Solve using Cross multiplication Method:
\[\begin{array}{l}
x + y = 7\\
2x - 3y = 11
\end{array}\]

Hint: If the two equation is in the form \[ax + by + c = 0\& dx + ey + f = 0\] then in cross multiplication method we write it as \[\dfrac{x}{{bf - ec}} = \dfrac{y}{{cd - af}} = \dfrac{1}{{ae - bd}}\] then use this to find the value of x and y respectively

Complete step by step answer:
The two equation we have are
\[\begin{array}{l}
x + y = 7\\
2x - 3y = 11
\end{array}\]
Which can also be written as
\[\begin{array}{l}
x + y - 7 = 0\\
2x - 3y - 11 = 0
\end{array}\]
Now if we compare this with \[ax + by + c = 0\& dx + ey + f = 0\]
We are getting \[a = 1,b = 1,c = - 7,d = 2,e = - 3,f = - 11\]
Now let us put all of these in
\[\dfrac{x}{{bf - ec}} = \dfrac{y}{{cd - af}} = \dfrac{1}{{ae - bd}}\]
So we will get
\[\begin{array}{l}
\therefore \dfrac{x}{{bf - ec}} = \dfrac{y}{{cd - af}} = \dfrac{1}{{ae - bd}}\\
 \Rightarrow \dfrac{x}{{ - 11 \times 1 - ( - 3) \times ( - 7)}} = \dfrac{y}{{( - 7) \times 2 - 1 \times ( - 11)}} = \dfrac{1}{{1 \times ( - 3) - 2 \times 1}}\\
 \Rightarrow \dfrac{x}{{ - 11 - 21}} = \dfrac{y}{{ - 14 - ( - 11)}} = \dfrac{1}{{ - 3 - 2}}\\
 \Rightarrow \dfrac{x}{{ - 32}} = \dfrac{y}{{ - 3}} = \dfrac{1}{{ - 5}}\\
 \Rightarrow \dfrac{x}{{32}} = \dfrac{y}{3} = \dfrac{1}{5}\\
 \Rightarrow x = \dfrac{{32}}{5},y = \dfrac{3}{5}
\end{array}\]

So We are getting the value of x and y as \[\dfrac{{32}}{5}\] and \[\dfrac{3}{5}\] respectively.

Note: For cross checking put the values of x and y in both the equations given, for the first equation \[x + y - 7 = 0\] we will get \[\dfrac{{32}}{5} + \dfrac{3}{5} - 7\] which is 7-7 and that is basically zero, similarly if we put it in the second equation we will again get zero, which itself proves that our answers are correct.