How do you solve the following system? $2x + 3y = 11$ , $ - x + y = - 3$

Hint: There are two ways to solve this question. We will answer this question by substitution method. In this method we find one variable in terms of another variable. Then, we will put the value of this variable in the second equation to find the answer.

Complete solution:
In the above question, it is given that
$ \Rightarrow 2x + 3y = 11 - - - - - \left( 1 \right)$
$ \Rightarrow - x + y = - 3 - - - - - \left( 2 \right)$
Now we will find the value of x in terms of variable y from the first equation.
$ \Rightarrow 2x + 3y = 11$
Now transpose $3y$ to the right hand side.
$ \Rightarrow 2x = 11 - 3y$
Now divide both sides by $2$.
$ \Rightarrow x = \dfrac{{11 - 3y}}{2}$
Now we will put this value in equation $1$.
\[ \Rightarrow - \dfrac{{11 - 3y}}{2} + y = - 3\]
Now taking LCM in left hand side
\[ \Rightarrow \dfrac{{ - 11 + 3y + 2y}}{2} = - 3\]
\[ \Rightarrow - 11 + 5y = - 6\]
$ \Rightarrow 5y = 11 - 6$
$ \Rightarrow 5y = 5$
Now divide both sides by $5$
$ \Rightarrow y = 1$
Now, put this value of y in equation $2$.
$ \Rightarrow - x + 1 = - 3$
On transposing x in the right hand side and $ - 3$ in the left hand side.
$ \Rightarrow x = 1 + 3$
$ \Rightarrow x = 4$

Therefore, the value of x is $4$ and the value of y is $1$.

Note:
There is one more method to answer this question. The method name is elimination method. In this method we make the coefficient of one variable in both the equations the same by multiplying both the equations with required constant value. Then we will add or subtract the equations in order to eliminate the variable.