Question

If $\frac{1}{2x}+\frac{2}{3y}=10,\frac{3}{2x}-\frac{5}{3y}=-3,$ then the value of $\frac{1}{y}-\frac{1}{x}$ is

Answer 1

$\frac{1}{2x}+\frac{2}{3y}=10\cdots \left(i\right)$
$\frac{3}{2x}-\frac{5}{3y}=-3\cdots \left(ii\right)$

Let $\frac{1}{2x}=u$ and $\frac{1}{3y}=v$
The given equation becomes:
$u+2v=10$
$3u-5v=-3$

By Cramer's rule:
$u=\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}=\frac{\left|\begin{array}{cc}10& 2\\ -3& -5\end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|}=\frac{-44}{-11}=4$

and
$v=\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}=\frac{\left|\begin{array}{cc}1& 10\\ 3& -3\end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|}=\frac{-33}{-11}=3$

$\because \frac{1}{2x}=u;\frac{1}{3y}=v$
$\frac{1}{2x}=4\Rightarrow x=\frac{1}{8}$
and$\frac{1}{3y}=3\Rightarrow y=\frac{1}{9}$
$\Rightarrow \frac{1}{y}-\frac{1}{x}=9-8=1$