Solve the following system of linear equations to find the values of x and y: 

$$2x+7y=11;3y-7x=44$$ 

The given linear equations are: 

$$2x+7y=11\dots \dots \dots \dots \dots . \left(i\right)$$ 

$$3y-7x=44 \dots \dots \dots \dots .. \left(ii\right)$$     

Multiplying equation (i) by 7, we get  

$$14x+49y=77\dots \dots \dots \dots .\left(iii\right)$$ 

Multiplying equation (ii) by 2, we get 

$$6y-14x=88\dots \dots \dots .\dots .\left(iv\right)$$   

Adding equations (iii) and (iv), we get 

$$55y=165$$ 

$$=>y=\frac{165}{55}=3$$   

Now, substituting $$y=3$$ in equation (i), we get 

$$2x+7\times 3=11$$ 

$$=> 2x+21=11$$ 

$$=>x=\frac{11-21}{2}=-\frac{10}{2}=-5$$ 

$$\therefore x=-5, y=3$$ satisfies the given linear equations.