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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 $y=2$. 
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