Can x, y be found to satisfy the following equations simultaneously?

$\dfrac{2}{y} + \dfrac{5}{x} = 19, \dfrac{5}{y} - \dfrac{3}{x} = 1$, 3x + 8y = 5.

If so, find them.

Given,

$\dfrac{2}{y} + \dfrac{5}{x} = 19$ ……(i)

$\dfrac{5}{y} - \dfrac{3}{x} = 1$ …….(ii)

3x + 8y = 5 …….(iii)

Multiplying eq. (i) by 5 and eq. (ii) by 2 we get,

$\dfrac{10}{y} + \dfrac{25}{x} = 95$ ……(iii)

$\dfrac{10}{y} - \dfrac{6}{x} = 2$ …….(iv)

Subtracting (iv) from (iii) we get,

$\Rightarrow \dfrac{10}{y} + \dfrac{25}{x} - \Big(\dfrac{10}{y} - \dfrac{6}{x}\Big) = 95 - 2 \\[1em] \Rightarrow \dfrac{25}{x} + \dfrac{6}{x} = 93 \\[1em] \Rightarrow \dfrac{31}{x} = 93 \\[1em] \Rightarrow x = \dfrac{31}{93} = \dfrac{1}{3}.$

Substituting value of x in (i) we get,

$\Rightarrow \dfrac{2}{y} + \dfrac{5}{\dfrac{1}{3}} = 19 \\[1em] \Rightarrow \dfrac{2}{y} + 15 = 19 \\[1em] \Rightarrow \dfrac{2}{y} = 4 \\[1em] \Rightarrow y = \dfrac{1}{2}.$

Substituting values of x and y in eq. (iii),

$\Rightarrow 3 \times \dfrac{1}{3} + 8 \times \dfrac{1}{2} = 5 \\[1em] \Rightarrow 1 + 4 = 5 \\[1em] \Rightarrow 5 = 5.$

Since, L.H.S. = R.H.S. hence, $x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{2}$ satisfies the equation.

Hence, equations can be satisfied simultaneously with $x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{2}$.