The solution of the system of equations $\dfrac{4}{x} + 5y = 7 \text{ and } \dfrac{3}{x} + 4y = 5$ is

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

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

3. x = 3, y = -1

4. x = -3, y = 1

Given,

$\dfrac{4}{x} + 5y = 7$ …….(i)

$\dfrac{3}{x} + 4y = 5$ …….(ii)

Multiplying eq. (i) by 4 and eq. (ii) by 5 we get,

$\dfrac{16}{x} + 20y = 28$ …….(iii)

$\dfrac{15}{x} + 20y = 25$ …….(iv)

Subtracting eq. (iv) from (iii) we get,

$\Rightarrow \dfrac{16}{x} + 20y - \Big(\dfrac{15}{x} + 20y\Big) = 28 - 25 \\[1em] \Rightarrow \dfrac{16}{x} - \dfrac{15}{x} = 3 \\[1em] \Rightarrow \dfrac{1}{x} = 3 \\[1em] \Rightarrow x = \dfrac{1}{3}.$

Substituting value of x in (ii),

$\Rightarrow \dfrac{3}{\dfrac{1}{3}} + 4y = 5 \\[1em] \Rightarrow 9 + 4y = 5 \\[1em] \Rightarrow 4y = 5 - 9 \\[1em] \Rightarrow 4y = -4 \\[1em] \Rightarrow y = -1.$

Hence, Option 1 is the correct option.