In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Exercise 3.2 Pdf, These solutions are solved subject experts from the latest edition books.

Table of Contents

## MP 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 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 boys = x and Number of girls = y

∴ x + y = 10 …. (1)

Also, Number of girls = [Number of boys] + 4

∴ y = x + 4 ⇒ x – y = – 4 … (2)

Now, from the equation (1), we have :

l_{1} : x + y = 10 ⇒ y = 10 – x

And from the equation (2), we have

l_{2} : x – y = -4 ⇒ y = x + 4

The lines l_{1} and l_{2} intersects at the point (3, 7)

∴ The solution of the pair of linear equations is : x = 3, y = 7

⇒ Number of boys = 3

and number of girls = 7 (10th Exercise 3.2 solutions)

Let the cost of a pencil = ₹ x

and cost of a pen = ₹ y

Since, cost of 5 pencils + Cost of 7 pens = ₹ 50

5 x + 7y = 50 … (1)

Also, cost of 7 pencils + cost of 5 pens = ₹ 46

7x + 5y = 46 …. (2)

Now, from equation (1), we have

Plotting the points (10, 0), (3, 5) and (-4, 10), we get a straight line l_{1} and plotting the points (8, -2), (3, 5) and (0, 9.2) we get a straight line l_{2}.

These two straight lines intersects at (3, 5).

∴ Cost of a pencil = ₹ 3 and cost of a pen = ₹ 5.

MP Board 10th Maths Exercise 3.2 solutions 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 coincident:

(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0

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

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

Solution: (10th maths exercise 3.2 solution important)

Comparing the given equations with

a_{1}x + b_{1}y + C_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0, we have

(i) For, 5x – 4y + 8 = 0, 7x + 6y – 9 = 0,

a_{1} = 5, b_{1} = -4, C_{1} = 8; a_{2} = 7, b_{2} = 6, c_{2} = -9

So, the lines are intersecting, i.e., they intersect at a point.

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

a_{1} = 9, b_{1} = 3, C_{1} = 12; a_{2} = 18, b_{2} = 6, c_{2} = 24

So, the lines are coincident.

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

So, the lines are parallel.

### MP Board Class 10th Maths Chapter 3 EX 3.2 Question 3.

On comparing the ratios a1a2,b1b2 and c1c2, 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) 32x+53y = 7; 9x-10y = 14

(iv) 5x – 3y = 11; -10x + 6y = -22

Solution:

Comparing the given equations with

So, lines are intersecting i.e., they intersect at a point.

∴ It is consistent pair of equations.

(ii) For 2x – 3y = 8, 4x – 6y = 9, 2x – 3y – 8 = 0 and 4x – 6y – 9 = 0

So, lines are parallel i.e., the given pair of linear equations has no solution.

∴ It is inconsistent pair of equations.

⇒ The given pair of linear equations has exactly one solution.

∴ It is a consistent pair of equations.

(iv) 5x – 3y = 11; -10x + 6y = -22

So, lines are coincident.

⇒ The given pair of linear equations has infinitely many solutions.

Thus, it is consistent pair of equations

(v)

So, the lines are coincident.

⇒ The given pair of linear equations have infinitely many solutions.

Thus it is consistent pair of equations.

### MP Board Class 10th Maths exercise 3.4 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 (10th Maths exercise 3.2 solutions)

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

**Solution:**

For any pair of linear equation,

a₁ x + b₁ y + c₁ = 0

a₂ x + b₂ y + c₂ = 0

a) a₁/a₂ ≠ b₁/b₂ (Intersecting Lines/uniqueSolution)

b) a₁/a₂ = b₁/b₂ = c₁/c₂ (Coincident Lines/Infinitely many Solutions)

c) a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (Parallel Lines/No solution)

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

a₁/a₂= 1/2

b₁/b₂= 1/2

c₁/c₂= -5/(-10) = 1/2

From the above,

a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, lines are coincident and have infinitely many solutions. Hence, they are consistent.

x + y – 5 = 0

y = – x + 5

y = 5 – x

x | 1 | 2 |

y = 5 – x | 4 | 3 |

2x + 2y – 10 = 0

2y = 10 – 2x

y = 5 – x

x | 3 | 4 |

y = 5- x | 2 | 1 |

All the points on coincident line are solutions for the given pair of equations.

(ii) x – y = 8, 3x – 3y =16

a₁/a₂ = 1/3

b₁/b₂ = -1/(-3) = 1/3

c₁/c₂ = – 8/(-16) = 1/2

From the above,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, lines are parallel and have no solution.

Hence, the pair of equations are inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 1/(-2) = -1/2

c₁/c₂ = -6/(-4) = 3/2

From the above,

a₁/a₂ ≠ b₁/b₂

Therefore, lines are intersecting and have a unique solution.

Hence, they are consistent. (mp board 10th maths exercise 3.2 solutions)

2x + y – 6 = 0

y = 6 – 2x

x | 0 | 2 |

y = 6 – 2x | 6 | 2 |

4x – 2y – 4 = 0

2y = 4x – 4

y = 2x – 2

x | 2 | 3 |

y = 2x – 2 | 2 | 4 |

x = 2 and y = 2 are solutions for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

a₁/a₂ = 2/4 = 1/2 (10th maths exercise 3.2 solutions)

b₁/b₂ = -2/(-4) = 1/2

c₁/c₂ = -2/(-5) = 2/5

From the above,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, lines are parallel and have no solution.

Hence, the pair of 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:**

Assuming the length of the garden as x and the width of the garden as y, two linear equations can be formed for the given data.

Perimeter of rectangle = 2(length + breadth)

Let the length of the garden be x and breadth be y

Then x = y + 4 [ Since its given that length is 4 m more than its width]

x – y = 4

y = x – 4

x | 8 | 16 |

y = x – 4 | 4 | 12 |

The half perimeter of the rectangle is x + y = 36 [Since, perimeter = 2(x + y)]

y = 36 – x

x | 16 | 26 |

y = 36 – x | 20 | 10 |

Thus, Length, x = 20 m and Breadth, y = 16 m

**Summary:**

Since, half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m, the length is 20 m and breadth is equal to 16 m.

MP Board 10th Maths Chapter 3 exercise 3.2 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:**

For any pair of linear equation,

a₁ x + b₁ y + c₁ = 0

a₂ x + b₂ y + c₂ = 0

a) If a₁/a₂ ≠ b₁/b₂ (Intersecting Lines)

b) If a₁/a₂ = b₁/b₂ = c₁/c₂ (Coincident Lines)

c) If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (Parallel Lines)

(i) Intersecting lines

Condition: a₁/a₂ ≠ b₁/b₂

2x + 3y – 8 = 0

a₁ = 2

b₁ = 3

So, considering a₂ = 3 and b₂ = 2 will satisfy the condition for intersecting lines. c₂ can be any value.

a₁/a₂ = 2/3

b₁/b₂ = 3/2

2/3 ≠ 3/2

Therefore, another linear equation is 3*x *+ 2*y *– 6 = 0

(ii) Parallel lines

Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2x + 3y – 8 = 0

a₁ = 2

b₁ = 3

c₁ = – 8

So, considering a₂ = 4, b₂ = 6, c₂ = 9 will satisfy the condition for parallel lines.

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = – 8/9

Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, another linear equation is 4*x *+ 6*y *+ 9 = 0

(iii) Coincident lines

Condition: a₁/a₂ = b₁/b₂ = c₁/c₂

2x + 3y – 8 = 0

Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2x + 3y – 8 = 0

We know that, a₁= 2, b₁= 3, c₁= – 8

So, considering a₂ = 4, b₂ = 6, c₂ = – 16 will satisfy the condition for coincident lines.

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = – 8/(-16) = 1/2

Thus, a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, linear equation is 4*x *+ 6*y *-16 = 0

**Summary:**

Given the linear equation 2x + 3y – 8 = 0, another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines is 3x + 2y – 6 = 0, for parallel lines is 4x + 6y + 9 = 0 and for the coincident lines is 4x + 6y – 16 = 0.

**More MP Board 10th Maths Solutions:**

- MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.3
- MP Board Class 10th Math Solutions Chapter 6 Triangles Ex 6.2
- MP Board Class 10th Math Solutions Chapter 6 Triangles Ex 6.1
- MP Board Class 10th Math Solutions Chapter 5 Arithmetic Progressions Ex 5.3
- MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2