How do you solve \[4\left( y-3 \right)=48\]?

Hint: The degree of an equation is the highest power of the variable in it. We can say whether an equation is linear or quadratic or any other polynomial equation from the degree of the equation. If the degree of the equation equals one, then the equation is linear. To solve a linear equation, we need to take all the variable terms to one side of the equation and constants to the other side of the equation. We will do the same for the given equation also.

Complete step by step solution:
The equation we are asked to solve is \[4\left( y-3 \right)=48\]. As we can see that the degree of this equation is one. Hence, it is a linear equation. We know that to solve a linear equation, we need to take all the variable terms to one side of the equation and constants to the other side of the equation. We will do the same for the given equation also.
\[4\left( y-3 \right)=48\]
Dividing both sides of the above equation by 4, we get
\[\Rightarrow \dfrac{4\left( y-3 \right)}{4}=\dfrac{48}{4}\]
\[\Rightarrow y-3=12\]
Adding 3 to both sides of the above equation, we get
\[\Rightarrow y=15\]

Hence, the solution of the given equation is \[y=15\].

Note: We can check if the solution is correct or not by substituting the value in the given equation. The left-hand side of the given equation is \[4\left( y-3 \right)\], and the right-hand side of the equation is 48.
Substituting \[y=15\] in the LHS of the equation, we get
\[\begin{align}
  & \Rightarrow LHS=4\left( 15-3 \right)=4\left( 12 \right) \\
 & \Rightarrow LHS=48=RHS \\
 & \therefore LHS=RHS \\
\end{align}\]
Thus, as the value satisfies the equation, the solution is correct.