CG Board Class 10 Maths Chapter 2 Solutions: Linear Equations in Two Variables (दो चरों का रैखिक समीकरण) (2026-27)

Let the cost of 1 bat = Rs. x and the cost of 1 ball = Rs. y.

According to the first condition: 3x + 6y = 3900

According to the second condition: x + 2y = 1300

• Equations: 3x + 6y = 3900 and x + 2y = 1300
• (b) The sum of two numbers is 16 and their difference is 8.

Let the first number = x and the second number = y.

According to the statement

• Equations: x + y = 16 and x – y = 8
• (c) The cost of 2 kg apples and 1 kg grapes is Rs. 160 in a fruit shop. In the same shop, the cost of 4 kg apples and 2 kg grapes is Rs. 300.

Let the cost of 1 kg apples = Rs. x and 1 kg grapes = Rs. y.

• Equations: 2x + y = 160 and 4x + 2y = 300
• (d) Naresh said to his daughter that seven years ago my age was seven times your age and three years from now I will be three times as old as you.

Let present age of Naresh = x years and daughter’s present age = y years.

7 years ago: Naresh’s age = (x – 7), Daughter’s age = (y – 7)

Condition 1: x – 7 = 7(y – 7) ⇒ x – 7 = 7y – 49 ⇒ x – 7y = -42

3 years from now: Naresh’s age = (x + 3), Daughter’s age = (y + 3)

Condition 2: x + 3 = 3(y + 3) ⇒ x + 3 = 3y + 9 ⇒ x – 3y = 6

• Equations: x – 7y = -42 and x – 3y = 6
• (e) A person travels 90 km by train and taxi to reach his office. The distance travelled by train is twice the distance travelled by taxi.

Let distance travelled by train = x km and by taxi = y km.

Total distance: x + y = 90

Relation: x = 2y ⇒ x – 2y = 0

• Equations: x + y = 90 and x – 2y = 0

#### *Exercise – 2

(i) x + y = 7**

(A) is a solution, (B) is not a solution.**

• (ii) 2x + 5y = 13

(B) is a solution, (A) is not a solution.**

• (iii) 2x – 3y = -11

Both (A) and (B) are solutions.**

• (iv) 5x + 3y = 4

(B) is a solution, (A) is not a solution.**

(i) 2x + 5y = 1; 2x + 3y = 3**

• Testing (A)

Eq 1: 2(3) + 5(-1) = 6 – 5 = 1 (True)

Eq 2: 2(3) + 3(-1) = 6 – 3 = 3 (True)

(A) is the solution.**

• (ii) x + y = 5xy; 3x + 2y = 13xy
• Testing (B): x = 1/2, y = 1/3

Eq 1: LHS = 1/2 + 1/3 = 5/6, RHS = 5(1/2)(1/3) = 5/6 (True)

Eq 2: LHS = 3(1/2) + 2(1/3) = 3/2 + 2/3 = 13/6, RHS = 13(1/2)(1/3) = 13/6 (True)

(B) is the solution.**

• (iii) 2x – 3/y = 9; 3x + 7/y = 2
• Testing (A): x = 3, y = -1

Eq 1: 2(3) – 3/-1 = 6 + 3 = 9 (True)

Eq 2: 3(3) + 7/-1 = 9 – 7 = 2 (True)

(A) is the solution.**

• (iv) 2x + 5y = 8/3; 3x – 2y = 5/6
• Testing (B): x = 1/2, y = 1/3

Eq 1: 2(1/2) + 5(1/3) = 1 + 5/3 = 8/3 (True)

Eq 2: 3(1/2) – 2(1/3) = 3/2 – 2/3 = 5/6 (True)

(B) is the solution.**

(i) x – y = -1; 3x – 2y = 12**

• From Eq 1: x = y – 1
• Substitute in Eq 2: 3(y – 1) – 2y = 12 ⇒ 3y – 3 – 2y = 12 ⇒ y = 15
• Then x = 15 – 1 = 14

x = 14, y = 15**

• (ii) x – 2y = 5; 2x – 4y = 6
• Ratio of coefficients: a₁/a₂ = 1/2, b₁/b₂ = -2/-4 = 1/2, c₁/c₂ = 5/6
• Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, lines are parallel.
• (iii) x + y = 6; x = y + 2

x = 4, y = 2**

• (iv) 5x – 8y = -1; 3x – 24/5y + 3/5 = 0
• Multiply Eq 2 by 5: 15x – 24y + 3 = 0 ⇒ 15x – 24y = -3 ⇒ 5x – 8y = -1
• Since both equations are identical (a₁/a₂ = b₁/b₂ = c₁/c₂), lines are coincident.
• (v) 3x – 4y – 1 = 0; 2x – 8/3y + 5 = 0
• (vi) x + 2y = 8; 2x + 4y = 16

(i) x + y = 7; x – y = -1**

• Adding both equations: 2x = 6 ⇒ x = 3
• Substituting x = 3 in Eq 1: 3 + y = 7 ⇒ y = 4

x = 3, y = 4**

• (ii) 2x + y = 8; x – 2y = -1
• From Eq 2: x = 2y – 1
• Substitute in Eq 1: 2(2y – 1) + y = 8 ⇒ 4y – 2 + y = 8 ⇒ 5y = 10 ⇒ y = 2
• Then x = 2(2) – 1 = 3

x = 3, y = 2**

• (iii) 4x + 3y = 5; 2x – y = 2
• From Eq 2: y = 2x – 2
• Substitute in Eq 1: 4x + 3(2x – 2) = 5 ⇒ 4x + 6x – 6 = 5 ⇒ 10x = 11 ⇒ x = 1.1
• Then y = 2(1.1) – 2 = 2.2 – 2 = 0.2

x = 1.1, y = 0.2**

• (iv) √7x + √11y = 0; √3x – √5y = 0
• From Eq 2: x = √5/√3y
• Substitute in Eq 1: √7(√5/√3y) + √11y = 0 ⇒ y(√35/√3 + √11) = 0 ⇒ y = 0
• Then x = 0

x = 0, y = 0**

Rs. 43.80*

30 Cows and 30 Pigeons*

25 coins of 50 paisa and 69 coins of 25 paisa*

Numbers are 20 and 5*

Numbers are 23 and 9*

Numbers are 9 and 5*

My age = 50 years, Daughter’s age = 20 years*

Speeds are 40 km/h and 30 km/h*

Room A = 100, Room B = 80*

Length = 40 units, Breadth = 30 units*

Comparing with a₁ x + b₁ y = c₁ and a₂ x + b₂ y = c₂

a₁ = 3, b₁ = 5, c₁ = 12

a₂ = 5, b₂ = 3, c₂ = 4

Ratios: a₁/a₂ = 3/5, b₁/b₂ = 5/3

Since a₁/a₂ ≠ b₁/b₂ (3/5 ≠ 5/3), the system has a unique solution.

a₁ = 2, b₁ = -3, c₁ = 5

a₂ = 6, b₂ = -9, c₂ = 15

Ratios

a₁/a₂ = 2/6 = 1/3, b₁/b₂ = -3/-9 = 1/3, c₁/c₂ = 5/15 = 1/3

Since a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/3, the system has infinitely many solutions.

(i) 8x + 5y = 9; kx + 10y = 15**

8/k = 5/10 ⇒ 8/k = 1/2 ⇒ k = 16

k = 16**

• (ii) kx + 3y = 3; 12x + ky = 6

k/12 = 3/k ≠ 3/6 ⇒ k² = 36 ⇒ k = ± 6

If k = 6, 6/12 = 3/6 = 3/6 (gives infinite solutions).

If k = -6, -6/12 = 3/-6 ≠ 3/6 ⇒ -1/2 = -1/2 ≠ 1/2 (no solution condition satisfied).

k = -6**

• (iii) kx – 5y = 2; 6x + 2y = 7

k/6 = -5/2 ⇒ 2k = -30 ⇒ k = -15

k = -15**

(i) kx + 2y = 5; 3x + y = 1**

k/3 ≠ 2/1 ⇒ k ≠ 6

k ≠ 6**

• (ii) x – 2y = 3; 3x + ky = 1

1/3 ≠ -2/k ⇒ k ≠ -6

k ≠ -6**

• (iii) kx + 3y = k – 3; 12x + ky = k

k/12 ≠ 3/k ⇒ k² ≠ 36 ⇒ k ≠ ± 6

k ≠ ± 6**

• (iv) 4x – 5y = k; 2x – 3y = 12

a₁/a₂ = 4/2 = 2, b₁/b₂ = -5/-3 = 5/3.

Since 2 ≠ 5/3, a₁/a₂ ≠ b₁/b₂ holds true for all real values of k.

k ≠ 10/3 (or all real values of k)**

(i) 2x + 3y = 7; (k – 1)x + (k + 2)y = 3k**

(2)/(k – 1) = (3)/(k + 2) = 7/3k

From (2)/(k – 1) = (3)/(k + 2) ⇒ 2k + 4 = 3k – 3 ⇒ k = 7

k = 7**

• (ii) kx + 2y – 4 = 0; 5x – 3y + 6 = 0

Rearranging: kx + 2y = 4 and 5x – 3y = -6

k/5 = 2/-3 = 4/-6 ⇒ k/5 = -2/3 ⇒ k = -10/3

k = -10/3**

• (iii) 3x + ky = 7; 2x – 5y = 1

Ratios: a₁/a₂ = 3/2, c₁/c₂ = 7/1.

Since 3/2 ≠ 7/1, no value of k can satisfy a₁/a₂ = b₁/b₂ = c₁/c₂.

No value of k exists*

• (iv) kx – 5y = 2; 2x – 3y = 12

Ratios: b₁/b₂ = 5/3, c₁/c₂ = 2/12 = 1/6.

Since 5/3 ≠ 1/6, no value of k can make all three ratios equal.

k = 24*

p = -2**

k = 7**

a₁ = 4, b₁ = 7, c₁ = 18

a₂ = 2, b₂ = 1, c₂ = 4

Ratios: a₁/a₂ = 4/2 = 2, b₁/b₂ = 7/1 = 7

Since a₁/a₂ ≠ b₁/b₂ (2 ≠ 7), the system has a unique solution.

Solving for x and y

From Eq 2: y = 4 – 2x

Substitute in Eq 1: 4x + 7(4 – 2x) = 18 ⇒ 4x + 28 – 14x = 18 ⇒ -10x = -10 ⇒ x = 1

Substitute x = 1 in y = 4 – 2x: y = 4 – 2(1) = 2