Page 50 - Maths Skills - 7

48 Maths

Exercise 3.2

1. Solve the following equations by trial and error method:

(i) 8p – 7 = 9 (ii) 7m = 49 (iii) 3r – 6 = 0
2. Write the steps to separate the variable and then solve the following equations:
4b
(i) 14b = – 56 (ii) – 6 = 18 (iii) 26 = 8 + v
(iv) 2m – 1 = 1 (v) – 15 + n = – 9 (vi) 7y – 3 = 46
4
3. Solve the following equations and verify your answer:
a 1 2 m
(i) 6m + 5 = 20 (ii) + = 1 (iii) = 12
2 2 4
(iv) 3n + 11 = 14 (v) 10b – 7 = 63 (vi) 4m = 25

4. Solve the following equations:
−b a 1 5m
(i) = 2 (ii) − = 1 (iii) = 15
3 2 2 8
(iv) 3x + 20 = 8 (v) 4p + 3 = 23 (vi) 8x – 6 = 50

Transposition Method
We can solve a linear equation by the method of transposition. Transposition means shifting a number or variable
from one side to other side of equality ("=") sign. While transposing a number to the other side of '=', we change
its sign, i.e., '+' sign changes into '–' sign, '–' sign changes into '+' sign. '×' sign changes into '÷ 'sign and '÷' sign
into '×' sign changes.

The above concept is illustrated with the help of examples that follows.

Let’s Attempt

Example 1: Solve the given equations by using transposition.
5 37
(i) 12x + 5 = 25 (ii) 2y + =
2 2
5 37
Solution: (i) 12x + 5 = 25 (ii) 2y + =
2 2

or 12x = 25 – 5 2y = 37 – 5
2 2

12x = 20 or 2y = 37 – 5 = 32 = 16
2 2
5
or x = 20 = 10 = or y = 16 = 8
12 6 3 2
y = 8

Example 2: Solve the given equations by using transposition.
1
(i) 6z + 10 = – 2 (ii) 3l = 2 (iii) 2b – 5 = 3 (iv) 2y – = 1
2 3 3 2 3