac − bc– b)}{(a^2 – b^2)} [/Tex]

x = [Tex]\frac{(c(a-b) –b)}{(a^2 – b^2)} [/Tex]

Now putting the value of x in eq(1), we get

ax + by = c

[Tex]a{\frac{(c(a-b) –b)}{(a^2 – b^2)}} [/Tex] +by = c

[Tex]\frac{(ac(a-b) –b)}{(a^2 – b^2)} [/Tex]+by = c

by = [Tex]\frac{c–ac(a−b)−ab}{(a^2 – b^2)} [/Tex]

by = [Tex]\frac{(c–ac(a−b)−ab)}{(a^2 – b^2)} [/Tex]

y = [Tex]\frac{(c(a-b)+a)}{(a^2 – b^2)} [/Tex]

### (iii) [Tex]\frac{x}{a} – \frac{y}{b} = 0 [/Tex]

### ax + by = a^{2}+ b^{2}.

**Solution:**

[Tex]\frac{x}{a} – \frac{y}{b} = 0 [/Tex] -(1)

Multiplying equation (1) with ab, we get

bx − ay = 0 -(new 1)

Multiplying a and b to equation (1) and (2) respectively, we get

b

^{2}x − aby = 0 -(3)a

^{2}x + aby = a^{3}+ ab^{3}-(4)Add eq(3) and (4), we get

b

^{2}x + a^{2}x = a^{3}+ ab^{2}x (b

^{2}+ a^{2}) = a (a^{2}+ b^{2})x = a

Now putting x = a in eq(1), we get

b(a) − ay = 0

ab − ay = 0

ay = ab,

y = b

### (iv) (a – b)x + (a + b) y = a^{2}– 2ab – b^{2}

### (a + b)(x + y) = a^{2}+ b^{2}

**Solution:**

(a – b)x + (a + b) y = a

^{2}– 2ab – b^{2}-(1)(a + b)(x + y) = a

^{2}+ b^{2}-(2)Subtract equation (2) from equation (1), we get

(a − b) x − (a + b) x = (a

^{2}− 2ab − b^{2}) − (a^{2}+ b^{2})x(a − b − a − b) = − 2ab − 2b

^{2}− 2bx = − 2b (b + a)

x = b + a

Substituting, x = b + a in equation (1), we get

(a + b)(a − b) +y (a + b) = a

^{2}− 2ab – b^{2}a

^{2}− b^{2}+ y(a + b) = a^{2}− 2ab – b^{2}-(Using the identity (a + b)(a – b) = a^{2}– b^{2})(a + b) y = − 2ab

y = [Tex]\frac{-2ab}{a+b} [/Tex]

### (v) 152x – 378y = – 74

### –378x + 152y = – 604

**Solution:**

x = [Tex]\frac{(189y-137)}{76} [/Tex] -(1)

− 378x + 152y = − 604

Dividing it by 2, we get

− 189x + 76y = − 302 -(2)

Substitute x in equation (2), we get

−189[Tex](\frac{(189y-137)}{76}) [/Tex]+76y=−302

− (189)

^{2}y + 189 × 37 + (76)^{2}y = − 302 × 76189 × 37 + 302 × 76 = (189)

^{2}y − (76)^{2}y6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation(1), we get

x = [Tex]\frac{189-37}{76} [/Tex]

x = 152/76

x = 2

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

**Solution:**

It is known that the sum of the opposite angles of a cyclic quadrilateral is 180°

Thus, we have

∠C +∠A = 180

4y + 20− 4x = 180

− 4x + 4y = 160

x − y = − 40 -(1)

And, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 -(2)

Multiplying 3 to equation (1), we get

3x − 3y = − 120 -(3)

Adding equation (2) to equation (3), we get

− 7x + 3x = 180 – 120

− 4x = 60

x = −15

Substituting this value in equation (1), we get

x − y = − 40

-y−15 = − 40

y = 40-15

y = 25

∠A = 4y + 20 = 20 + 4(25) = 120°

∠B = 3y − 5 = − 5 + 3(25) = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = 5 – 7x

∠D= 5 − 7(−15) = 110°

Hence, ∠A, ∠B, ∠C and ∠D are equal to 120°, 70°, 60° and 110° respectively.

## Key Features of NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

- These NCERT solutions are developed by the GfG team, with a focus on students’ benefit.
- These solutions are entirely accurate and can be used by students to prepare for their board exams.
- Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

**Also Check:**

__NCERT Solutions for Class 9 Maths Chapter 1 Number Systems____NCERT Solutions for Class 9 Maths Chapter 2 Polynomials____NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry____NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry____NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles____NCERT Solutions for Class 9 Maths Chapter 7 Triangles____NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals__

## NCERT Solutions for Class 10 Maths **Chapter 3 Pair of Linear Equations in Two Variables – **FAQs

**Chapter 3 Pair of Linear Equations in Two Variables –**

### Q1: Why is it important to learn **Chapter 3 Pair of Linear Equations in Two Variables** of NCERT Class 10?

**Chapter 3 Pair of Linear Equations in Two Variables**

Firstly, linear equations are fundamental in mathematics and have broad applications in various fields. They help us model and solve real-world problems, such as calculating distances, determining rates of change, and analyzing patterns. Secondly, understanding linear equations enhances critical thinking and problem-solving skills. It develops the ability to analyze and interpret relationships between variables and make predictions based on mathematical models.

### Q2: What topics are covered in NCERT Solutions for Class 10 Maths **Chapter 3 – Pair of Linear Equations in Two Variables?**

**Chapter 3 – Pair of Linear Equations in Two Variables?**

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables, include topics such as Introduction to linear equations, Pair of linear equations, Graphical representation of linear equations, Graphical method of solution ,Algebraic methods of solution (substitution, elimination), Equations reducible to linear form, and Application of linear equations in real-life problems.

### Q3: How can NCERT Solutions for Class 10 Maths **Chapter 3 Pair of Linear Equations in Two Variables **help me?

**Chapter 3 Pair of Linear Equations in Two Variables**

NCERT Solutions for Class 10 Maths

can help you solve the NCERT exercise and free yourself from the frustration of being stuck on some question.Chapter 3 Pair of Linear Equations in Two Variables

### Q4: How many exercises are there in Class 10 Maths **Chapter 3 Pair of Linear Equations in Two Variables**?

**Chapter 3 Pair of Linear Equations in Two Variables**

There are 7 exercises in the Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables which covers all the important topics and sub-topics.

### Q5: Where can I find NCERT Solutions for Class 10 Maths **Chapter 3 Pair of Linear Equations in Two Variables**?

**Chapter 3 Pair of Linear Equations in Two Variables**

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.