In each of the following, find the co-ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation :

(i) 3 - 2x = 7; 2y + 1 = 10 - $2 \dfrac{1}{2}y$.

(ii) $\dfrac{2a}{3} - 1 = \dfrac{a}{2}; \dfrac{15 - 4b}{7} = \dfrac{2b - 1}{3}$

(iii) $5x -(5 - x) = \dfrac{1}{2}(3 - x); 4 - 3y = \dfrac{4 + y}{3}$

(i) 3 - 2x = 7

⇒ 2x = 3 - 7

⇒ 2x = - 4

⇒ x = - $\dfrac{4}{2}$

⇒ x = - 2

2y + 1 = 10 - $2 \dfrac{1}{2}y$

⇒ 2y + $\dfrac{5}{2}y$ = 10 - 1

⇒ $\dfrac{4 + 5}{2}y$ = 9

⇒ $\dfrac{9}{2}y$ = 9

⇒ y = $\dfrac{2 \times 9}{9}$

⇒ y = 2

Hence, the co-ordinates of the point = (-2, 2).

(ii)

$\dfrac{2a}{3} - 1 = \dfrac{a}{2}\\[1em] ⇒ \dfrac{2a}{3} - \dfrac{a}{2} = 1\\[1em] ⇒ \dfrac{4a - 3a}{6} = 1\\[1em] ⇒ \dfrac{a}{6} = 1\\[1em] ⇒ a = 6\\[1em]$

$\dfrac{15 - 4b}{7} = \dfrac{2b - 1}{3}\\[1em] ⇒ 3(15 - 4b) = 7(2b - 1)\\[1em] ⇒ 45 - 12b = 14b - 7\\[1em] ⇒ 45 + 7 = 14b + 12b\\[1em] ⇒ 52 = 26b\\[1em] ⇒ b = \dfrac{52}{26}\\[1em] ⇒ b = 2\\[1em]$

Hence, the co-ordinates of the point = (6, 2).

(iii)

$5x -(5 - x) = \dfrac{1}{2}(3 - x)\\[1em] ⇒ 5x - 5 + x = \dfrac{1}{2}(3 - x) \\[1em] ⇒ 6x - 5 = \dfrac{1}{2}(3 - x) \\[1em] ⇒ 2(6x - 5) = 3 - x\\[1em] ⇒ 12x + x = 3 + 10 \\[1em] ⇒ 13x = 13 \\[1em] ⇒ x = \dfrac{13}{13}\\[1em] ⇒ x = 1 \\[1em]$

$4 - 3y = \dfrac{4 + y}{3}\\[1em] ⇒ 3(4 - 3y) = 4 + y\\[1em] ⇒ 12 - 9y = 4 + y\\[1em] ⇒ 12 - 4 = 9y + y\\[1em] ⇒ 8 = 10y\\[1em] ⇒ y = \dfrac{8}{10}\\[1em] ⇒ y = \dfrac{4}{5}\\[1em]$

Hence, the co-ordinates of the point = $\Big(1, \dfrac{4}{5}\Big)$.