Equations in One Variable – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)
MathematicsGeneral ScienceMaharashtra Board Solutions
Exercise 38:
Solution 1(1):
y – 2 = 9
[Addition Property of Equality: If the same number is added to both sides of the equality, the sums so obtained are equal.]
∴ y – 2 + 2 = 9 + 2
∴ y = 11
Solution 1(2):
p + 3 = 12
[Subtraction Property of Equality: If the same number is subtracted from both sides of an equality the sums so obtained are equal]
∴ p + 3 – 3 = 12 – 3
∴ p = 9
Solution 1(3):
Solution 1(4):
Solution 1(5):
Solution 1(6):
Solution 1(7):
Solution 1(8):
Solution 1(9):
Solution 1(10):
Exercise 39:
Solution 1(1):
2p = p + 7
∴ 2p – p = p + 7 – p
∴ p = 7
Solution 1(2):
Solution 1(3):
Solution 1(4):
4x – 5 = 3(x + 2)
∴ 4x – 5 = 3x + 6
∴ 4x – 5 – 3x = 3x + 6 – 3x
∴ x – 5 = 6
∴ x – 5 + 5 = 6 + 5
∴ x = 11
Solution 1(5):
Solution 1(6):
Solution 1(7):
Solution 1(8):
Solution 1(9):
Solution 1(10):
Solution 1(11):
Solution 1(12):
Exercise 40:
Solution 1(1):
Let x be the given number.
4 less than another number = x – 4
It is given that 4 less than one number is equal to 11.
∴ x – 4 = 11
Solution 1(2):
Let Soham’s age be x years.
Sagar is 2 years younger than Soham.
∴ Sagar’s age = (x – 2) years
It is given that the sum of their ages is 38.
∴ x + (x – 2) = 38
Solution 1(3):
Let x be the given number.
Now, twice the number = 2x
9 less than twice the number = 2x – 9
It is given that, 9 less than twice the number is 15.
∴ 2x – 9 = 15
Solution 1(4):
Let the breadth of the given rectangle be x cm.
So the length of the rectangle = (x + 3) cm
Perimeter of the rectangle = 2(Length + Breadth)
∴ Perimeter of the rectangle = 2(x + 3 + x)
It is given that the perimeter of the rectangle is 30 cm.
∴ 2(x + 3 + x) = 30
Solution 1(5):
Let x be the amount with Kiran.
∴ Amount with Sultana = x – 5
It is given that the two of them together have Rs. 51.
∴ (x – 5) + x = 51
Exercise 41:
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Solution 6:
Solution 7:
