Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Set 12.1 Solutions

Question 1. Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.
(i) x - 4 = 3, x = - 1, 7, - 7
(ii) 9m = 81, m = 3, 9, -3
(iii) 2a + 4 = 0, a = 2, - 2, 1
(iv) 3 - y = 4, y = - 1, 1, 2
Answer:
(i) x - 4 = 3 ....(i)
Substituting x = - 1 in L.H.S. of equation (i),
L.H.S. = (-1) - 4
= -5
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = - 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i),
L.H.S. = (7) - 4
= 3
R.H.S. = 3
∴ L.H.S. = R.H.S.
∴ x = 7 is the solution of the given equation.

Substituting x = - 7 in L.H.S. of equation (i),
L.H.S. = (-7) - 4
= -11
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = - 7 is not the solution of the given equation.

(ii) 9m = 81 ....(i)
Substituting m = 3 in L.H.S. of equation (i),
L.H.S. = 9 × (3)
= 27
R.H.S. = 81
∴ L.H.S. ≠ R.H.S.
∴ m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i),
L.H.S. = 9 × (9)
= 81
R.H.S. = 81
∴ L.H.S. = R.H.S.
∴ m = 9 is the solution of the given equation.

Substituting m = - 3 in L.H.S. of equation (i),
L.H.S. = 9 × (-3)
= -27
R.H.S. = 81
∴ L.H.S. ≠ R.H.S.
∴ m = - 3 is not the solution of the given equation.

(iii) 2a + 4 = 0 .....(i)
Substituting a = 2 in L.H.S. of equation (i),
L.H.S. = 2 (2) + 4
= 4 + 4
= 8
R.H.S. = 0
∴ L.H.S. ≠ R.H.S.
∴ a = 2 is not the solution of the given equation.

Substituting a = - 2 in L.H.S. of equation (i),
L.H.S. = 2 (-2)+ 4
= -4 + 4
= 0
R.H.S. = 0
∴ L.H.S. = R.H.S.
∴ a = - 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),
L.H.S. = 2(1)+ 4
= 2 + 4
= 6
R.H.S. = 0
∴ L.H.S. ≠ R.H.S.
∴ a = 1 is not the solution of the given equation.

(iv) 3 - y = 4 ...(i)
Substituting y = -1 in L.H.S. of equation (i),
L.H.S. = 3 - (-1)
= 3 + 1
= 4
R.H.S. = 4
∴ L.H.S. = R.H.S.
∴ y = - 1 is the solution of the given equation.

Substituting y = 1 in L.H.S. of equation (i),
L.H.S. = 3-(1)
= 2
R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ y = 1 is not the solution of the given equation.

Substituting y = 2 in L.H.S. of equation (i),
L.H.S. = 3-(2)
= 1
R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ y = 2 is not the solution of the given equation.
In simple words: To check if a value is a solution to an equation, substitute the value into the equation's Left Hand Side (LHS) and Right Hand Side (RHS). If LHS equals RHS, the value is a solution; otherwise, it is not.

🎯 Exam Tip: Always show the steps for calculating both LHS and RHS clearly. Incorrect calculations on either side will lead to wrong conclusions about the solution.

Question 2. Solve the following equations:
(i) 17p - 2 = 49
(ii) 2m + 7 = 9
(iii) 3x + 12 = 2x - 4
(iv) 5 (x - 3) = 3 (x + 2)
(v) \( \frac{9x}{8} \) + 1 = 10
(vi) \( \frac{y}{7} + \frac{y-4}{3} \) = 2
(vii) 13x - 5 = \( \frac{3}{2} \)
(viii) 3 (y + 8) = 10 (y - 4) + 8
(ix) \( \frac{x-9}{5} = \frac{x-5}{7} \)
(x) \( \frac{y-4}{3} \) + 3y = 4
(xi) \( \frac{b+(b+1)+(b+2)}{4} \) = 21
Answer:
(i) 17p - 2 = 49
∴ 17p - 2 + 2 = 49 + 2 ...[Adding 2 on both the sides]
∴ 17p = 51
∴ \( \frac{17p}{17} = \frac{51}{17} \) ...[Dividing both the sides by 17]
p = 3

(ii) 2m + 7 = 9
∴ 2m + 7 - 7 = 9 - 7 ...[Subtracting 7 from both the sides]
∴ 2m = 2
∴ \( \frac{2m}{2} = \frac{2}{2} \) [Dividing both the sides by 2]
∴ m = 1

(iii) 3x + 12 = 2x - 4
∴ 3x + 12 - 12 = 2x - 4 - 12 ...[Subtracting 12 from both the sides]
∴ 3x = 2x - 16
∴ 3x - 2x = 2x - 16 - 2x ...[Subtracting 2x from both the sides]
∴ x = - 16

(iv) 5 (x - 3) = 3 (x + 2)
∴ 5x - 15 = 3x + 6
∴ 5x - 15 + 15 = 3x + 6 + 15 ...[Adding 15 on both the sides]
∴ 5x = 3x + 21
∴ 5x - 3x = 3x + 21 - 3x ...[Subtracting 3x from both the sides]
∴ 2x = 21
∴ \( \frac{2x}{2} = \frac{21}{2} \) ...[Dividing both the sides by 2]
∴ x = \( \frac{21}{2} \)

(v) \( \frac{9x}{8} \) + 1 = 10
∴ \( \frac{9x}{8} \) + 1 - 1 = 10 - 1 ...[Subtracting 1 from both the sides]
∴ \( \frac{9x}{8} \) = 9
∴ \( \frac{9x}{8} \) × 8 = 9 × 8 ...[Multiplying both the sides by 8]
∴ 9x = 72
∴ \( \frac{9x}{9} = \frac{72}{9} \) ...[Dividing both the sides by 9]
∴ x = 8

(vi) \( \frac{y}{7} + \frac{y-4}{3} \) = 2
∴ \( \frac{y \times 3}{7 \times 3} + \frac{(y-4) \times 7}{3 \times 7} \) = 2
∴ \( \frac{3y}{21} + \frac{7y-28}{21} \) = 2
∴ \( \frac{3y+7y-28}{21} \) = 2
∴ \( \frac{10y-28}{21} \) = 2
∴ \( \frac{10y-28}{21} \times 21 \) = 2 × 21 ...[Multiplying both the sides by 21]
∴ 10y - 28 = 42
∴ 10y - 28 + 28 = 42 + 28 ...[Adding 28 on both the sides]
∴ 10y = 70
∴ \( \frac{10y}{10} = \frac{70}{10} \) ...[Dividing both the sides by 10]
∴ y = 7

(vii) 13x - 5 = \( \frac{3}{2} \)
∴ (13x - 5) × 2 = \( \frac{3}{2} \) × 2 ...[Multiplying both the sides by 2]
∴ 26x - 10 = 3
∴ 26x - 10 + 10 = 3 + 10 ...[Adding 10 on both the sides]
∴ 26x = 13
∴ \( \frac{26x}{26} = \frac{13}{26} \) ...[Dividing both the sides by 26]
∴ x = \( \frac{1}{2} \)

(viii) 3 (y + 8) = 10 (y - 4) + 8
∴ 3y + 24 = 10y - 40 + 8
∴ 3y + 24 = 10y - 32
∴ 3y + 24 - 24 = 10y - 32 - 24 ...[Subtracting 24 from both the sides]
∴ 3y = 10y - 56
∴ 3y - 10y = 10y - 56 - 10y ...[Subtracting 10y from both the sides]
∴ - 7y = - 56
∴ \( \frac{-7y}{-7} = \frac{-56}{-7} \) ...[Dividing both the sides by - 7]
∴ y = 8

(ix) \( \frac{x-9}{5} = \frac{x-5}{7} \)
∴ \( \frac{x-9}{5} \times 5 \times 7 = \frac{x-5}{7} \times 5 \times 7 \) ...[Multiplying both the sides by 7 (x - 5)]
∴ 7 (x - 9) = 5 (x - 5)
∴ 7x - 63 = 5x - 25
∴ 7x - 63 + 63 = 5x - 25 + 63 ...[Adding 63 on both the sides]
∴ 7x = 5x + 38
∴ 7x - 5x = 5x + 38 - 5x ...[Subtracting 5x from both the sides]
∴ 2x = 38
∴ \( \frac{2x}{2} = \frac{38}{2} \) ...[Dividing both the sides by 2]
∴ x = 19

(x) \( \frac{y-4}{3} \) + 3y = 4
∴ \( \frac{y-4}{3} \) × 3 + 3y × 3 = 4 × 3 ...[Multiplying both the sides by 3]
∴ y - 4 + 9y = 12
∴ 10y - 4 = 12
∴ 10y - 4 + 4=12 + 4 ...[Adding 4 on both the sides]
∴ 10y = 16
∴ \( \frac{10y}{10} = \frac{16}{10} \) ...[Dividing both the sides by 10]
∴ y = \( \frac{8}{5} \)

(xi) \( \frac{b+(b+1)+(b+2)}{4} \) = 21
∴ \( \frac{b+(b+1)+(b+2)}{4} \times 4 \) = 21 × 4 ...[Multiplying both the sides by 4]
∴ b + b + 1 + b + 2 = 84
∴ 3b + 3 = 84
∴ 3b + 3 - 3 = 84 - 3 ...[ Subtracting 3 from both the sides]
∴ 3b = 81
∴ \( \frac{3b}{3} = \frac{81}{3} \) ...[Dividing both the sides by 3]
∴ b = 27
In simple words: To solve linear equations, isolate the variable by applying inverse operations (addition/subtraction, multiplication/division) to both sides of the equation, maintaining equality at each step until the variable stands alone.

🎯 Exam Tip: Always verify your solution by substituting the found value back into the original equation to ensure it satisfies the equality. This helps catch computational errors.

Maharashtra Board Class 8 Maths Chapter 12 Equations In One Variable Practice Set 12.1 Intext Questions And Activities

Question 1. Fill in the boxes to solve the following equations. (Textbook pg. no. 75)
(i) x + 4 = 9
∴ x + 4 - _ = 9 - _
...[Subtracting 4 from both the sides]
∴ x = _
(ii) x - 2 = 7
∴ x - 2 + _ = 7 + _
[Adding 2 on both the sides]
∴ x = _
(iii) \( \frac{x}{3} \) = 4
\( \frac{x}{3} \) = 4 x_
∴ x = _
(iv) 4x = 24
∴ \( \frac{\text{_}}{\text{_}} = \frac{\text{_}}{\text{_}} \)
∴ x = _
Answer:
(i) x + 4 = 9
∴ x + 4 - 4 = 9 - 4
...[Subtracting 4 from both the sides]
∴ x = 5

(ii) x - 2 = 7
∴ x - 2 + 2 = 7 + 2
[Adding 2 on both the sides]
∴ x = 9

(iii) \( \frac{x}{3} \) = 4
\( \frac{x}{3} \) × 3 = 4 × 3
[Multiplying both the sides by 3]
∴ x = 12

(iv) 4x = 24
∴ \( \frac{4x}{4} = \frac{24}{4} \)
[Dividing both the sides by 4]
∴ x = 6
In simple words: These are basic steps for solving one-variable linear equations by isolating the variable using inverse operations: subtraction for addition, addition for subtraction, multiplication for division, and division for multiplication.

🎯 Exam Tip: Understanding these fundamental inverse operations is crucial as they form the bedrock for solving more complex equations in algebra.