Contents
Equations with One Variable – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)
MathematicsGeneral ScienceMaharashtra Board Solutions
Exercise-58
Solution 1:
Exercise-59
Solution 1:
Exercise-60
Solution 1:
- Equation.
- Equation.
- Equality.
- Equation.
- Equation.
- Equality.
Exercise-61
Solution 1:
Exercise-62
Solution 1(1):
Let us see, for which value of x, the value of 7x becomes 14.
When x = 1, 7x = 7 × 1 = 7
When x = 2, 7x = 7 × 2 = 14
∴ The solution of the equation 7x = 14 is 2.
Solution 1(2):
Let us see, for which value of x, the value of x – 10 becomes 2.
When x = 11, 11 – 10 = 1
When x = 12, 12 – 10 = 2
∴ The solution of the equation x – 10 = 2 is 12.
Solution 1(3):
Let us see, for which value of p, the value of p + 6 becomes 10.
When p = 1, p + 6 = 1 + 6 = 7
When p = 2, p + 6 = 2 + 6 = 8
When p = 3, p + 6 = 3 + 6 = 9
When p = 4, p + 6 = 4 + 6 = 10
∴ The solution of the equation p + 6 = 10 is 4.
Solution 1(4):
Let us see, for which value of p, the value of 7 – p becomes 5.
When p = 1, 7 – p = 7 – 1 = 6
When p = 2, 7 – p = 7 – 2 = 5
∴ The solution of the equation 5 = 7 – p is 2.
Solution 1(5):
Let us see, for which value of p, the value of 13 + y becomes 18.
When y = 3, 13 + y = 13 + 3 = 16
When y = 4, 13 + y = 13 + 4 = 17
When y = 5, 13 + y = 13 + 5 = 18
∴ The solution of the equation 13 + y = 18 is 5.
Solution 1(6):
Solution 1(7):
Solution 1(8):
Let us see, for which value of p, the value of 2m becomes 16.
When m = 6, 2m = 2 × 6 = 12
When m = 7, 2m = 2 × 7 = 14
When m = 8, 2m = 2 × 8 = 16
∴ The solution of the equation 16 = 2m is 8.
Solution 1(9):
Solution 1(10):
Let us see, for which value of n, the value of 2 + n becomes 8.
When n = 4, 2 + n = 2 + 4 = 6
When n = 5, 2 + n = 2 + 5 = 7
When n = 6, 2 + n = 2 + 6 = 8
∴ The solution of the equation 2 + n = 8 is 6.
Solution 1(11):
Let us see, for which value of x, the value of 9 – x becomes 6.
When x = 1, 9 – x = 9 – 1 = 4
When x = 2, 9 – x = 9 – 2 = 5
When x = 3, 9 – x = 9 – 3 = 6
∴ The solution of the equation 9 – x = 6 is 3.
Solution 1(12):
Let us see, for which value of y, the value of y – 4 becomes 0.
When y = 1, y – 4 = 1 – 4 = -3
When y = 2, y – 4 = 2 – 4 = -2
When y = 3, y – 4 = 3 – 4 = -1
When y = 4, y – 4 = 4 – 4 = 0
∴ The solution of the equation y – 4 = 0 is 4.
Exercise-63
Solution 1(1):
m – 4 = 1
∴ m – 4 + 4 = 1 + 4 (Addition property of equality)
∴ m + 0 = 5 [∵(-5) + 5 = 0 and 4 +1 = 5]
∴ m = 5 (∵ Any number + 0 = the same number.)
When the value of m is 5, the two sides of the equation become equal.
∴ 5 is the solution of the given equation.
Solution 1(2):
p + 4 = 11
∴ p + 4 – 4 = 11 – 4 (Subtraction property of equality)
∴ p + 0 = 7 [∵(-4) + 4 = 0 and 11 – 4 = 7]
∴ p = 7 (∵ Any number + 0 = the same)
When the value of p is 7, the two sides of the equation become equal.
∴ 7 is the solution of the given equation.
Solution 1(3):
Solution 1(4):
Solution 1(5):
6 = k – 2
∴ 6 + 2 = k – 2 + 2 (Addition property of equality)
∴ 8 = k + 0 [∵(-2) + 2 = 0 and 6 + 2 = 8]
∴ 8 = k (∵ Any number + 0 = the same)
When the value of k is 8, the two sides of the equation become equal.
∴ 8 is the solution of the given equation.
Solution 1(6):
25 = t + 16
∴ 25 – 16 = t + 16 – 16 (Subtraction property of equality)
∵ 9 = t + 0 [∵(-16) + 16 = 0 and 25 – 16 = 9]
∵ 9 = t (∵ Any number + 0 = the same)
When the value of t is 9, the two sides of the equation become equal.
∴ 9 is the solution of the given equation.
Solution 1(7):
Solution 1(8):
Solution 1(9):
8 = z + 5
∵ 8 – 5 = z + 5 – 5 (Subtraction property of equality)
∵ 3 = z + 0 [∵(-5) + 5 = 0 and 8 – 5 = 3]
∵ 3 = z (∵ Any number + 0 = the same number)
When the value of z is 3, the two sides of the equation become equal.
∴ 3 is the solution of the given equation.
Solution 1(10):
n – 6 = 6
∴ n – 6 + 6 = 6 + 6 (Addition property of equality)
∴ n + 0 = 12 [∵(-6) + 6 = 0 and 6 + 6 = 12]
∴ n = 12 (∵ Any number + 0 = the same number)
When the value of n is 12, the two sides of the equation become equal.
∴ 12 is the solution of the given equation.
Solution 1(11):
Solution 1(12):
