Maharashtra Board 8th Class Maths Practice Set 12.1 Solutions Chapter 12 Equations in One Variable
Maharashtra Board 8th Class Maths Practice Set 12.1 Solutions Chapter 12 Equations in One Variable
Practice Set 12.1 8th Std Maths Answers Chapter 12 Equations in One Variable
Equation in One Variable Practice Set 12.1 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
Solution:
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.
Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.1 Intext Questions and Activities
Std 8 Maths Practice Set 12.1 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. [latex]\frac { x }{ 3 }=4\)
∴x/3 × __ = 4 ×__
∴x = __
iv. 4x = 24
∴ __ = __
∴x = __
Solution:
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. x/3=4
∴x/3 × 3 = 4 × 3
… [Multiplying both the sides by 3]
∴x = 12
iv. 4x = 24
∴ 4x/[4]=24/[4]
… [Dividing both the sides by 4]
∴x = 6