For solving each pair of equations, use the method of elimination by equating coefficients :

$\dfrac{1}{5}(x - 2) = \dfrac{1}{4}(1 - y)$

26x + 3y + 4 = 0

Simplifying first equation :

⇒ $\dfrac{1}{5}(x - 2) = \dfrac{1}{4}(1 - y)$

⇒ 4(x - 2) = 5(1 - y)

⇒ 4x - 8 = 5 - 5y

⇒ 4x + 5y - 8 - 5 = 0

⇒ 4x + 5y - 13 = 0 …….(1)

⇒ 26x + 3y + 4 = 0 …….(2)

Multiplying equation (1) by 3, we get :

⇒ 3(4x + 5y - 13) = 0

⇒ 12x + 15y - 39 = 0 …….(3)

Multiplying equation (2) by 5, we get :

⇒ 5(26x + 3y + 4) = 0

⇒ 130x + 15y + 20 = 0 …….(4)

Subtracting equation (3) from (4), we get :

⇒ 130x + 15y + 20 - (12x + 15y - 39) = 0

⇒ 130x - 12x + 15y - 15y + 20 - (-39) = 0

⇒ 118x + 59 = 0

⇒ 118x = -59

⇒ x = $-\dfrac{59}{118} = -\dfrac{1}{2}$

Substituting value of x in equation (1), we get :

$\Rightarrow 4x + 5y - 13 = 0 \\[1em] \Rightarrow 4 \times -\dfrac{1}{2} + 5y - 13 = 0 \\[1em] \Rightarrow -2 + 5y - 13 = 0 \\[1em] \Rightarrow 5y - 15 = 0 \\[1em] \Rightarrow 5y = 15 \\[1em] \Rightarrow y = \dfrac{15}{5} = 3.$

Hence, $x = -\dfrac{1}{2}$ and y = 3.