Find the value of x and y by using the cross-multiplication Method: 

$4x-7y+28=0$ 

$-7x+5y+9=0$ 

 $$4x-7y+28=0$$       — (i) 

 $$-7x+5y+9=0$$     — (ii) 

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 

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

Now, substituting the value of a, b and c from equation (i) and (ii) 

 $$\frac{x}{\left\{\right(-7)\times 9-5\times 28\}}=\frac{y}{\{28\times (-7)-9\times 4\}}=\frac{1}{\{4\times 5-(-7)\times (-7\left)\right\}}$$ 

 $$\frac{x}{-63-140}=\frac{y}{-196-36}=\frac{1}{20-49}$$ 

 $$\frac{x}{-203}=\frac{y}{-232}=\frac{1}{-29}$$ 

 $$\frac{x}{-203}=\frac{1}{-29}$$ 

 $$-29x=-203$$ 

 $$x=7$$ 

And, 

 $$\frac{y}{-232}=\frac{1}{-29}$$ 

 $$-29y=-232$$ 

 $$y=8$$ 

Hence, the required value of x is 7 and y is 8.