NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Pair of Linear Equations in Two Variables |

Exercise |
Ex 3.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

### NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

**Class 10th Exercise 3.2 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.

**(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:**

**(i)** Let the number of girls be x and number of boys be y.

**A.T.Q.**

**1st Condition :**

x +y =10

**Table :
**

**2nd Condition :**

x = y + 4 ⇒ x – y = 4

**Table :**

Solving

Both the lines cut at (7, 3)

Hence, solutions is (7, 3), i.e.x = 7,y = 3

Number of girls – 7 and Number of boys = 3

Solving

**(i)**and**(ii)**graphicallyBoth the lines cut at (7, 3)

Hence, solutions is (7, 3), i.e.x = 7,y = 3

Number of girls – 7 and Number of boys = 3

**(ii)** Let cost of 1 pencil = ₹ x and cost of 1 pen = ₹ y.

**A.T.Q.**

**1st Condition :**

5x + 7y = 50

**Table :
**

**2nd Condition :**

7x + 5y = 46

**Table :**

**Class 10 Maths Ex 3.2 Question 2.**

On comparing the ratios and find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

**(i)** 5x – 4y + 8 = 7x + 6y – 9 = 0

**(ii)** 9x + 3y + 12 = 0, 18x + 6y + 24 = 0

**(iii)** 6 – 3y + 10 = 0, Zx – v + 9 = 0

**Solution:**

**(i)** Equations are 5x – 4y + 8 = 7x + 6y – 9 = 0

∴ Pair of lines represented by given equations intersect at one point. So, the system has exactly one solution.

**
(ii)** 9x + 3y + 12 = 0, 18 + 6y + 24 = 0

∴ Pair of equations represents coincident lines and having infinitely many solutions.

**
(iii)** 6x – 3y + 10 = 0, 2x -y + 9 = 0

∴ It represents parallel lines and having no solution.

**Ex 3.2 NCERT Class 10 Question 3.**

On comparing the ratios and find out whether the following pair of linear equations are consistent, or inconsistent,

**(i)** 3x + 2y = 5; 2x – 3y = 7

**(ii)** 2x – 3y = 8; 4x – 6y = 9

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

**(iv)** 5x-3y = 11; -10c + 6y = -22

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

**Solution:**

∴ Pair of equations is consistent with infinitely many solutions.

**Ex 3.2 Class 10 NCERT Solutions 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

**(ii)** x – y = 8, 3x – 3y = 16

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

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

**Solution:**

**(i)** x + y – 5, 2x + 2y = 10

Here,

∴ System of equations is consistent and the graph gr represents coincident lines.

Table for equation (i),

Table for equation (ii),

(ii) x – y = 8, 3x – 3y = 16

Here \(\frac { a_{ 1 } }{ a_{ 2 } } =\frac { 1 }{ 3 } ,\)

Pair of equations is inconsistent. Hence, lines are parallel and

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

Pair of equations is consistent.

Table for equation 2x + y – 6 = ()

Table for equation 4x – 2y – 4 = 0

**2x – 2y – 2 = 0, 4x – 4y – 5 = 0**

(iv)

(iv)

∴ Pair of equations is inconsistent. Hence, lines are parallel and system has no solution.

**Class 10 Maths Chapter 3 Exercise 3.2 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 length of garden = x m and width of garden = y m

Perimeter of rectangular garden = 2(x + y)

**A.T.Q.**

**1st Condition :**

\(\frac { 2(x+y) }{ 2 }\) = 36 ⇒ x + y = 36

**2nd Condition :**

x = y + 4 ⇒ x -y = 4 … (ii)

Adding equation (i) and (ii), we get

2x = 40 ⇒ x = 20

Putting x = 20 in equation (i), we get

20 + y = 36

y = 16

Hence, dimensions of the garden are 20 m and 16 m.

**Exercise 3.2 Class 10 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

**(ii)** parallel lines

**(iii)** coincident lines

**Solution:**

Given equation is 2x + 3y – 8 – 0

We have 2x + 3y = 8

Let required equation be ax + by = c

**Condition :**

**(i)** For intersecting lines

\(\frac { 2 }{ a } \neq \frac { 3 }{ b } \neq \frac { 8 }{ c }\) where a, b, c can have any value which satisfy the above condition.

Let a = 3, b = 2, c = 4

so, \(\frac { 2 }{ 3 } \neq \frac { 3 }{ 2 } \neq \frac { 8 }{ 4 }\)

∴ Equations are 2v + 3y = 8 and 3A + 2y = 4 have unique solution and their geometrical representation shows intersecting lines.

**(ii)** Given equation is 2x + 3y = 8 Required equation be ax + by = c

**Condition :**

For parallel lines

\(\frac { 2 }{ a } \neq \frac { 3 }{ b } \neq \frac { 8 }{ c }\)

where a, b, c can have any value which satisfy the above condition.

Let, a = 2, b = 3, c = 4

Required equation will be 2x + 3y = 4

Equations 2x + 3y = 8 and 2x + 3y = 4 have no solution and their geometrical representation shows parallel lines.

**
(iii)** For coincident lines:

Given equation is 2x + 3y = 8 Let required equation be ax + by = c For coincident lines.

\(\frac { 2 }{ 3 } =\frac { 3 }{ 2 } =\frac { 8 }{ 4 } \) where a, b, c can have any a b c possible value which satisfy the above condition.

Let a = 4, b = 6, c = 16 Required equation will be 4x + 6y = 16.

Equations 2x + 3y = 8 and 4x + 6y = 16 have infinitely many solutions and their geometrical representation shows coincident lines.

**NCERT Ex 3.2 Class 10 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:**

First equation is x – y + 1 = 0.

Table for 1st equation x = y – 1

Second equation is 3x + 2y = 12

⇒ 3x = 12 – 2y ⇒ x = \(x=\frac { 12-2y }{ 3 }\)

Required triangle is ABC. Coordinates of its vertices are A(2, 3), B(-1, 0), C(4, 0).

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