Solve the following systems of simultaneous linear equations by the elimination method

$\dfrac{3}{x} + 4y = 7$

$\dfrac{5}{x} + 6y = 13$

Given,

$\dfrac{3}{x} + 4y = 7$ …….(i)

$\dfrac{5}{x} + 6y = 13$ …….(ii)

Multiplying eq. (i) by 3 and eq. (ii) by 2 we get,

$\dfrac{9}{x} + 12y = 21$ ……(iii)

$\dfrac{10}{x} + 12y = 26$ …….(iv)

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

$\Rightarrow \dfrac{10}{x} - \dfrac{9}{x} + 12y - 12y = 26 - 21 \\[1em] \Rightarrow \dfrac{1}{x} = 5 \\[1em] \Rightarrow x = \dfrac{1}{5}.$

Substituting value of x in eq. (i) we get,

$\Rightarrow \dfrac{3}{x} + 4y = 7 \\[1em] \Rightarrow \dfrac{3}{\dfrac{1}{5}} + 4y = 7 \\[1em] \Rightarrow 15 + 4y = 7 \\[1em] \Rightarrow 4y = -8 \\[1em] \Rightarrow y = -2.$

Hence, x = $\dfrac{1}{5}$ and y = -2.