Pair of Linear Equations in Two Variables l Exercise 3.1 l Class 10 l NCERT Solutions

NCERT Class 10 Maths – Chapter 3

Pair of Linear Equations in Two Variables

Exercise 3.1 – Complete Solutions

Class: 10
Subject: Mathematics
Chapter: Pair of Linear Equations in Two Variables
Exercise: 3.1


Important Concepts Used

  • Standard form: ax + by + c = 0
  • Condition for intersection:
    • a₁/a₂ ≠ b₁/b₂ → Intersecting lines (Unique solution)
    • a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines (No solution)
    • a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines (Infinite solutions)

Question 1

(i) 10 students took part. Girls are 4 more than boys.

Let number of boys = x
Number of girls = y

x + y = 10
y = x + 4

Substitute: x + (x + 4) = 10
2x + 4 = 10
2x = 6
x = 3

y = 3 + 4 = 7

Boys = 3, Girls = 7


(ii) Cost of pencils and pens

Let cost of one pencil = x
Cost of one pen = y

5x + 7y = 50
7x + 5y = 46

Multiply first by 7: 35x + 49y = 350
Multiply second by 5: 35x + 25y = 230

Subtract: 24y = 120
y = 5

Substitute: 5x + 7(5) = 50
5x + 35 = 50
5x = 15
x = 3

Pencil = ₹3, Pen = ₹5


Question 2

(i) 5x − 4y + 8 = 0
7x + 6y − 9 = 0

a₁/a₂ = 5/7
b₁/b₂ = −4/6 = −2/3

Ratios unequal → Intersecting lines (Unique solution)


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

9/18 = 3/6 = 12/24

All ratios equal → Coincident lines (Infinite solutions)


(iii) 6x − 3y + 10 = 0
2x − y + 9 = 0

6/2 = −3/−1 ≠ 10/9

Intersecting lines (Unique solution)


Question 3

(i) 3x + 2y = 5 ; 2x − 3y = 7

a₁/a₂ ≠ b₁/b₂ → Consistent (Unique solution)

(ii) 2x − 3y = 8 ; 4x − 6y = 9

a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Inconsistent

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

Ratios unequal → Consistent

(iv) 5x − 3y = 11 ; −10x + 6y = −22

All ratios equal → Infinite solutions

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

Ratios unequal → Consistent


Question 4

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

Coincident lines → Infinite solutions

(ii) x − y = 8 ; 3x − 3y = 16

Parallel lines → No solution

(iii) 2x + y − 6 = 0 ; 4x − 2y − 4 = 0

Intersecting → Unique solution

(iv) 2x − 2y − 2 = 0 ; 4x − 4y − 5 = 0

Parallel → No solution


Question 5

Let width = x
Length = x + 4

Half perimeter = 36
So, x + (x + 4) = 36

2x + 4 = 36
2x = 32
x = 16

Length = 20

Width = 16 m, Length = 20 m


Question 6

Given equation: 2x + 3y − 8 = 0
  • Intersecting: 2x − y + 1 = 0
  • Parallel: 4x + 6y + 5 = 0
  • Coincident: 4x + 6y − 16 = 0

Question 7

Equations: x − y + 1 = 0
3x + 2y − 12 = 0

Solve simultaneously: x − y = −1
3x + 2y = 12

Multiply first by 2: 2x − 2y = −2

Add: 5x = 10
x = 2

Substitute: 2 − y = −1
y = 3

Intersection point = (2, 3)

Vertices of triangle: (2,3), (0,6), (−1,0)

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