Solve the following pairs of equations:
$ \frac{1}{2 x}-\frac{1}{y}=-1 $
$ \frac{1}{x}+\frac{1}{2 y}=8, x, y ≠ 0 $

Given:

\( \frac{1}{2 x}-\frac{1}{y}=-1 \)
\( \frac{1}{x}+\frac{1}{2 y}=8,  x, y ≠ 0 \)

To do:

We have to solve the given pairs of equations.

Solution:

$\frac{1}{2 x}-\frac{1}{y}=-1$........(i)

$\frac{1}{x}+\frac{1}{2 y}=8$........(ii)

Put $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in equations (i) and (ii), we get,

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

$\Rightarrow \frac{u-2 v}{2}=-1$

$\Rightarrow u-2 v=-2$.....(iii)

From equation (ii),

$u+\frac{1}{2} v=8$

$\Rightarrow \frac{2 u+v}{2}=8$

$2u+v=16$......(iv)

Multiplying (iv) by 2 and adding the result with (iii), we get,

$u-2v+2(2u+v)=-2+2(16)$

$u+4u-2v+2v=-2+32$

$5u=30$

$u=6$

This implies,

$6-2v=-2$

$2v=6+2$

$v=\frac{8}{2}$

$v=4$

Therefore,

$x=\frac{1}{u}=\frac{1}{6}$

$y=\frac{1}{v}=\frac{1}{4}$

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Updated on: 10-Oct-2022

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