How do you solve it \[6 = 2\left( {y + 2} \right)\] ?

Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, we need to know how to multiply constant term with variable, variable with the variable, and constant term with constant term to solve these types of questions. First, we try to solve the operation inside the parenthesis or try to eliminate the parenthesis by using arithmetic operations. We need to find the value \[y\] to solve this question.

Complete step-by-step answer:
The given equation is shown below,
  \[6 = 2\left( {y + 2} \right) \to \left( 1 \right)\]
To solve the given equation we have to solve the RHS part of the above equation.
Let’s solve the RHS of the equation \[\left( 1 \right)\] ,
We know that
  \[a\left( {b + c} \right) = ab + bc\]
By using this formula we can solve the RHS part as shown below,
  \[2\left( {y + 2} \right) = \left( {2 \cdot y} \right) + \left( {2 \cdot 2} \right)\]
We know that,
  \[
  2 \cdot y = 2y \\
  2 \cdot 2 = 4 \;
 \]
So, we get
  \[2\left( {y + 2} \right) = 2y + 4 \to \left( 2 \right)\]
Let’s substitute the equation \[\left( 2 \right)\] in the equation \[\left( 1 \right)\] , we get
  \[\left( 1 \right) \to 6 = 2\left( {y + 2} \right)\]
  \[6 = 2y + 4\]
Let’s move the term \[4\] from RHS to LHS, we get
  \[
  6 - 4 = 2y \\
  2 = 2y \;
 \]
Let’s move the term \[2\] from RHS to LHS, we get
  \[
  \dfrac{2}{2} = y \\
  1 = y \;
 \]
So, the final answer is,
  \[y = 1\]
So, the correct answer is “ \[y = 1\] ”.

Note: This question describes the arithmetic operation like addition/ subtraction/ multiplication/ division. Note that when we move one term from LHS to RHS or RHS to LHS, the operations can be modified as given below,
The addition operation can be converted into the subtraction process.
The subtraction process can be converted into the addition process.
The multiplication process can be converted into the division process.
The division process can be converted into the multiplication process.
Also, remember the basic algebraic formula to make easy calculations.