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

$\mathrm{x}-3\mathrm{y}-8=0$ 

$\mathrm{x}-9\mathrm{y}-14=0$

The given equations are: 

 $\mathrm{x}-3\mathrm{y}-8=0...\left(\mathrm{i}\right)$ 

 $x-9y-14=0$...(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{\mathrm{x}}{({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1)}=$$\frac{\mathrm{y}}{({{\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{c}}_2{\mathrm{a}}_1)}_{ }}=$ $\frac{1}{({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1)}$ 

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.