Solve the following pairs of equations:
$ 4 x+\frac{6}{y}=15 $
$ 6 x-\frac{8}{y}=14, y ≠ 0 $

Given: 

The given pair of equations is:

\( 4 x+\frac{6}{y}=15 \)

\( 6 x-\frac{8}{y}=14, y ≠ 0 \)

To do: 

We have to solve the given pair of equations.

Solution: 

$4 x+\frac{6}{y}=15$

Let $\frac{1}{y}=u$

This implies,

$4x+6u=15$......(i)

$6 x-\frac{8}{y}=14$

$6x-8u=14$.......(ii)

Multiplying (i) by 8 and (ii) by 6, we get,

$8(4x+6u)=8(15)$

$32x+48u=120$.......(iii)

$6(6x-8u)=6(14)$

$36x-48u=84$.........(iv)

Adding (iii) and (iv), we get,

$32x+36x+48u-48u=120+84$

$68x=204$

$x=\frac{204}{68}$

$x=3$

This implies,

$4(3)+6u=15$

$6u=15-12$

$u=\frac{3}{6}$

$u=\frac{1}{2}$

This implies,

$y=\frac{1}{\frac{1}{2}}$

$y=2$

Hence, the solution of the given pair of equations is $x=3$ and $y=2$.