Solve the following linear equations:

(i) $2x - 1\dfrac{1}{2} = 4\dfrac{1}{2}$

(ii) 3(y - 1) = 2(y + 1)

(iii) n - 3 = 5n - 5

(iv) $\dfrac{1}{3}(7x - 1) = \dfrac{1}{4}$

(i) We have:

$\phantom{=} 2x - 1\dfrac{1}{2} = 4\dfrac{1}{2} \\[1em] \Rightarrow 2x - \dfrac{3}{2} = \dfrac{9}{2} \\[1em] \Rightarrow 2x = \dfrac{9}{2} + \dfrac{3}{2} \quad\text{[Transposing } -\dfrac{3}{2} \text{ to RHS]} \\[1em] \Rightarrow 2x = \dfrac{9 + 3}{2} \\[1em] \Rightarrow 2x = \dfrac{12}{2} \\[1em] \Rightarrow 2x = 6 \\[1em] \Rightarrow x = \dfrac{6}{2}$

Hence, x = 3.

(ii) We have:

3(y - 1) = 2(y + 1)

⇒ 3y - 3 = 2y + 2 $\quad$[Removing brackets]

⇒ 3y - 2y = 2 + 3 $\quad$[Transposing -3 to RHS and +2y to LHS]

⇒ y = 5

Hence, y = 5.

(iii) We have:

n - 3 = 5n - 5

⇒ n - 5n = -5 + 3 $\quad$[Transposing -3 to RHS and +5n to LHS]

⇒ -4n = -2

⇒ n = $\dfrac{-2}{-4}$

⇒ n = $\dfrac{1}{2}$

Hence, n = $\dfrac{1}{2}$.

(iv) We have:

$\phantom{=} \dfrac{1}{3}(7x - 1) = \dfrac{1}{4} \\[1em] \Rightarrow 7x - 1 = \dfrac{1}{4} \times 3 \quad\text{[Multiplying both sides by 3]} \\[1em] \Rightarrow 7x - 1 = \dfrac{3}{4} \\[1em] \Rightarrow 7x = \dfrac{3}{4} + 1 \quad\text{[Transposing -1 to RHS]} \\[1em] \Rightarrow 7x = \dfrac{3 + 4}{4} \\[1em] \Rightarrow 7x = \dfrac{7}{4} \\[1em] \Rightarrow x = \dfrac{7}{4} \times \dfrac{1}{7}$

Hence, x = $\dfrac{1}{4}$.