Solve the following pair of linear equations by using the cross multiplication Method: 

$$x-3y-8=0$$ 

$$x-9y-14=0$$

Given equations are 

      $x-3y-8=0...\left(i\right)$  $$x-9y-14=0$$...(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) 

 $$\frac{x}{\left\{\right(-3)\times (-14)-(-9)\times (-8\left)\right\}}=\frac{y}{\left\{\right(-8)\times 1-(-14)\times 1\}}=\frac{1}{\{1\times (-9)-1\times (-3\left)\right\}}$$ 

 $$\frac{x}{(42-72)}=\frac{y}{(-8+14)}=\frac{1}{(-9+{3)}_{ }}$$ 

 $$\frac{x}{-30}=\frac{y}{6}=\frac{1}{-6}$$ 

 $$\frac{x}{-30}=\frac{1}{-6}$$ 

 $$-6x=-30$$ 

$$x=5$$ 

And, 

$$\frac{y}{6}=\frac{1}{-6}$$ 

 $$-6y=6$$ 

 $$y=-1$$ 

Hence, the required value of x is 5 and y is -1.