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

$$2x_{ }+3y=17$$ 

$$3x-2y=6$$ 

Given equations are 

   $$2x_{ }+3y=17$$     

   $$3x-2y=6$$       

We may also write the equations as 

$$2x_{ }+3y-17=0$$..(i)

$$3x-2y-6=0$$..(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) 

 $$\frac{x}{\{3\times (-6)-(-2)\times (-17\left)\right\}}=\frac{y}{\left\{\right(-17)\times 3-(-6)\times 2\}}=\frac{1}{\{2\times (-2)-3\times 3\}}$$ 

 $$\frac{x}{-18-34}=\frac{y}{-51+12}=\frac{1}{-4-9}$$ 

 $$\frac{x}{-52}=\frac{y}{-39}=\frac{1}{-13}$$ 

 $$\frac{x}{-52}=\frac{1}{-13}$$ 

 $$-13x=-52$$ 

 $$x=4$$ 

And, 

 $$\frac{y}{-39}=\frac{1}{-13}$$ 

 $$-13y=-39$$ 

 $$y=3$$ 

Hence, the required value of x is 4 and y is 3.