Find the value of x and y, where x$\ne$0 and y$\ne$0. 

$\frac{2}{x}+\frac{3}{y}= 13$ 

$\frac{5}{x}-\frac{4}{y}=-2$

Given equations are 

   $\frac{2}{x}+\frac{3}{y}= 13$       — (i) 

   $\frac{5}{x}-\frac{4}{y}=-2$    — (ii) 

Let, $$\frac{1}{x}=u$$ and $$\frac{1}{y}=v$$ 

$$2u+3v-13=0$$     — (iii) 

$$5u-4v+2=0$$       — (iv) 

The form of the equations is 

$$a_{1}x+b_{1}y+c_1=0, a_{2}x+b_{2}y+c_2=0$$ 

By cross multiplication method, we know that 

$$\frac{u}{(b_1{c}_2-b_2{c}_1)}=$$$$\frac{v}{({c_1{a}_2-c_2{a}_1)}_{ }}=$$ $$\frac{1}{(a_1{b}_2-a_2{b}_1)}$$ 

Now, substituting the value of a, b and c from equation (iii) and (iv) 

 $$\frac{u}{\{3\times 2-(-4)\times (-13\left)\right\}}=\frac{v}{\left\{\right(-13)\times 5-2\times 2\}}=\frac{1}{\{2\times (-4)-5\times 3\}}$$ 

 $$\frac{u}{6-52}=\frac{v}{-65-4}=\frac{1}{-8-15}$$ 

 $$\frac{u}{-46}=\frac{v}{-69}=\frac{1}{-23}$$ 

 $$\frac{u}{-46}=\frac{1}{-23}$$ 

 $$-23u=-46$$ 

 $$u=2$$ 

And, 

 $$\frac{v}{-69}=\frac{1}{-23}$$ 

 $$-23v=-69$$ 

 $$v=3$$ 

Now, 

 $\frac{1}{x}=2$ , $\frac{1}{y}=3$ 

 $$x=\frac{1}{2}, y=\frac{1}{3}$$  

Hence, the required value of x is 1/2 and y is 1/3.