Solve the following pairs of linear equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{3}$

$\dfrac{2}{x} - \dfrac{1}{y} = 2$

Substituting $\dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b$ in above equations we get,

$2a + \dfrac{2}{3}b = \dfrac{1}{3}$

Multiply by 3:

6a + 2b = 1 …….(i)

2a - b = 2 …….(ii)

⇒ -b = 2 - 2a

⇒ b = 2a - 2

Substituting this value of b in equation (i) we get,

6a + 2(2a - 2) = 1

⇒ 6a + 4a - 4 = 1

⇒ 10a - 4 = 1

⇒ 10a = 1 + 4

⇒ 10a = 5

⇒ a = $\dfrac{5}{10}$

⇒ a = $\dfrac{1}{2}$

⇒ $\dfrac{1}{x} = \dfrac{1}{2}$

⇒ $x = 2$

From (ii) we have,

b = 2a - 2

Substituting value of a in above equation we get,

b = 2 x $\dfrac{1}{2}$ - 2

⇒ b = 1 - 2

⇒ b = -1

⇒ $\dfrac{1}{y}$ = -1

⇒ $y$ = -1

Hence, x = 2 and y = -1.