CG Board Class 8 Mathematics Chapter 12 Solutions: Equation (समीकरण) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 8 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 8 Mathematics Chapter 12 Solutions: Equation (समीकरण) in Bilingual Format (English & Hindi Medium).

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  • Balanced Operations / : Adding, subtracting, multiplying, or dividing by the same non-zero number on both sides of an equation does not change its solution.
  • Cross Multiplication Method / : An equation of the form (ax + b)/(cx + d) = k or (ax + b)/(cx + d) = p/q can be converted into a linear equation by cross-multiplying: q(ax + b) = p(cx + d).
  • Transposition / : Moving terms with variables to one side and constants to the other side changes their signs.
SECTION / EXERCISE

EXERCISE 12.1 / 12.1

Question 1

Solve 4x 185x 2 and verify your answer…


📝 Question:

Solve (4x + 18)/(5x) = 2 and verify your answer. / (4x + 18)/(5x) = 2

💡 Solution:

Solving the Equation:**

Given equation

(4x + 18)/(5x) = 2

On cross-multiplying

4x + 18 = 2 × 5x

4x + 18 = 10x

Transposing 4x to the RHS

10x – 4x = 18

6x = 18 ⇒ x = 18/6 = 3

  • Verification:

Substituting x = 3 in LHS

LHS = (4(3) + 18)/(5(3)) = (12 + 18)/(15) = 30/15 = 2 = RHS

Since LHS = RHS, verified!

Question 2

Solve 5x 22x 3 75 and verify your…


📝 Question:

Solve (5x + 2)/(2x + 3) = 7/5 and verify your answer. / (5x + 2)/(2x + 3) = 7/5

💡 Solution:

Solving the Equation:**

On cross-multiplying

5(5x + 2) = 7(2x + 3)

25x + 10 = 14x + 21

Transposing variables to LHS and constants to RHS

25x – 14x = 21 – 10

11x = 11 ⇒ x = 11/11 = 1

  • Verification:

Substituting x = 1 in LHS

LHS = (5(1) + 2)/(2(1) + 3) = (5 + 2)/(2 + 3) = 7/5 = RHS

Verified!

Question 3

Solve 7m 64m 2 2 and verify your…


📝 Question:

Solve (7m + 6)/(4m + 2) = 2 and verify your answer. / (7m + 6)/(4m + 2) = 2

💡 Solution:

Solving the Equation:**

On cross-multiplying

7m + 6 = 2(4m + 2)

7m + 6 = 8m + 4

Transposing

8m – 7m = 6 – 4 ⇒ m = 2

  • Verification:

Substituting m = 2 in LHS

LHS = (7(2) + 6)/(4(2) + 2) = (14 + 6)/(8 + 2) = 20/10 = 2 = RHS

Verified!

Question 4

Solve x 32x 7 98 and verify your…


📝 Question:

Solve (x – 3)/(2x + 7) = -9/8 and verify your answer. / (x – 3)/(2x + 7) = -9/8

💡 Solution:

Solving the Equation:**

On cross-multiplying

8(x – 3) = -9(2x + 7)

8x – 24 = -18x – 63

Transposing variables to LHS and constants to RHS

8x + 18x = -63 + 24

26x = -39 ⇒ x = -39/26 = -3/2

  • Verification:

Substituting x = -3/2 in LHS

LHS = (-3/2 – 3)/(2(-3/2) + 7) = (-9/2)/(-3 + 7) = (-9/2)/(4) = -9/8 = RHS

Verified!

Question 5

Solve 2y 53y 1 313 and verify your…


📝 Question:

Solve (2y – 5)/(3y + 1) = 3/13 and verify your answer. / (2y – 5)/(3y + 1) = 3/13

💡 Solution:

Solving the Equation:**

On cross-multiplying

13(2y – 5) = 3(3y + 1)

26y – 65 = 9y + 3

Transposing

26y – 9y = 3 + 65

17y = 68 ⇒ y = 68/17 = 4

  • Verification:

Substituting y = 4 in LHS

LHS = (2(4) – 5)/(3(4) + 1) = (8 – 5)/(12 + 1) = 3/13 = RHS

Verified!

Question 6

Solve 8 3y5y 2 16 and verify your…


📝 Question:

Solve (8 – 3y)/(5y + 2) = 1/6 and verify your answer. / (8 – 3y)/(5y + 2) = 1/6

💡 Solution:

Solving the Equation:**

On cross-multiplying

6(8 – 3y) = 1(5y + 2)

48 – 18y = 5y + 2

Transposing

48 – 2 = 5y + 18y

23y = 46 ⇒ y = 46/23 = 2

  • Verification:

Substituting y = 2 in LHS

LHS = (8 – 3(2))/(5(2) + 2) = (8 – 6)/(10 + 2) = 2/12 = 1/6 = RHS

Verified!

Question 7

Solve 17 2kk 5 3 and verify your…


📝 Question:

Solve (17 – 2k)/(k – 5) = -3 and verify your answer. / (17 – 2k)/(k – 5) = -3

💡 Solution:

Solving the Equation:**

On cross-multiplying

17 – 2k = -3(k – 5)

17 – 2k = -3k + 15

Transposing

-2k + 3k = 15 – 17 ⇒ k = -2

  • Verification:

Substituting k = -2 in LHS

LHS = (17 – 2(-2))/(-2 – 5) = (17 + 4)/(-7) = 21/-7 = -3 = RHS

Verified!

Question 8

Solve 4x x 73x 5x 9 23 and…


📝 Question:

Solve (4x – (x + 7))/(3x – (5x – 9)) = 2/3 and verify your answer. / (4x – (x + 7))/(3x – (5x – 9)) = 2/3

💡 Solution:

Solving the Equation:**

Simplifying numerator and denominator first

Numerator: 4x – x – 7 = 3x – 7

Denominator: 3x – 5x + 9 = -2x + 9

Equation becomes

(3x – 7)/(-2x + 9) = 2/3

On cross-multiplying

3(3x – 7) = 2(-2x + 9)

9x – 21 = -4x + 18

Transposing

9x + 4x = 18 + 21

13x = 39 ⇒ x = 39/13 = 3

  • Verification:

Substituting x = 3 in LHS

LHS = (4(3) – (3 + 7))/(3(3) – (5(3) – 9)) = (12 – 10)/(9 – (15 – 9)) = (2)/(9 – 6) = 2/3 = RHS

Verified!

Question 9

Solve 15x 033x 310 and verify your answer…


📝 Question:

Solve (1.5x + 0.3)/(3x) = 3/10 and verify your answer. / (1.5x + 0.3)/(3x) = 3/10

💡 Solution:

Solving the Equation:**

On cross-multiplying

10(1.5x + 0.3) = 3(3x)

15x + 3 = 9x

Transposing

15x – 9x = -3

6x = -3 ⇒ x = -3/6 = -0.5

  • Verification:

Substituting x = -0.5 in LHS

LHS = (1.5(-0.5) + 0.3)/(3(-0.5)) = (-0.75 + 0.3)/(-1.5) = (-0.45)/(-1.5) = 0.3 = 3/10 = RHS

Verified!

SECTION / EXERCISE

EXERCISE 12.2 / 12.2

Question 1

The sum of two numbers is 42 If…


📝 Question:

The sum of two numbers is 42. If second number is twice the first one, find the numbers. / 42 ,

💡 Solution:

Let the first number = x.

Then the second number = 2x.

According to the condition

x + 2x = 42

3x = 42 ⇒ x = 42/3 = 14

First number = 14, Second number = 2 × 14 = 28.

Question 2

Neeraj has mangoes three times that of Mayank…


📝 Question:

Neeraj has mangoes three times that of Mayank. If Mayank is given 8 mangoes more and Neeraj is given 6 mangoes more, the ratio of mangoes with Mayank and Neeraj will become 1 : 2. Then how many mangoes does each of them have? / 8 6 , 1 : 2 – ?

💡 Solution:

Let the number of mangoes with Mayank = x.

Then the number of mangoes with Neeraj = 3x.

After giving 8 more mangoes to Mayank = x + 8.

After giving 6 more mangoes to Neeraj = 3x + 6.

According to the question

(x + 8)/(3x + 6) = 1/2

On cross-multiplying

2(x + 8) = 1(3x + 6)

2x + 16 = 3x + 6

3x – 2x = 16 – 6 ⇒ x = 10

Mayank has 10 mangoes, Neeraj has 3 × 10 = 30 mangoes.

Question 3

Out of 2 equilateral triangles the side of…


📝 Question:

Out of 2 equilateral triangles, the side of the first triangle is 3 cms. more than that of the second. The ratio of the perimeters of both the triangles is 5 : 2, then find out the sides of the triangles. / 3 5 : 2 ,

💡 Solution:

Let the side of the second equilateral triangle = x cm.

Side of the first equilateral triangle = (x + 3) cm.

Perimeter of an equilateral triangle = 3 × side.

Perimeter of first triangle = 3(x + 3) cm.

Perimeter of second triangle = 3x cm.

According to the question

(3(x + 3))/(3x) = 5/2 ⇒ (x + 3)/(x) = 5/2

On cross-multiplying

2(x + 3) = 5x

2x + 6 = 5x ⇒ 3x = 6 ⇒ x = 2 cm

Side of second triangle = 2 cm, Side of first triangle = 2 + 3 = 5 cm.

Question 4

In a two digit number the tens place…


📝 Question:

In a two digit number, the ten’s place digit is three times the digit in unit’s place. If the digits are reversed, the new number becomes 36 less than the original number. Find the number. / 36

💡 Solution:

Let the unit’s place digit = x.

Then the ten’s place digit = 3x.

Original number = 10(3x) + x = 30x + x = 31x.

On reversing the digits

New unit’s digit = 3x, New ten’s digit = x.

New number = 10(x) + 3x = 13x.

According to the question

New Number = Original Number – 36

13x = 31x – 36

31x – 13x = 36

18x = 36 ⇒ x = 2

Unit’s digit = 2, Ten’s digit = 3(2) = 6.

Original number = 62.

Question 5

The sum of the digits of a two…


📝 Question:

The sum of the digits of a two digit number is 7. On reversing the digits, the new number obtained is 9 more than the original number. Find the number. / 7 9

💡 Solution:

Let the unit’s place digit = x.

Then the ten’s place digit = 7 – x.

Original number = 10(7 – x) + x = 70 – 10x + x = 70 – 9x.

On reversing the digits

New number = 10(x) + (7 – x) = 9x + 7.

According to the question

9x + 7 = (70 – 9x) + 9

9x + 7 = 79 – 9x

9x + 9x = 79 – 7

18x = 72 ⇒ x = 4

Unit’s digit = 4, Ten’s digit = 7 – 4 = 3.

Original number = 34.

Question 6

The denominator of a rational number is 2…


📝 Question:

The denominator of a rational number is 2 more than its numerator. If the numerator is increased 4 times, and 8 is added to the denominator, the new number would be 4/3. Find the original number. / 2 4 8 , 4/3

💡 Solution:

Let the numerator = x.

Then the denominator = x + 2.

Original rational number = (x)/(x + 2).

According to the condition

New numerator = 4x, New denominator = (x + 2) + 8 = x + 10.

(4x)/(x + 10) = 4/3

Dividing numerators on both sides by 4

(x)/(x + 10) = 1/3

On cross-multiplying

3x = x + 10 ⇒ 2x = 10 ⇒ x = 5

Numerator = 5, Denominator = 5 + 2 = 7.

Original rational number = 5/7.

Question 7

The ages of Arun and Aakash are in…


📝 Question:

The ages of Arun and Aakash are in the ratio of 7 : 5. After 6 years, their ages would be in the ratio of 5 : 4. Find their present ages. / 7 : 5 6 5 : 4

💡 Solution:

Let Arun’s present age = 7x years.

Let Aakash’s present age = 5x years.

After 6 years

Arun’s age = (7x + 6) years, Aakash’s age = (5x + 6) years.

According to the question

(7x + 6)/(5x + 6) = 5/4

On cross-multiplying

4(7x + 6) = 5(5x + 6)

28x + 24 = 25x + 30

28x – 25x = 30 – 24

3x = 6 ⇒ x = 2

Arun’s present age = 7(2) = 14 years.

Aakash’s present age = 5(2) = 10 years.

Question 8

Manishas mother is thrice the age of Manisha…


📝 Question:

Manisha’s mother is thrice the age of Manisha. After 4 years, mother’s age would be two and a half times that of Manisha. Find their present ages. / 4

💡 Solution:

Let Manisha’s present age = x years.

Then Mother’s present age = 3x years.

After 4 years

Manisha’s age = (x + 4) years, Mother’s age = (3x + 4) years.

According to the question

3x + 4 = 2.5(x + 4) = 5/2(x + 4)

Multiplying both sides by 2

2(3x + 4) = 5(x + 4)

6x + 8 = 5x + 20

6x – 5x = 20 – 8 ⇒ x = 12 years

Manisha’s present age = 12 years, Mother’s present age = 3(12) = 36 years.

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