Question

Question 9 (iv)
Solve the following pairs of equation
$\frac{1}{2x}-\frac{1}{y}=-1,\frac{1}{x}+\frac{1}{2y}=8,x,y\ne 0$

Answer 1

Given pair of linear equations is

$\frac{1}{2x}-\frac{1}{y}=-1and\frac{1}{x}+\frac{1}{2y}=8,x,y\ne 0$
Let $u=\frac{1}{x}andv=\frac{1}{y}$.

Then the above equations can be rewritten as:
$\frac{u}{2}-v=-1$

$\Rightarrow u-2v=-\mathrm{2....}\left(iii\right)$

and $u+\frac{v}{2}=8\Rightarrow 2u+v=16.....iv$
On, multiplying Eq. (iv) by 2 and then adding with Eq. (iii), we get
$4u+2v=32u-2v=-2\underset{–}{}5u=30$
$\Rightarrow u=6$

Now, put the value of u in Eq. (iv), we get
$2\times 6+v=16$
$\Rightarrow v=16-12=4$

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

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