How do you solve $ - 4x - 6y = - 28$ and $2x + 3y = 14$

Hint: We are given a pair of linear equations and we have to find the value of x and y by using the given equation. We can solve the equation by the method of elimination or by using the method of substitution for the method of substitution. First we will find the value of one variable in the form of another for example we will find the value of x in terms of y then substitute that value in another equation. Then we will solve the equation and find the value of that variable. After that substitute the value of that variable in the other equation and find the value of the remaining one variable.

Complete step-by-step answer:
Step 1: We are given a pair of linear equations $ - 4x - 6y = - 28$ and $2x + 3y = 14$ by applying the method of substitution we will find the value of both variables. We will solve the first equation for $x$:
$ \Rightarrow - 4x - 6y = - 28$
Adding $6y$ to both sides:
$ \Rightarrow - 4x - 6y + 6y = - 28 + 6y$
On proper rearrangement we will get:
$ \Rightarrow - 4x = - 28 + 6y$
Divide both sides by $ - 4$:
$ \Rightarrow x = \dfrac{{ - 28}}{{ - 4}} + \dfrac{{6y}}{{ - 4}}$
$ \Rightarrow x = 7 - \dfrac{{3y}}{2}$
Step2: Now we will substitute the value of $x$ in the second equation and solve for $y$:
$ \Rightarrow 2\left( {7 - \dfrac{{3y}}{2}} \right) + 3y = 14$
$ \Rightarrow 14 - 3y + 3y = 14$
On proper rearrangement we will get:
$ \Rightarrow 14 = 14$
Therefore, the statement $14 = 14$ is true for all real numbers which means the system of equations has infinitely many solutions.

Hence the system of equations has infinitely many solutions.

Note:
This type of question we can solve by two methods first is substitution and the second one is elimination. In this method, the main thing is to find the value of one variable in terms of other students mainly doing the mistakes here.
Alternate method:
We are given two equations i.e.
$ - 4x - 6y = - 28$ …(1)
$2x + 3y = 14$ ….(2)
Multiply equation $\left( 2 \right)$ by $2$
$4x + 6y = 28$ …(3)
Add equations $\left( 3 \right)$and $\left( 1 \right)$
$
{{ - 4x}} -{{6y}} ={{ - 28}} \\
  \underline {{\text{ }}{{4x}} +{{6y}} = {\text{ }}{{{\text{ }}28}}} \\
  {\text{ }}0 = 0 \\
$
Here also we will get the same solution i.e. infinitely many solutions.