Exercise 2.3

1. Solve the following equations

i. 7x - 5 = 2x

Solution:

Subtract from both sides of the equation: 7x - 5 - 2x = 2x - 2x i.e. - 5 = 0

5x - 5 + 5 = 0 + 5

5x =

5x5 = 55

x =

Therefore, the solution is x = 1.

ii. 5x – 12 = 2x – 6

Solution:

Subtract from both sides: 5x - 12 - 2x = 2x - 6 - 2x

- 12 = -6

3x - 12 + 12 = -6 + 12

3x =

3x3 = 63

x =

Therefore, the solution is x = 2.

iii. 7p - 3 = 3p + 8

Solution:

Subtract from both sides: 7p - 3 - 3p = 3p + 8 - 3p

4p - 3 =

4p - 3 + 3 = 8 + 3

4p =

4p4 = 114

p =

Therefore, the solution is p = 114.

iv. 7z + 13 = 2z + 4

Solution:

Subtract from both sides: 7z + 13 - 2z = 2z + 4 - 2z

+ 13 = 4

5z + 13 - 13 = 4 - 13

5z =

5z5 = −95

z =

Therefore, the solution is z = −95.

v. 8m + 9 = 7m +8

Solution: Subtract from both sides: 8m + 9 - 7m = 7m + 8 - 7m

+ 9 = 8

m + 9 - 9 = 8 - 9

m =

Therefore, the solution is m = -1.

vi. 9y + 5 = 15y – 1

Solution: Subtract from both sides:

9y + 5 - 15y = 15y - 1 - 15y

+ 5 = -1

-6y + 5 - 5 = -1 - 5

-6y =

−6y−6 = −6−6

y =

Therefore, the solution is y = 1.

vii. 3x + 4 = 5 (x–2)

Solution: Distribute 5 on the right side: 3x + 4 = -

3x + 4 - 3x = 5x - 10 - 3x

4 = - 10

4 + 10 = 2x - 10 + 10

= 2x

142 = 2x2

= x

Therefore, the solution is x = 7.

viii. 3 (t – 3) = 5 (2t – 1)

Solution: Expanding, - 9 = - 5

3t - 9 - 3t = 10t - 5 - 3t

-9 = - 5

-9 + 5 = 7t - 5 + 5

= 7t

−47 = 7t7

= t

Therefore, the solution is t = −47.